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Simultaneous Equations

Before starting with simultaneous equations, let’s recall what are equations in maths and the types of equations. In Mathematics, an equation is a mathematical statement in which two things should be equal to each other. An equation consists of two expressions on each side of an equal sign (=). It consists of two or more variables. In short, the L.H.S value should be equal to the R.H.S value. While substituting the values of the variables in an equation, it should prove its equality. There are different types of equations in Maths, such as:

  • Linear Equation
  • Quadratic Equation
  • Polynomial Equations 

and so on. In this article, we are going to discuss the simultaneous equations which involve two variables along with different methods to solve.

What are Simultaneous Equations?

The simultaneous equation is an equation that involves two or more quantities that are related using two or more equations. It includes a set of few independent equations. The simultaneous equations are also known as the system of equations, in which it consists of a finite set of equations for which the common solution is sought. To solve the equations, we need to find the values of the variables included in these equations.

The system of equations or simultaneous equations can be classified as:

  • Simultaneous linear equations (Or) System of linear equations
  • Simultaneous non-linear equations
  • System of bilinear equations
  • Simultaneous polynomial equations
  • System of differential equations

Here, you will learn the methods of solving simultaneous linear equations along with examples.

The general form of simultaneous linear equations is given as:

dx + ey = f

Methods for Solving Simultaneous Equations

The simultaneous linear equations can be solved using various methods. There are three different approaches to solve the simultaneous equations such as substitution, elimination, and augmented matrix method. Among these three methods, the two simplest methods will effectively solve the simultaneous equations to get accurate solutions. Here we are going to discuss these two important methods, namely,

  • Elimination Method
  • Substitution Method

Apart from those methods, we can also the system of linear equations using Cramer’s rule .

If the simultaneous linear equations contain only two variables, we may also use the cross-multiplication method to find their solution.

Simultaneous Equation Example

Let us now understand how to solve simultaneous equations through the above-mentioned methods. We will get the value of a and b to find the solution for the same. x and y are the two variables in these equations. Go through the following problems which use substitution and elimination methods to solve the simultaneous equations.

Try Out:   Simultaneous Equation Solver

Solving Simultaneous Linear Equations Using Elimination Method

Go through the solved example given below to understand the method of solving simultaneous equations by the elimination method along with steps.

Example: Solve the following simultaneous equations using the elimination method.

4a + 5b = 12,

3a – 5b = 9

The two given equations are

4a + 5b = 12 …….(1)

3a – 5b = 9……….(2)

Step 1: The coefficient of variable ’b’ is equal and has the opposite sign to the other equation. Add equations 1 and 2 to eliminate the variable ‘b’.

Step 2:  The like terms will be added.

(4a+3a) +(5b – 5b) = 12 + 9

Step 3:  Bring the coefficient of a to the R.H.S of the equation

Step 4:  Dividing the R.H. S of the equation, we get a = 3

Step 5: Now, substitute the value a=3 in the equation (1), it becomes

4(3) + 5b = 12,

12 + 5b = 12

b = 0/5 = 0

Step 6:  Hence, the solution for the given simultaneous equations is a = 3 and b = 0.

Solving Simultaneous Linear Equations Using Substitution Method

Below is the solved example with steps to understand the solution of simultaneous linear equations using the substitution method in a better way.

Example: Solve the following simultaneous equations using the substitution method.

b = a + 2 ————–(1)

a + b = 4 ————–(2)

We will solve it step-wise:

Step 1:  Substitute the value of b into the second equation. We will get,

a + (a + 2) = 4

Step 2: Solve for a

a +a + 2 = 4

2a = 4 – 2

a = 2/2 = 1

Step 3: Substitute this value of a in equation 1

step 4:  Hence, the solution for the given simultaneous equations is: a = 1 and b = 3

Practice Problems

  • Solve: 5x + 3y = 7 and -3x + 5y = 23
  • Solve for a and b: 10a – 8b = 6 10a – 9b = -2

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Simultaneous Equations

Here is everything you need to know about simultaneous equations for GCSE maths (Edexcel, AQA and OCR).

You’ll learn what simultaneous equations are and how to solve them algebraically. We will also discuss their relationship to graphs and how they can be solved graphically.

Look out for the simultaneous equations worksheets and exam questions at the end.

What are simultaneous equations?

Simultaneous equations are two or more algebraic equations that share variables such as x and y .

They are called simultaneous equations because the equations are solved at the same time.

The number of variables in simultaneous equations must match the number of equations for it to be solved.

An example of simultaneous equations is 2 x + 4 y = 14 4 x − 4 y = 4

Here are some more:

Each of these equations on their own could have infinite possible solutions.

However when we have at least as many equations as variables we may be able to solve them using methods for solving simultaneous equations.

Representing simultaneous equations graphically

We can consider each equation as a function which, when displayed graphically, may intersect at a specific point. This point of intersection gives the solution to the simultaneous equations.

When we draw the graphs of these two equations, we can see that they intersect at (1,5).

So the solutions to the simultaneous equations in this instance are:

x = 1 and y = 5

What are simultaneous equations?

Solving simultaneous equations

When solving simultaneous equations you will need different methods depending on what sort of simultaneous equations you are dealing with. There are two sorts of simultaneous equations you will need to solve:

  • linear simultaneous equations
  • quadratic simultaneous equations

A linear equation contains terms that are raised to a power that is no higher than one.

Linear simultaneous equations are usually solved by what’s called the elimination method (although the substitution method is also an option for you ) .

Solving simultaneous equations using the elimination method requires you to first eliminate one of the variables, next find the value of one variable, then find the value of the remaining variable via substitution. Examples of this method are given below.

A quadratic equation contains terms that are raised to a power that is no higher than two.

Quadratic simultaneous equations are solved by the substitution method.

See also: 15 Simultaneous equations questions

What are linear and quadratic simultaneous equations?

What are linear and quadratic simultaneous equations?

Simultaneous equations worksheets

Get your free simultaneous equations worksheet of 20+ questions and answers. Includes reasoning and applied questions.

How to solve simultaneous equations

To solve pairs of simultaneous equations you need to:

  • Use the elimination method to get rid of one of the variables.
  • Find the value of one variable.
  • Find the value of the remaining variables using substitution.
  • Clearly state the final answer.
  • Check your answer by substituting both values into either of the original equations.

How do you solve pairs of simultaneous equations?

How do you solve pairs of simultaneous equations?

See the examples below for how to solve the simultaneous linear equations using the three most common forms of simultaneous equations.

See also: Substitution

Quadratic simultaneous equations

Quadratic simultaneous equations have two or more equations that share variables that are raised to powers up to 2 e.g.  x^{2} and y^{2} . Solving quadratic simultaneous equations algebraically by substitution is covered, with examples, in a separate lesson.

Step-by-step guide: Quadratic simultaneous equations

Simultaneous equations examples

For each of the simultaneous equations examples below we have included a graphical representation.

Step-by-step guide : Solving simultaneous equations graphically

Example 1: Solving simultaneous equations by elimination (addition)

  • Eliminate one of the variables.

By adding the two equations together we can eliminate the variable y .

2 Find the value of one variable.

3 Find the value of the remaining variable via substitution.

We know x = 3 so we can substitute this value into either of our original equations.

4 Clearly state the final answer.

5 Check your answer by substituting both values into either of the original equations.

This is correct so we can be confident our answer is correct.

Graphical representation of solving by elimination (addition)

When we draw the graphs of these linear equations they produce two straight lines. These two lines intersect at (1,5). So the solution to the simultaneous equations is x = 3 and y = 2 .

Example 2: Solving simultaneous equations by elimination (subtraction)

By subtracting the two equations we can eliminate the variable b .

NOTE: b − b = 0 so b is eliminated

3 Find the value of the remaining variable/s via substitution.

We know a = 2 so we can substitute this value into either of our original equations.

Graphical representation of solving by elimination (subtraction)

When graphed these two equations intersect at (1,5). So the solution to the simultaneous equations is a = 2 and b = 6 .

Example 3: Solving simultaneous equations by elimination (different coefficients)

Notice that adding or subtracting the equations does not eliminate either variable (see below).

This is because neither of the coefficients of h or i are the same. If you look at the first two examples this was the case.

So our first step in eliminating one of the variables is to make either coefficients of h or i the same.

We are going to equate the variable of h .

Multiply every term in the first equation by 2 .

Multiply every term in the second equation by 3 .

Now the coefficients of h are the same in each of these new equations, we can proceed with our steps from the first two examples. In this example, we are going to subtract the equations.

Note: 6h − 6h = 0 so h is eliminated

Careful : 16 − − 6 = 22

We know i = − 2 so we can substitute this value into either of our original equations.

Graphical representation of solving by elimination (different coefficients)

When graphed these two equations intersect at (1,5). So the solution to the simultaneous equations is h = 4 and i = − 2 .

Example 4: Worded simultaneous equation

David buys 10 apples and 6 bananas in a shop. They cost £5 in total. In the same shop, Ellie buys 3 apples and 1 banana. She spends £1.30 in total. Find the cost of one apple and one banana.

Additional step: conversion

We need to convert this worded example into mathematical language. We can do this by representing apples with a and bananas with b .

Notice we now have equations where we do not have equal coefficients (see example 3).

We are going to equate the variable of b .

Multiply every term in the first equation by 1 .

Multiply every term in the second equation by 6 .

Now the coefficients of b are the same in each equation we can proceed with our steps from the previous examples. In this example, we are going to subtract the equations.

NOTE: 6b − 6b = 0 so b is eliminated

16 − − 6 = 22

Note : we ÷ (− 8) not 8

We know a = 0.35 so we can substitute this value into either of our original equations.

1 apple costs £0.35 (or 35p ) and 1 banana costs £0.25 (or 25p ).

Graphical representation of the worded simultaneous equatio

When graphed these two equations intersect at (1,5). So the solution to the simultaneous equations is a = 0.35 and b = 0.25 .

Common misconceptions

  • Incorrectly eliminating a variable. Using addition to eliminate one variable when you should subtract (and vice-versa).
  • Errors with negative numbers. Making small mistakes when +, −, ✕, ÷ with negative numbers can lead to an incorrect answer. Working out the calculation separately can help to minimise error. Step by step guide: Negative numbers (coming soon)
  • Not multiplying every term in the equation. Mistakes when multiplying an equation. For example, forgetting to multiply every term by the same number.
  • Not checking the answer using substitution. Errors can quickly be spotted by substituting your solutions in the original first or second equations to check they work.

Practice simultaneous equations questions

1. Solve the Simultaneous Equation

6x +3y = 48 6x +y =26

GCSE Quiz True

Subtracting the second equation from the first equation leads to a single variable equation. Use this equation to determine the value of y , then substitute this value into either equation to determine the value of x .

solving problems using simultaneous equations

2. Solve the Simultaneous Equation x -2y = 8 x -3y =3

Subtracting the second equation from the first equation leads to a single variable equation, which determines the value of y . Substitute this value into either equation to determine the value of x .

solving problems using simultaneous equations

3. Solve the Simultaneous Equation 4x +2y = 34 3x +y =21

In this case, a good strategy is to multiply the second equation by 2 . We can then subtract the first equation from the second to leave an equation with a single variable. Once this value is determined, we can substitute it into either equation to find the value of the other variable.

solving problems using simultaneous equations

4. Solve the Simultaneous Equation:

15x -4y = 82 5x -9y =12

In this case, a good strategy is to multiply the second equation by 3 . We can then subtract the second equation from the first to leave an equation with a single variable. Once this value is determined, we can substitute it into either equation to find the value of the other variable.

solving problems using simultaneous equations

Simultaneous equations GCSE questions

1. Solve the simultaneous equations

\begin{array}{l} 5x=-10 \\ x=-2 \end{array}      or correct attempt to find y

One unknown substituted back into either equation

2. Solve the simultaneous equations

Correct attempt to multiple either equation to equate coefficients e.g.

Correct attempt to find y or x ( 16y=56 or 16x = 24 seen)

3. Solve the simultaneous equations

Correct attempt to find y or x ( 13x=91 or 13y=-39 seen)

Learning checklist

  • Solve two simultaneous equations with two variables (linear/linear) algebraically
  • Derive two simultaneous equations, solve the equation(s) and interpret the solution

The next lessons are

  • Maths formulas
  • Types of graphs
  • Interpreting graphs

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If you have two different equations with the same two unknowns in each, you can solve for both unknowns. There are three common methods for solving: addition/subtraction, substitution, and graphing.

Addition/subtraction method

This method is also known as the elimination method.

To use the addition/subtraction method, do the following:

  • Multiply one or both equations by some number(s) to make the number in front of one of the letters (unknowns) the same or exactly the opposite in each equation.
  • Add or subtract the two equations to eliminate one letter.
  • Solve for the remaining unknown.
  • Solve for the other unknown by inserting the value of the unknown found in one of the original equations.

Solve for x and y . 

solving problems using simultaneous equations

Adding the equations eliminates the y ‐terms. 

solving problems using simultaneous equations

Now inserting 5 for x in the first equation gives the following:

solving problems using simultaneous equations

Answer: x = 5, y = 2 

By replacing each x with a 5 and each y with a 2 in the original equations, you can see that each equation will be made true. 

In Example and Example , a unique answer existed for x and y that made each sentence true at the same time. In some situations you do not get unique answers or you get no answers. You need to be aware of these when you use the addition/subtraction method. 

Solve for x and y.

solving problems using simultaneous equations

First multiply the bottom equation by 3. Now the y is preceded by a 3 in each equation. 

solving problems using simultaneous equations

The equations can be subtracted, eliminating the y terms. 

solving problems using simultaneous equations

Insert x = 5 in one of the original equations to solve for y . 

solving problems using simultaneous equations

Answer: x = 5, y = 3 

Of course, if the number in front of a letter is already the same in each equation, you do not have to change either equation. Simply add or subtract.

To check the solution, replace each x in each equation with 5 and replace each y in each equation with 3. 

solving problems using simultaneous equations

Solve for a and b . 

solving problems using simultaneous equations

Multiply the top equation by 2. Notice what happens.

solving problems using simultaneous equations

Now if you were to subtract one equation from the other, the result is 0 = 0.

This statement is always true . 

When this occurs, the system of equations does not have a unique solution. In fact, any a and b replacement that makes one of the equations true, also makes the other equation true. For example, if a = –6 and b = 5, then both equations are made true. 

[3(– 6) + 4(5) = 2 AND 6(– 6) + 8(5) = 4]

What we have here is really only one equation written in two different ways. In this case, the second equation is actually the first equation multiplied by 2. The solution for this situation is either of the original equations or a simplified form of either equation.

solving problems using simultaneous equations

Now if you were to subtract the bottom equation from the top equation, the result is 0 = 1. This statement is never true . When this occurs, the system of equations has no solution. 

In Examples 1–4, only one equation was multiplied by a number to get the numbers in front of a letter to be the same or opposite. Sometimes each equation must be multiplied by different numbers to get the numbers in front of a letter to be the same or opposite.

solving problems using simultaneous equations

Notice that there is no simple number to multiply either equation with to get the numbers in front of x or y to become the same or opposites. In this case, do the following:

  • Select a letter to eliminate.
  • Use the two numbers to the left of this letter. Find the least common multiple of this value as the desired number to be in front of each letter.
  • Determine what value each equation needs to be multiplied by to obtain this value and multiply the equation by that number.

Suppose you want to eliminate x . The least common multiple of 3 and 5, the number in front of the x , is 15. The first equation must be multiplied by 5 in order to get 15 in front of x . The second equation must be multiplied by 3 in order to get 15 in front of x . 

solving problems using simultaneous equations

Eliminate y and solve for x . 

The least common multiple of 4 and 6 is 12. Multiply the top equation by 3 and the bottom equation by 2.

solving problems using simultaneous equations

Now add the two equations to eliminate y . 

solving problems using simultaneous equations

Substitution method

Sometimes a system is more easily solved by the substitution method. This method involves substituting one equation into another. 

solving problems using simultaneous equations

From the first equation, substitute ( y + 8) for x in the second equation. 

( y + 8) + 3 y = 48 

Now solve for y. Simplify by combining y 's. 

solving problems using simultaneous equations

Now insert y 's value, 10, in one of the original equations. 

solving problems using simultaneous equations

Answer: y = 10, x = 18 

Check the solution.

solving problems using simultaneous equations

Solve for x and y using the substitution method. 

solving problems using simultaneous equations

First find an equation that has either a “1” or “ – 1” in front of a letter. Solve for that letter in terms of the other letter.

Then proceed as in example 6.

In this example, the bottom equation has a “1” in front of the y . 

Solve for y in terms of x . 

solving problems using simultaneous equations

Substitute 4 x – 17 for y in the top equation and then solve for x . 

solving problems using simultaneous equations

Replace x with 4 in the equation y – 4 x = –17 and solve for y . 

solving problems using simultaneous equations

The solution is x = 4, y = –1. 

solving problems using simultaneous equations

Graphing method

Another method of solving equations is by graphing each equation on a coordinate graph. The coordinates of the intersection will be the solution to the system. If you are unfamiliar with coordinate graphing, carefully review the articles on coordinate geometry before attempting this method. 

Solve the system by graphing.

solving problems using simultaneous equations

First, find three values for x and y that satisfy each equation. (Although only two points are necessary to determine a straight line, finding a third point is a good way of checking.) Following are tables of x and y values:

x y  | 4 | 0  | 2 | –2  | 5 | 1

x y  | 1 | -1  | 4 | 0  | 7 | 1

Now graph the two lines on the coordinate plane, as shown in Figure 1. 

The point where the two lines cross (4, 0) is the solution of the system.

If the lines are parallel, they do not intersect, and therefore, there is no solution to that system.

Figure 1. The graph of lines x = 4 + y and x – 3 y = 4 indicating the solution.

solving problems using simultaneous equations

Find three values for x and y that satisfy each equation. 

3 x + 4 y = 2 6 x + 8 y = 4 

Following are the tables of x and y values. See Figure 2. 

solving problems using simultaneous equations

Notice that the same points satisfy each equation. These equations represent the same line.

Therefore, the solution is not a unique point. The solution is all the points on the line.

Therefore, the solution is either equation of the line since they both represent the same line.

This is like Example when it was done using the addition/subtraction method. 

Figure 2. The graph of lines 3 x + 4 y = 2 and 6 x + 8 y = 4 indicating the solution.

solving problems using simultaneous equations

Find three values for x and y that satisfy each equation. See the following tables of x and y values:

solving problems using simultaneous equations

In Figure 3, notice that the two graphs are parallel. They will never meet. Therefore, there is no solution for this system of equations. 

No solution exists for this system of equations.

This is like Example done using the addition/subtraction method. 

Figure 3. The graph of lines 3 x + 4 y = 4 and 6 x + 8 y = 16, indicating the solution.

solving problems using simultaneous equations

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Unit 1: Linear simultaneous equations

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Solution of simultaneous equations by use of graphs

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Simultaneous Linear Equations

Simultaneous linear equations in two variables involve two unknown quantities to represent real-life problems. It helps in establishing a relationship between quantities, prices, speed ,  time , distance , etc results in a better understanding of the problems. We all use simultaneous linear equations in our daily life without knowing it.

Simultaneous linear equation

In this mini-lesson, we will learn in detail about the solutions of linear equations, consistent and inconsistent equations, homogenous linear equations, simultaneous equation solver, example of simultaneous equation, etc. in this mini-lesson.

Lesson Plan

What do you mean by simultaneous linear equation.

Two linear equations in two or three variables solved together to find a common solution are called simultaneous linear equations.

For example, We can visualize the solution of the linear system of equations by drawing 2 linear graphs and finding out their intersection point.

Example image of simultaneous linear equations

The red dot represents the solutions for equation 1, and equation 2. The intersection is the unique point (2,1) is the solution that we are looking for which will satisfy both the equations

Linear Equation Calculator

This simultaneous equation solver will find the non-trivial solution on entering the coefficients of x and y and constant.

How To Solve Simultaneous Linear Equations?

The following methods can be used to find the solution of linear system of equations, let's see some example of the simultaneous equation.

1. Substitution Method

Consider the following pair of linear equations:

\[\begin{array}{l}x + 2y = 6\;\;\;\;  & ...(1)\\ x - y = 3\;\;\;\; & ...(2)\end{array}\]Let’s rearrange the first equation to express \(x\) in terms of \(y\), as follows:\[\begin{array}{l}x + 2y = 6\\ \Rightarrow \;\;\;x = 6 - 2y\end{array}\]

This expression for \(x\) can now be substituted in the second equation, so that we will be left with an equation in \(y\) alone:

\[\begin{align}& x - y = 3\\ &\Rightarrow \;\;\;6 - 2y -y = 3\\ &\Rightarrow \;\;\;-3y = 3 - 6\\ &\Rightarrow \;\;\;y = \frac{-3}{-3}\\ &\Rightarrow \;\;\;y = 1\end{align}\]

Once we have the value of \(y\), we can plug this back into any of the two equations to find out \(x\). Lets plug it into the first equation:

\[\begin{array}{l}x + 2y = 6\\ \Rightarrow \;\;\;x + 2 \times 1 = 6\\ \Rightarrow \;\;\;x = 6 - 2 = 4\\ \Rightarrow \;\;\;x = 4\end{array}\]

The final non-trivial solution is:

\[x = 4,\;y = 1\]

It should be clear why this process is called substitution. We express one variable in terms of another using one of the pair of equations, and substitute that expression into the second equation.

2. Elimination Method

\[\begin{array}{l} 2x + 3y - 7 = 0\\ 3x + 2y - 3 = 0 \end{array}\]

The coefficients of x in the two equations are 2 and 3 respectively. Let us multiply the first equation by 3 and the second equation by 2, so that the coefficients of x in the two equations become equal:

\[\begin{array}{l}\left\{ \begin{array}{l}3 \times \left( {2x + 3y - 7 = 0} \right)\\2 \times \left( {3x + 2y - 3 = 0} \right)\end{array} \right.\\ \Rightarrow \;\;\;6x + 9y - 21 = 0\\\qquad6x + 4y - 6 = 0\end{array}\]

Now, let us subtract the two equations, which means that we subtract the left hand sides of the two equations, and the right hand side of the two equations, and the equality will still be preserved (this should be obvious: if I = II and III = IV, then I – III will be equal to II – IV):

\[\begin{array}{l}\left\{ \begin{array}{l}\,\,\,\,6x + 9y - 21 = 0\\\,\,\,\,6x + 4y - 6 = 0\\\underline { - \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\,0 + 5y\,\,\, - 15 = 0\end{array} \right.\\ \Rightarrow \;\;\;5y = 15\\ \Rightarrow \;\;\;y = 3\end{array}\]

Note how x gets eliminated, and we are left with an equation in y alone. Once we have the value of y, we proceed as earlier – we plug this into any of the two equations. Let us put this into the first equation:

\[\begin{array}{l}2x + 3y - 7 = 0\\ \Rightarrow \;\;\;2x + 3\left( 3 \right) - 7 = 0\\ \Rightarrow \;\;\;2x + 9 - 7 = 0\\ \Rightarrow \;\;\;2x = -2\\ \Rightarrow \;\;\;x = -1\end{array}\]

Thus, the solution is:

\[x = -1,\;y = 3\]

3. Graphical Method

As an example, consider the following pair of linear equations:

\[\begin{array}{l}  x - y = 0\\ x + y - 4 = 0\end{array}\]

We draw the corresponding lines on the same axes:

graphical interpretation of simultaneous linear equation

The point of intersection is \(A\left( {2,\,\,2} \right)\), which means that \(x = 2,\;\;y = 2\) is a solution to the pair of linear equations given by (2). In fact, it is the only solution to the pair, as two non-parallel lines cannot intersect in more than one point.

tips and tricks

You can check directly about the types of solution using the following conditions: Unique solution( Consistent and independent)  \[\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\] No solution( Inconsistent and independent)       \[\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\] Infinite Many Solutions (Consistent and Dependent) \[\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\]

Solved Examples

The sum weights of Fabia and Valerian is 60 pounds and the difference is 2. Find the weights of Fabia and Valerian.

Fabia and Valerian measuring weight

Let the weight of Fabia and Valerian be \(x\) pounds and \(y\) pounds respectively. Therefore, the simultaneous equations are \[\begin{array}{l}x + y = 60\;\;\;\;  & ...(1)\\ x - y = 2\;\;\;\; & ...(2)\end{array}\] Let’s rearrange the first equation to express \(x\) in terms of \(y\), as follows:\[\begin{array}{l}x + y = 60\\ \Rightarrow \;\;\;x = 60 - y\end{array}\]

\[\begin{align}& x - y = 2\\ &\Rightarrow \;\;\;60 - y -y = 2\\ &\Rightarrow \;\;\;-2y = 2 - 60\\ &\Rightarrow \;\;\;y = \frac{-58}{-2}\\ &\Rightarrow \;\;\;y = 29\end{align}\]

\[\begin{array}{l}x + y = 60\\ \Rightarrow \;\;\;x + 29 = 60\\ \Rightarrow \;\;\;x = 60 - 29\\ \Rightarrow \;\;\;x = 31\end{array}\]

The final solution is:

\[x = 31,\;y = 29\]

Can you help Alex to find a two-digit number whose units digit is thrice the tens digit and if 36 is added to the number, the digits interchange their place.

Let the digit in the units place is \(x\). And the digit in the tens place be \(y\) Then \(x = 3y\) and the number \(= 10y + x\) The number obtained by reversing the digits is \(10x + y\).

If \(36\) is added to the number, digits interchange their places, Therefore, we have \[10y + x + 36 = 10x + y\] \[10y – y + x + 36 = 10x + y - y\] \[9y + x – 10x + 36 = 10x - 10x\] \[9y - 9x + 36 = 0 \] \[9(x - y) = 36\] \[ x - y = \frac{36}{9}\] \[x - y += 4 .... (i) \]

Substituting the value of x = 3y in equation (i), we get \[3y - y = 4\] \[ 2y = 4\] \[ y = \frac{4}{2}\] \[ y = 2\]

Substituting the value of\(y = 2\)in equation (i),we get \[ x - 2 = 4 \] \[ x = 4 + 2 \] \[ x = 6\]

Challenge your math skills

Solved Crossword Puzzle on Simultaneous linear equations

crossword puzzle

Interactive Questions

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Let's Summarize

The mini-lesson targeted the fascinating concept of Simultaneous Linear Equations. The math journey around Simultaneous Equations starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. Here lies the magic with Cuemath.

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Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic.

Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.

FAQs on  Simultaneous linear equations

1. what is the degree of a linear equation.

The degree of a linear equation is 1

2. How many types of linear equations are there?

The three major forms of linear equations are point-slope form, standard form, and intercept form.

3. What are linear equations?

A linear equation is an equation in which the variable(s) is(are) with the exponent 1 Example: \[2 x = 23 \] \[ x - y = 5\]

4. How does one solve the system of linear equations?

We have different methods to solve the system of linear equations:

Graphical Method Substitution Method Cross Multiplication Method Elimination Method Determinants Method

5. What is the formula of linear equations in two variables?

The general equation of linear equation in two variables is \[ ax + by + c = 0 \]

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  • x+y+z=25,\:5x+3y+2z=0,\:y-z=6
  • x+2y=2x-5,\:x-y=3
  • 5x+3y=7,\:3x-5y=-23
  • x^2+y=5,\:x^2+y^2=7
  • xy+x-4y=11,\:xy-x-4y=4
  • 3-x^2=y,\:x+1=y
  • xy=10,\:2x+y=1
  • substitution\:x+2y=2x-5,\:x-y=3
  • elimination\:x+2y=2x-5,\:x-y=3
  • How to solve linear Simultaneous equations with two variables by graphing?
  • To solve linear simultaneous equations with two variables by graphing, plot both equations on the same set of axes. The coordinates of the points at which the two lines intersect are the solutions to the system.
  • What are Simultaneous Equations?
  • Simultaneous equations are a set of equations that are solved at the same time. These equations are used to define the relationships between variables and can have multiple solutions, a single solution, or no solution.
  • What are Simultaneous Equations used for?
  • Simultaneous equations can be used to solve a wide range of problems in finance, science, engineering, and other fields. They are often used to find the values of variables that make multiple equations or expressions true at the same time.
  • What are the methods for solving Simultaneous Equations?
  • The common methods for solving simultaneous equations are Graphing, Substitution, and Elimination. The choice of method depends on the specific equations and the desired solution.

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  • High School Math Solutions – Systems of Equations Calculator, Nonlinear In a previous post, we learned about how to solve a system of linear equations. In this post, we will learn how... Read More

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Simultaneous Equations

The aim of this section is to understand what are simultaneous equations and how we can solve them? After reading this section we will be able to write down a word problem in the form of simultaneous equations and be able to find out the solution.

Introduction

Types of simultaneous equations, methods for solving simultaneous equations..

Who to verify the solution?

  • Problems on Simultaneous Equations

Mathematics plays a vital role in our life; without mathematics many situations go wrong. There is no problem in physical science that can be solved without converting it into mathematics. Whenever a problem is converted to mathematics it gives an equation in some variable form. Like this situation if a person buys two cupcakes in £3, then what will be the price of one cupcake?

Here the problem in mathematical form is

Where x represents the price of cupcake. We can find the value of x by dividing 2 on both sides, but sometimes problems give the two or more equations. These equations involve two or more unknown variables, as x is an unknown value in above equation, which we have to determine.  These types of equations are called simultaneous equations .

Word simultaneous represents “at a same time”. So simultaneous equations are those equations which are correct for the certain values of unknown variables at a same time.

For examples

Above equations are simultaneous equations in unknown variables ‘x’   and ‘y’ . Both equations are true for x = 1 and y = 4 . Let’s check that these equations are true or not?

Put x = 1 and y = 4

Now the other equation

Now put x = 1 and y = 4

See both the equations are true at a same time for single value of x and y .

There are two types of simultaneous equations which we will see in this section.

1) Linear simultaneous equations

Linear simultaneous equations are called those equations in which power of each unknown variable is one. i.e.

x = 2y        5x – y = 15

2) Nonlinear simultaneous equations

Nonlinear simultaneous equations are those equations in which power of at least one unknown variable must be greater than one. i.e.

x^{2}+y = 5

In above set first equation comes with second degree so this set will be called nonlinear simultaneous equations.

Note: An equation involves a variable with second degree is known as quadratic equation. In addition, above nonlinear equation is also quadratic simultaneous equations.

There are well known three methods we use to solve simultaneous equations, as are listed below.

1) Elimination Method : In this method we eliminate one variable to find the value of other variable.

In this method first we multiply both equations with different numbers to make coefficient same of any one variable and then subtract these equations, after subtraction one variable vanishes out so that we can find the value of another unknown variable easily. After finding out the value of one unknown variable we put this in any one equation and find out the other equations. We will see this method in examples.

Example 1: Solve the simultaneous equations 2x + 3y = 8   and 3x + 2y = 7

First give the name to both equations.

2x + 3y = 8       (1)

3x + 2y = 7       (2)

We will solve these equations by elimination method. To eliminate the one unknown variable, we make the coefficient same of one variable, here we are going to eliminate the unknown variable ‘x’ first multiply equation (1) by 3 and equation (2) by 2. (multiplying by coefficient of ‘x’ in equation (1) with equation (2) and coefficient of ‘x’ in equation (2) by equation (1) is an easy way to make coefficient same

3×1⇒  6x+9y=24

2×2⇒  6x+4y=14

Now subtract equation (2) from equation (1)

solving problems using simultaneous equations

In this step ‘x’ eliminates, we get the equation in term of ‘y’ only. Now divide by ‘5’ on both side.

\frac{5y}{5}=\frac{10}{5}

Now put the value of ‘y’ in any equation, we get the same results, let’s put the value of ‘y’ in equation (1)

2x + 3(2) = 8

⇒  2x + 6 = 8

Subtracting ‘6’ on both sides

2x + 6 – 6 = 8 – 6

Dividing by ‘2’ on both sides of the above equation, we get:

\frac{2x}{2}=\frac{2}{2}

The solution of simultaneous equation is x = 1   and y = 2 .

2) Substitution Method: In this method first we write any one unknown variable in terms of second unknown variable from one equation. Then substitute the value of this variable in the other equation. Then one variable vanishes out and we find out the value of other one. After it, the procedure is same as discussed in elimination method. We will see this method in the following example.

x^{2}+y^{2}=10

Since it is nonlinear simultaneous equations, we first list them by numbering.

x + y = 4          (2)

We solve them by substitution method, firstly write down the equation two in term of ‘x’ only,

x = 4 – y     (3)

Now substitute the value of ‘x’ from equation (1) to equation (2)

(4-y)^{2}+y^{2}=10

Adding same terms and rearranging above equation

2y^{2}-8y+16=10

Subtracting ’10’ on both side

2y^{2}-8y+16-10=10-10

Dividing by two ‘2’ on both sides

\frac{2y^{2}}{2}-\frac{8y}{2}+\frac{6}{2}=0

Since it is a quadratic equation in term of ‘y’ , we can solve it by factorization.

y^{2}-3y-y+3=0

Taking common similar terms

y(y-3)-1(y-3)=0

From above equation we can write as

y – 3 = 0   or   y-  1 = 0

Rearranging above equations we get

y = 3 or y = 1

So, here is the values of ‘y’ , now put these values in equation (3) one by one

x = 4 – 3 = 1

Now put y = 1

x = 4 – 1 = 3

So, there are two different answers, one is x = 1 , when y = 3 and the other is x = 3 , when y = 1 .

3) Graphical Method: In this method we draw the graph of each line and trace out the intersection of these lines. Basically, this intersection is the solution of these equation. Normally we use graphically method for linear. In nonlinear simultaneous equations graphically, method is not so effective because its solution out in surd form.

Example 3: Solve the simultaneous equations by graphical method. 6x + y = 40;   4x + 3y = 36

First of all, we draw the graph of both equation one by one and then trace out the intersection of lines, which will be the our required solution.

In this graph point A representing the point of intersection. At point A the value of x-axis is 6 and y-axis is 4, so point of intersection is 6,4, which is the required solution.

solving problems using simultaneous equations

Many times, we find the solution but forget to check that even it is true or false. In mathematics it is very important to find out the collect solution for carrying good grades. So, for checking the solution that it is true of false we put the answer or values of unknown variables in both equation and see that either both sides are same or not, if same then our solution is correct, if that then we have to check our calculation again.

For example, in the last section we find out the solution of 6x + y = 40 ;   4x + 3y = 36 , which is x = 6 and y = 4 . Put the values in both equation one by one and see that either it is correct or not.

Put in first equation.

66 + 4 = 36 + 4 = 40

Now in second equation

46 + 34 = 24 + 12 = 36

Since values of x and y satisfy both equations, so our solution is correct.

Word Problems for Simultaneous Equations

Problem 1: If Jon bought three packets of chips and 2 packets of biscuits in £29, and Charlie bought one packet of chips and seven packets of biscuits in £54. Then what is the price of each packet of chips and biscuits?

Here we first give the name to the objects for which we have to determine the price.

Let ‘x’ is the price of one packets of chips and ‘y’ is the price of one packet of biscuit.

Now write down the statement in mathematical form step by step.

Mathematical expression for Jon:

Jon bought three packets of chips so, amount for chips will be ‘3x’ and he bought two packets of biscuits, amount for biscuits will be ‘2y’ . According to statement he spends £29 on chips and biscuits. Mathematical expression for Jon is written below.

3x + 2y = 29       (1)

Mathematical expression for Charlie:

Jon bought one packets of chips so, amount for chips will be ‘x’ and he bought seven packets of biscuits, amount for biscuits will be ‘7y’ . According to statement he spends £54 on chips and biscuits. Mathematical expression for Charlie is written below.

x + 7y = 54       (2)

From equation (1) and equation (2) we will determine the value of x and y . Since these are the simultaneous equations. So, we can solve them elimination method.

According to method first make the coefficient same of a one variable, here we make the same coefficient of x .

3×(2)⇒   3x + 21y = 162

Subtract the equation (1) from the above equation.

solving problems using simultaneous equations

Dividing on both sides by ’19’

\frac{19y}{19}=\frac{133}{19}

Put the value of y = 7 , in equation (2)

x + 7(7) = 54

x + 49 = 54

x = 54 – 49

So, the price of one packet of chips is £5 and price of one packet of biscuit is £7.

Problem 2: If the sum money in the pocket of two person A and B is $8 and sum of square their amount is $34, then how much amount each person has?

Let consider Person A have x and Person B have y in their pockets.

Then mathematical form for first condition, which is the sum of their amount is 8

x + y = 8      (1)

Then mathematical form for first condition, which is the sum of square their amount is 34

x^{2}+y^{2}=34

Rearranging equation (1), we get

Put this value of y in equation (2)

x^{2}+(8-x)^{2}=34

Collecting same terms

2x^{2}-16x+64-34=0

Dividing by 2 on both side

x^{2}-8x+15=0

Factorize the above quadratic equation

x^{2}-5x-3x+15=0

x – 5 = 0    or     x – 3 = 0

⇒   x=5   or    x=3

Put the value of x , in y

y = 8 – 5 = 3

y = 8 – 3 = 5

Here is the solution x = 5 , when y = 3 and x = 3 , when y = 5 .

Its mean one of them have $3 and one of them have $5.

Problem 3: Solve the following Nonlinear Systems of Equations.

x + y = 2         (1)

6x^{2}+3y^{2}=12

Rearrange the equation (1), we will get

Put this value in equation (2)

6x^{2}+3(2-x)^{2}=12

3x = 0     or       3x – 4 = 0

x = 0        or      x = 43

Now put the values of x into y = 2 – x , one by one

x = 0   ⇒ y = 2 – 0 = 2

x=\frac{4}{3}

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4.4 Solving simultaneous equations

4.4 solving simultaneous equations (ema38).

Up to now we have solved equations with only one unknown variable. When solving for two unknown variables, two equations are required and these equations are known as simultaneous equations. The solutions are the values of the unknown variables which satisfy both equations simultaneously. In general, if there are \(n\) unknown variables, then \(n\) independent equations are required to obtain a value for each of the \(n\) variables.

An example of a system of simultaneous equations is:

We have two independent equations to solve for two unknown variables. We can solve simultaneous equations algebraically using substitution and elimination methods. We will also show that a system of simultaneous equations can be solved graphically.

Solving by substitution (EMA39)

Use the simplest of the two given equations to express one of the variables in terms of the other.

Substitute into the second equation. By doing this we reduce the number of equations and the number of variables by one.

We now have one equation with one unknown variable which can be solved.

Use the solution to substitute back into the first equation to find the value of the other unknown variable.

The following video shows how to solve simultaneous equations using substitution.

Video: 2FD5

Worked example 6: Simultaneous equations

Solve for \(x\) and \(y\):

Use equation \(\left(1\right)\) to express \(x\) in terms of \(y\)

Substitute \(x\) into equation \(\left(2\right)\) and solve for \(y\), substitute \(y\) back into equation \(\left(1\right)\) and solve for \(x\), check the solution by substituting the answers back into both original equations, write the final answer, worked example 7: simultaneous equations.

Solve the following system of equations:

Use either equation to express \(x\) in terms of \(y\)

Solving by elimination (ema3b), worked example 8: simultaneous equations, make the coefficients of one of the variables the same in both equations.

The coefficients of \(y\) in the given equations are \(\text{1}\) and \(-\text{1}\). Eliminate the variable \(y\) by adding equation \(\left(1\right)\) and equation \(\left(2\right)\) together:

Simplify and solve for \(x\)

Substitute \(x\) back into either original equation and solve for \(y\), check that the solution \(x=3\) and \(y=-7\) satisfies both original equations, worked example 9: simultaneous equations.

By multiplying equation \(\left(1\right)\) by \(\text{3}\) and equation \(\left(2\right)\) by \(\text{2}\), both coefficients of \(a\) will be \(\text{6}\).

(When subtracting two equations, be careful of the signs.)

Simplify and solve for \(b\)

Substitute value of \(b\) back into either original equation and solve for \(a\), check that the solution \(a=10\) and \(b=5\) satisfies both original equations, solving graphically (ema3c).

This section can be included in the chapter on functions and graphs with graphs of linear equations. Before beginning this section it may be necessary to revise plotting graphs of linear equations with your learners.

It is also important that learners are either given the graphs or are encouraged to draw accurate graphs on graph paper to help them solve simultaneous equations graphically. Graph sketching software can be used in this section to ensure that graphs are accurate.

Simultaneous equations can also be solved graphically. If the graphs of each linear equation are drawn, then the solution to the system of simultaneous equations is the coordinates of the point at which the two graphs intersect.

For example:

The graphs of the two equations are shown below.

4e6a26f9cb3dae7bb899566dda1f06fe.png

The intersection of the two graphs is \((2;1)\). So the solution to the system of simultaneous equations is \(x=2\) and \(y=1\). We can also check the solution using algebraic methods.

Substitute equation \((1)\) into \((2)\):

Then solve for \(y\):

Substitute the value of \(y\) back into equation \((1)\):

Notice that both methods give the same solution.

You can use an online tool such as graphsketch to draw the graphs and check your solution.

Worked example 10: Simultaneous equations

Solve the following system of simultaneous equations graphically:

Write both equations in form \(y=mx + c\)

Sketch the graphs on the same set of axes.

87973ba835ca238d1caa651d5e178a71.png

Find the coordinates of the point of intersection

The two graphs intersect at \(\left(4;22\right)\)

Look at the graph below

948fea85d5e2333ce402b8e391f975ed.png

Solve the equations \(y = 2x + 1\) and \(y = -x - 5\) simultaneously

From the graph we can see that the lines intersect at \(x = -2\) and \(y = -3\)

9fc4dcc33ce4c3a24c9bcaf9a7b5189c.png

Solve the equations \(y = 2x - 1\) and \(y = 2x + 1\) simultaneously

The lines are parallel. Therefore there is no solution to \(x\) and \(y\).

4679ae2a15aac4ffc857f4b6c150ad89.png

Solve the equations \(y = -2x + 1\) and \(y = -x - 1\) simultaneously

From the graph we can see that the lines intersect at \(x = 2\) and \(y = -3\)

\(- 10 x = -1\) and \(- 4 x + 10 y = -9\).

Solve for \(x\):

Substitute the value of \(x\) into the second equation and solve for \(y\):

Therefore \(x = \frac{1}{10} \text{ and } y = - \frac{43}{50}\).

\(3x - 14y = 0\) and \(x - 4y + 1 = 0\)

Write \(x\) in terms of \(y\):

Substitute value of \(x\) into second equation:

Substitute value of \(y\) back into first equation:

Therefore \(x = -7 \text{ and } y = -\frac{3}{2}\).

\(x + y = 8\) and \(3x + 2y = 21\)

Therefore \(x = 5 \text{ and } y = 3\).

\(y = 2x + 1\) and \(x + 2y + 3 = 0\)

Write \(y\) in terms of \(x\):

Substitute value of \(y\) into second equation:

Substitute value of \(x\) back into first equation:

Therefore \(x = -1 \text{ and } y = -1\).

\(5x-4y = 69\) and \(2x+3y = 23\)

Make \(x\) the subject of the first equation:

Therefore \(x =13 \text{ and } y = -1\).

\(x + 3y = 26\) and \(5x + 4y = 75\)

Therefore \(x =11 \text{ and } y = 5\).

\(3x - 4y = 19\) and \(2x - 8y = 2\)

If we multiply the first equation by 2 then the coefficient of \(y\) will be the same in both equations:

Now we can subtract the second equation from the first:

Substitute the value of \(x\) into the first equation and solve for \(y\):

Therefore \(x = 9 \text{ and } y = 2\).

\(\dfrac{a}{2} + b = 4\) and \(\dfrac{a}{4} - \dfrac{b}{4} = 1\)

Make \(a\) the subject of the first equation:

Substitute value of \(a\) into second equation:

Substitute value of \(b\) back into first equation:

Therefore \(a = \frac{16}{3} \text{ and } b = \frac{4}{3}\).

\(-10x + y = -1\) and \(-10x - 2y = 5\)

If we subtract the second equation from the first then we can solve for \(y\):

Solve for \(y\):

Substitute the value of \(y\) into the first equation and solve for \(x\):

Therefore \(x = \frac{-1}{10} \text{ and } y = -2\).

\(- 10 x - 10 y = -2\) and \(2 x + 3 y = 2\)

Substitute the value of \(y\) in the first equation:

Therefore \(x = - \frac{7}{5} \text{ and } y = \frac{8}{5}\).

\(\dfrac{1}{x} + \dfrac{1}{y} = 3\) and \(\dfrac{1}{x} - \dfrac{1}{y} = 11\)

Rearrange both equations by multiplying by \(xy\):

Add the two equations together:

Therefore \(x = \frac{1}{7} \text{ and } y = -\frac{1}{4}\).

\(y = \dfrac{2(x^2 + 2) - 3}{x^2 + 2}\) and \(y = 2 - \dfrac{3}{x^2 + 2}\)

Since this is true for all \(x\) in the real numbers, \(x\) can be any real number.

Look at what happens to \(y\) when \(x\) is very small or very large:

The smallest \(x\) can be is 0. When \(x=0\), \(y=2-\frac{3}{2}=\frac{1}{2}\).

If \(x\) gets very large, then the fraction \(\dfrac{3}{x^{2}+2}\) becomes very small (think about what happens when you divide a small number by a very large number). Then \(y = 2 - 0 = 2\).

From this we can see that \(\frac{1}{2}\leq y \leq 2\).

Therefore \(x\) can be any real number, \(\frac{1}{2} \leq y < 2\).

\(3a + b=\dfrac{6}{2a}\) and \(3a^2 = 3 - ab\)

Note \(a \neq 0\)

Look at the first equation

Note that this is the same as the second equation

\(a\) and \(b\) can be any real number except for \(\text{0}\).

Solve graphically and check your answer algebraically:

\(y + 2x = 0\) and \(y - 2x - 4 = 0\)

First write the equations in standard form:

Draw the graph:

e194090d42ef57a5db1cebf151da5fab.png

The graphs intersect at \((-1;2)\) so \(x = -1\) and \(y=2\).

Checking algebraically we get:

Substitute the value of \(x\) back into the first equation:

\(x + 2y = 1\) and \(\dfrac{x}{3} + \dfrac{y}{2} = 1\)

ecb9f42b122b2cb9b6c4241da8adfc91.png

The graphs intersect at \((9;-4)\) so \(x = 9\) and \(y=-4\).

Substitute value of \(x\) into first equation:

Substitute the value of \(y\) back into the first equation:

\(y - 2 = 6x\) and \(y - x = -3\)

6984bf4df5bf31405940d1a65a41b5b0.png

The graphs intersect at \((-1;-4)\) so \(x = -1\) and \(y=-4\).

Substitute value of \(y\) into first equation:

\(2x + y = 5\) and \(3x - 2y = 4\)

c282bbfd5d6580c30cc193affa9394b7.png

The graphs intersect at \((2;1)\) so \(x = 2\) and \(y=1\).

\(5 = x + y\) and \(x = y - 2\)

dca55ce0fac2faec33a2077033d2340f.png

The graphs intersect at \((\text{1,5};\text{3,5})\) so \(x = \text{1,5}\) and \(y=\text{3,5}\).

Word Problems on Simultaneous Linear Equations

Solving the solution of two variables of system equation that leads for the word problems on simultaneous linear equations is the ordered pair (x, y) which satisfies both the linear equations.

Problems of different problems with the help of linear simultaneous equations:

We have already learnt the steps of forming simultaneous equations from mathematical problems and different methods of solving simultaneous equations.

In connection with any problem, when we have to find the values of two unknown quantities, we assume the two unknown quantities as x, y or any two of other algebraic symbols.

Then we form the equation according to the given condition or conditions and solve the two simultaneous equations to find the values of the two unknown quantities. Thus, we can work out the problem.

Worked-out examples for the word problems on simultaneous linear equations: 1. The sum of two number is 14 and their difference is 2. Find the numbers. Solution: Let the two numbers be x and y.

x + y = 14 ………. (i)

x - y = 2 ………. (ii)

Adding equation (i) and (ii), we get 2x = 16

or, 2x/2 = 16/2 or, x = 16/2

or, x = 8 Substituting the value x in equation (i), we get

or, 8 – 8 + y = 14 - 8

or, y = 14 - 8

or, y = 6 Therefore, x = 8 and y = 6

Hence, the two numbers are 6 and 8.

2. In a two digit number. The units digit is thrice the tens digit. If 36 is added to the number, the digits interchange their place. Find the number. Solution:

Let the digit in the units place is x

And the digit in the tens place be y.

Then x = 3y and the number = 10y + x

The number obtained by reversing the digits is 10x + y. If 36 is added to the number, digits interchange their places,

Therefore, we have 10y + x + 36 = 10x + y

or, 10y – y + x + 36 = 10x + y - y

or, 9y + x – 10x + 36 = 10x - 10x

or, 9y - 9x + 36 = 0 or, 9x - 9y = 36

or, 9(x - y) = 36

or, 9(x - y)/9 = 36/9

or, x - y = 4 ………. (i) Substituting the value of x = 3y in equation (i), we get

or, y = 4/2

or, y = 2 Substituting the value of y = 2 in equation (i),we get

or, x = 4 + 2

Therefore, the number becomes 26. 

3.  If 2 is added to the numerator and denominator it becomes 9/10 and if 3 is subtracted from the numerator and denominator it become 4/5. Find the fractions. 

Solution:   Let the fraction be x/y. 

If 2 is added to the numerator and denominator fraction becomes 9/10 so, we have

(x + 2)/(y + 2) = 9/10

or, 10(x + 2) = 9(y + 2) 

or, 10x + 20 = 9y + 18

or, 10x – 9y + 20 = 9y – 9y + 18

or, 10x – 9x + 20 – 20 = 18 – 20 

or, 10x – 9y = -2 ………. (i)  If 3 is subtracted from numerator and denominator the fraction becomes 4/5 so, we have 

(x – 3)/(y – 3) = 4/5

or, 5(x – 3) = 4(y – 3) 

or, 5x – 15 = 4y – 12

or, 5x – 4y – 15 = 4y – 4y – 12 

or, 5x – 4y – 15 + 15 = – 12 + 15

or, 5x – 4y = 3 ………. (ii) 

So, we have 10x – 9y = – 2 ………. (iii) 

and 5x – 4y = 3 ………. (iv)  Multiplying both sided of equation (iv) by 2, we get

10x – 8y = 6 ………. (v) 

Now, solving equation (iii) and (v) , we get

10x – 9y = -2

10x – 8y =  6         - y = - 8

          y = 8 

Substituting the value of y in equation (iv) 

5x – 4 × (8) = 3

5x – 32 = 3

5x – 32 + 32 = 3 + 32

Therefore, fraction becomes 7/8. 4.  If twice the age of son is added to age of father, the sum is 56. But if twice the age of the father is added to the age of son, the sum is 82. Find the ages of father and son.  Solution:  Let father’s age be x years

Son’s ages = y years

Then 2y + x = 56 …………… (i) 

And 2x + y = 82 …………… (ii)  Multiplying equation (i) by 2, (2y + x = 56 …………… × 2)we get

linear equations

or, 3y/3 = 30/3

or, y = 30/3

or, y = 10 (solution (ii) and (iii) by subtraction) Substituting the value of y in equation (i), we get;

2 × 10 + x = 56

or, 20 + x = 56

or, 20 – 20 + x = 56 – 20

or, x = 56 – 20

5. Two pens and one eraser cost Rs. 35 and 3 pencil and four erasers cost Rs. 65. Find the cost of pencil and eraser separately. Solution: Let the cost of pen = x and the cost of eraser = y

Then 2x + y = 35 ……………(i)

And 3x + 4y = 65 ……………(ii) Multiplying equation (i) by 4,

problems on simultaneous equations

Subtracting (iii) and (ii), we get;

or, 5x/5 = 75/5

or, x = 75/5

or, x = 15 Substituting the value of x = 15 in equation (i) 2x + y = 35 we get;

or, 2 × 15 + y = 35

or, 30 + y = 35

or, y = 35 – 30

Therefore, the cost of 1 pen is Rs. 15 and the cost of 1 eraser is Rs. 5.

●   Simultaneous Linear Equations

Simultaneous Linear Equations

Comparison Method

Elimination Method

Substitution Method

Cross-Multiplication Method

Solvability of Linear Simultaneous Equations

Pairs of Equations

Practice Test on Word Problems Involving Simultaneous Linear Equations

●   Simultaneous Linear Equations - Worksheets

Worksheet on Simultaneous Linear Equations

Worksheet on Problems on Simultaneous Linear Equations

8th Grade Math Practice   From Word Problems on Simultaneous Linear Equations to HOME PAGE

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How to Solve Simultaneous Equations Using Elimination Method

Last Updated: November 28, 2022 Fact Checked

wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, volunteer authors worked to edit and improve it over time. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 110,462 times. Learn more...

Have you ever had a simultaneous problem equation you needed to solve? When you use the elimination method, you can achieve a desired result in a very short time. This article can explain how to perform to achieve the solution for both variables.

Step 1 Write down both of the equations that you'll need to solve.

  • 3x - y = 12
  • 2x + y = 13

Step 2 Number the equations.

  • + 2x + y = 13
  • -------------

Step 4 Look for signs of the unknown variables or terms.

  • 3(5) - 3 = 12
  • 15 - 3 = 12

Community Q&A

Donagan

  • You can solve it horizontally as well. Thanks Helpful 1 Not Helpful 1
  • Remember that if the unknown terms have the same signs you must subtract and if they have different signs you must add the equations. Thanks Helpful 3 Not Helpful 0
  • The easiest of all is the elimination method and fastest as well. It is simple and can be learned much more easily then the other two. It is usually used when the equations have the same variable/unknown term disregarding the sign. Thanks Helpful 3 Not Helpful 1

solving problems using simultaneous equations

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Solve Simultaneous Equations Using Substitution Method

  • ↑ https://flexbooks.ck12.org/cbook/ck-12-cbse-math-class-10/section/3.6/primary/lesson/solving-simultaneous-linear-equations-by-elimination/
  • ↑ https://www.youtube.com/watch?v=O-rrOPPmFgM
  • ↑ https://www.youtube.com/watch?v=7Ja_H6a8ltY
  • ↑ https://www.mathsteacher.com.au/year10/ch04_simultaneous/03_elimination_method/elim.htm
  • ↑ https://www.bbc.co.uk/bitesize/guides/z9y9jty/revision/1
  • ↑ https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:systems-of-equations/x2f8bb11595b61c86:solving-systems-elimination/a/elimination-method-review

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Mathematics LibreTexts

6: Gaussian Elimination Method for Solving Simultaneous Linear Equations

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  • Page ID 104131

Learning Objectives

After successful completion of this lesson, you should be able to

  • write the algorithm to solve a set of simultaneous linear equations using Naïve Gauss elimination method
  • solve a set of simultaneous linear equations using Naïve Gauss elimination.
  • use the forward elimination steps of Gauss elimination method to find determinant of a square matrix,
  • enumerate theorems related to determinant of matrices,
  • relate the zero and non-zero value of the determinant of a square matrix to the existence or non-existence of the matrix inverse.
  • enumerate the pitfalls of the Naïve Gauss elimination method
  • show the pitfalls of Naïve Gauss elimination method through examples
  • write the algorithm to solve a set of simultaneous linear equations using Gaussian elimination with Partial Pivoting.
  • solve a set of simultaneous linear equations using Gauss elimination method with partial pivoting

How is a set of equations solved numerically by Gaussian elimination method?

One of the most popular techniques for solving simultaneous linear equations is the Gaussian elimination method. The approach is designed to solve a general set of \(n\) equations and \(n\) unknowns

\[\begin{split} &a_{11}x_{1} + a_{12}x_{2} + a_{13}x_{3} + \ldots + a_{1n}x_{n} = b_{1}\\ &a_{21}x_{1} + a_{22}x_{2} + a_{23}x_{3} + \ldots + a_{2n}x_{n} = b_{2}\\ &\vdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots\\ &a_{n1}x_{1} + a_{n2}x_{2} + a_{n3}x_{3} + \ldots + a_{nn}x_{n} = b_{n} \end{split} \nonumber \]

Gaussian elimination consists of two steps

  • Forward Elimination of Unknowns: In this step, the unknown is eliminated in each equation starting with the first equation. This way, the equations are reduced to one equation and one unknown in each equation.
  • Back Substitution: In this step, starting from the last equation, each of the unknowns is found.

Forward Elimination of Unknowns:

In the first step of forward elimination, the first unknown, \(x_{1}\) is eliminated from all rows below the first row. The first equation is selected as the pivot equation to eliminate \(x_{1}\) . So, to eliminate \(x_{1}\) in the second equation, one divides the first equation by \(a_{11}\) (hence called the pivot element) and then multiplies it by \(a_{21}\) . This is the same as multiplying the first equation by \(a_{21}/a_{11}\) to give

\[a_{21}x_{1} + \frac{a_{21}}{a_{11}}a_{12}x_{2} + \ldots + \frac{a_{21}}{a_{11}}a_{1n}x_{n} = \frac{a_{21}}{a_{11}}b_{1} \nonumber \]

Now, this equation can be subtracted from the second equation to give

\[\left( a_{22} - \frac{a_{21}}{a_{11}}a_{12} \right)x_{2} + \ldots + \left( a_{2n} - \frac{a_{21}}{a_{11}}a_{1n} \right)x_{n} = b_{2} - \frac{a_{21}}{a_{11}}b_{1} \nonumber \]

\[{a^\prime }_{22}x_{2} + \ldots + {a^\prime }_{2n}x_{n} = {b^\prime }_{2} \nonumber \]

\[\begin{split} &{a^\prime }_{22} = a_{22} - \frac{a_{21}}{a_{11}}a_{12}\\ &\vdots\\ &{a^\prime }_{2n} = a_{2n} - \frac{a_{21}}{a_{11}}a_{1n}\end{split} \nonumber \]

This procedure of eliminating \(x_{1}\) , is now repeated for the third equation to the \(n^{th}\) equation to reduce the set of equations as

\[\begin{split} &a_{11}x_{1} + a_{12}x_{2} + a_{13}x_{3} + \ldots + a_{1n}x_{n} = b_{1}\\ &{a^\prime }_{22}x_{2} + {a^\prime }_{23}x_{3} + \ldots + {a^\prime }_{2n}x_{n} = {b^\prime }_{2}\\ &{a^\prime }_{32}x_{2} + {a^\prime }_{33}x_{3} + \ldots + {a^\prime }_{3n}x_{n} = {b^\prime }_{3}\\ &\vdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots\\ &{a^\prime }_{n2}x_{2} + {a^\prime }_{n3}x_{3} + \ldots + {a^\prime }_{\text{nn}}x_{n} = {b^\prime }_{n} \end{split} \nonumber \]

This is the end of the first step of forward elimination. Now for the second step of forward elimination, we start with the second equation as the pivot equation and \({a^\prime }_{22}\) as the pivot element. So, to eliminate \(x_{2}\) in the third equation, one divides the second equation by \({a^\prime }_{22}\) (the pivot element) and then multiply it by \({a^\prime }_{32}\) . This is the same as multiplying the second equation by \(\displaystyle {a^\prime }_{32}/{a^\prime }_{22}\) and subtracting it from the third equation. This makes the coefficient of \(x_{2}\) zero in the third equation. The same procedure is now repeated for the fourth equation till the \(n^{\text{th}}\) equation to give

\[\begin{split} &a_{11}x_{1} + a_{12}x_{2} + a_{13}x_{3} + \ldots + a_{1n}x_{n} = b_{1}\\ &{a^\prime }_{22}x_{2} + {a^\prime }_{23}x_{3} + \ldots + {a^\prime }_{2n}x_{n} = {b^\prime }_{2}\\ &{a^{\prime\prime}}_{33}x_{3} + \ldots + {a^{\prime\prime}}_{3n}x_{n} = {b^{\prime\prime}}_{3}\\ &\vdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots\\ &{a^{\prime\prime}}_{n3}x_{3} + \ldots + {a^{\prime\prime}}_{\text{nn}}x_{n} = {b^{\prime\prime}}_{n} \end{split} \nonumber \]

The next steps of forward elimination are conducted by using the third equation as a pivot equation and so on. That is, there will be a total of \(n - 1\) steps of forward elimination. At the end of \(n - 1\) steps of forward elimination, we get a set of equations that look like

\[\begin{split} &a_{11}x_{1} + a_{12}x_{2} + a_{13}x_{3} + \ldots + a_{1n}x_{n} = b_{1}\\ &\ \ \ \ \ \ \ \ \ \ \ \ \ {a^\prime }_{22}x_{2} + {a^\prime }_{23}x_{3} + \ldots + {a^\prime }_{2n}x_{n} = {b^\prime }_{2}\\ &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {a^{\prime\prime}}_{33}x_{3} + \ldots + {a^{\prime\prime}}_{3n}x_{n} = {b{\prime\prime}}_{3}\\ &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots\\ &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a_{nn}^{\left( n - 1 \right)}x_{n} = b_{n}^{\left( n - 1 \right)}\end{split} \nonumber \]

Back Substitution:

Now the equations are solved starting from the last equation as it has only one unknown.

\[x_{n} = \frac{b_{n}^{(n - 1)}}{a_{nn}^{(n - 1)}} \nonumber \]

Then the second last equation, that is the \((n - 1)^{th}\) equation, has two unknowns: \(x_{n}\) and \(x_{n - 1}\) , but \(x_{n}\) is already known. This reduces the \((n - 1)^{th}\) equation also to one unknown. Back substitution hence can be represented for all equations by the formula

\[x_{i} = \frac{b_{i}^{\left( i - 1 \right)} - \displaystyle\sum_{j = i + 1}^{n}{a_{ij}^{\left( i - 1 \right)}x_{j}}}{a_{ii}^{\left( i - 1 \right)}}\ \text{for }i = n - 1,\ \ n - 2,\ldots\ ,\ 1 \nonumber \]

The upward velocity of a rocket is given at three different times in Table 1.

The velocity data is approximated by a polynomial as

\[v\left( t \right) = a_{1}t^{2} + a_{2}t + a_{3},5 \leq t \leq 12 \nonumber \]

The coefficients \(a_{1},a_{2},anda_{3}\) for the above expression are given by

\[\begin{bmatrix} 25 & 5 & 1 \\ 64 & 8 & 1 \\ 144 & 12 & 1 \\ \end{bmatrix}\begin{bmatrix} a_{1} \\ a_{2} \\ a_{3} \\ \end{bmatrix} = \begin{bmatrix} 106.8 \\ 177.2 \\ 279.2 \\ \end{bmatrix} \nonumber \]

Find the values of \(a_{1},a_{2},and\ a_{3}\) using the Naïve Gauss elimination method. Find the velocity at \(t = 6,7.5,9,11\) seconds.

The augmented matrix is

\[\begin{bmatrix} 25 & 5 & 1 & | & 106.8 \\ 64 & 8 & 1 & | & 177.2 \\ 144 & 12 & 1 & | & 279.2 \\ \end{bmatrix} \nonumber \]

Forward Elimination of Unknowns

Since there are three equations, there will be two steps of forward elimination of unknowns.

Divide Row \(1\) by \(25\) and then multiply it by \(64\) , that is, multiply Row \(1\) by \(64/25 = 2.56\) .

\[\left( \begin{matrix} \left\lbrack 25 \right.\ & \ 5 & \ \ \ \ \ \ \ 1\ \ \ \ \ \ \ |\ \ \ \ \ \\ \end{matrix}106.8\rbrack \right) \times 2.56\ \text{gives Row 1 as} \nonumber \]

\[\begin{matrix} \left\lbrack 64 \right.\ & 12.8 & 2.56 \\ \end{matrix}\ \ \ \ \ |\ \ 273.408\rbrack \nonumber \]

Subtract the result from Row \(2\)

\[\frac{\begin{matrix} \ \ \ \lbrack\begin{matrix} 64 & \ \ \ \ 8 & \ \ \ \ \ 1 \\ \end{matrix}\ \ \ \ \ |\ \ & 177.2\rbrack \\- \lbrack\begin{matrix} 64 & 12.8 & 2.56 \\ \end{matrix}\ \ | & 273.408\rbrack \\ \end{matrix}}{\ \ \ \begin{matrix} \begin{matrix} 0 & - 4.8 & - 1.56 \\ \end{matrix} & - 96.208 \\ \end{matrix}} \nonumber \]

to get the resulting equations as

\[\left\lbrack \begin{matrix} 25 & 5 & 1 \\ 0 & - 4.8 & - 1.56 \\ 144 & 12 & 1 \\ \end{matrix} \ \ \ \ \ \ \ \ \begin{matrix} | \ \ \ \ \ \ \ 106.8 \\ \ |\ - 96.208 \\ |\ \ \ \ \ \ \ 279.2 \\ \end{matrix} \right\rbrack \nonumber \]

Divide Row \(1\) by \(25\) and then multiply it by \(144\) , that is, multiply Row \(1\) by \(144/25 = 5.76\) .

\[\left( \begin{matrix} \left\lbrack 25 \right.\ & \ \ \ \ 5 & \ \ \ 1 \\ \end{matrix} \ \ \ \ \ \ |\ \ \ \ \ \ \ \ \ 106.8\rbrack \right) \times 5.76\ \text{gives Row 1 as} \nonumber \]

\[\left\lbrack \begin{matrix} 144 & 28.8 & 5.76 \\ \end{matrix} \ \ \ \ \right|\ \ \ \ \ \ 615.168\rbrack \nonumber \]

Subtract the result from Row \(3\)

\[\frac{\begin{matrix} \ \ \ \ \lbrack\begin{matrix} 144 & 12 & \ \ \ \ \ \ 1 \\ \end{matrix}\ \ \ \ | & 279.2\rbrack \\ - \begin{matrix} \lbrack 144 & 28.8 & 5.76 \\ \end{matrix}\ \ \ | & 615.168\rbrack \\ \end{matrix}}{\ \ \ \ \begin{matrix} \begin{matrix} 0 & - 16.8 & - 4.76 \\ \end{matrix} & - 335.968 \\ \end{matrix}} \nonumber \]

\[\left\lbrack \begin{matrix} 25 & 5 & 1 \\ 0 & - 4.8 & - 1.56 \\ 0 & - 16.8 & - 4.76 \\ \end{matrix}\ \ \ \ \ \ \ \ \ \begin{matrix} |\ \ \ \ \ \ \ \ \ 106.8 \\ |\ \ - 96.208 \\| - 335.968 \\ \end{matrix} \right\rbrack \nonumber \]

Second step

We now divide Row \(2\) by \(-4.8\) and then multiply by \(-16.8\) , that is, multiply Row \(2\) by \(-16.8/-4.8 = 3.5\) .

\[\left( \left\lbrack 0 - 4.8\ \ - 1.56\ \ \ \ \ \ \ \ \ \ \right|\ \ \ - 96.208\rbrack\ \right) \times 3.5\ \text{gives Row 2 as} \nonumber \]

\[\left\lbrack 0 - 16.8\ - 5.46\ \ \ \ \ \ \ \ \ \ \right| - 336.728\rbrack \nonumber \]

\[\frac{\begin{matrix} \ \ \ \ \ \lbrack\begin{matrix} 0 & -16.8 & -4.76\ \\ \end{matrix} \ \ \ \ \ | & -335.968\rbrack \\ \ -\ \lbrack \begin{matrix} 0 & -16.8 & -5.46\ \\ \end{matrix}\ \ \ \ \ | & -336.728\rbrack \\ \end{matrix}}{\begin{matrix} \ \ \ \ \ \begin{matrix} 0 & \ \ \ \ \ \ \ 0 & \ \ \ \ \ \ 0.7 \end{matrix} & \ \ \ \ \ \ \ \ \ \ \ 0.76 \ \ \ \end{matrix}} \nonumber \]

\[\left\lbrack \begin{matrix} 25 & 5 & 1 \\ 0 & - 4.8 & - 1.56 \\ 0 & 0 & 0.7 \\ \end{matrix} \ \ \ \ \ \ \ \begin{matrix} |\ \ \ \ \ \ 106.8 \\ \ \ |\ - 96.208 \\ |\ \ \ \ \ \ \ \ 0.76 \\ \end{matrix} \right\rbrack \nonumber \]

\[\begin{bmatrix} 25 & 5 & 1 \\ 0 & - 4.8 & - 1.56 \\ 0 & 0 & 0.7 \\ \end{bmatrix}\begin{bmatrix} a_{1} \\ a_{2} \\ a_{3} \\ \end{bmatrix} = \begin{bmatrix} 106.8 \\ - 96.208 \\ 0.76 \\ \end{bmatrix} \nonumber \]

Back substitution

From the third equation

\[0.7a_{3} = 0.76 \nonumber \]

\[\begin{split} a_{3} &= \frac{0.76}{0.7}\\ &= 1.08571 \end{split} \nonumber \]

Substituting the value of \(a_{3}\) in the second equation,

\[- 4.8a_{2} - 1.56a_{3} = - 96.208 \nonumber \]

\[\begin{split} a_{2} &= \frac{- 96.208 + 1.56a_{3}}{- 4.8}\\ &= \frac{- 96.208 + 1.56 \times 1.08571}{- 4.8}\\ &= 19.6905\end{split} \nonumber \]

Substituting the value of \(a_{2}\) and \(a_{3}\) in the first equation,

\[25a_{1} + 5a_{2} + a_{3} = 106.8 \nonumber \]

\[\begin{split} a_{1} &= \frac{106.8 - 5a_{2} - a_{3}}{25}\\ &= \frac{106.8 - 5 \times 19.6905 - 1.08571}{25}\\ &= 0.290472 \end{split} \nonumber \]

Hence the solution vector is

\[\begin{bmatrix} a_{1} \\ a_{2} \\ a_{3} \\ \end{bmatrix} = \begin{bmatrix} 0.290472 \\ 19.6905 \\ 1.08571 \\ \end{bmatrix} \nonumber \]

The polynomial that passes through the three data points is then

\[\begin{split} v\left( t \right) &= a_{1}t^{2} + a_{2}t + a_{3}\\ &= 0.290472t^{2} + 19.6905t + 1.08571,\ \ 5 \leq t \leq 12 \end{split} \nonumber \]

Since we want to find the velocity at \(t = 6,7.5,9\) and \(11\) seconds, we could simply substitute each value of \(t\) in \(v\left( t \right) = 0.290472t^{2} + 19.6905t + 1.08571\) and find the corresponding velocity. For example, at \(t = 6\)

\[\begin{split} v\left( 6 \right) &= 0.290472\left( 6 \right)^{2} + 19.6905\left( 6 \right) + 1.08571\\ &= 129.686\ \ m/s \end{split} \nonumber \]

However, we could also find all the needed values of velocity at \(t\) = \(6, 7.5, 9, 11\) seconds using matrix multiplication.

\[v\left( t \right) = \left\lbrack 0.290472\ \ 19.6905 \ \ 1.08571 \right\rbrack\begin{bmatrix} t^{2} \\ t \\ 1 \\ \end{bmatrix} \nonumber \]

So, if we want to find \(v\left( 6 \right),v\left( 7.5 \right),v\left( 9 \right),v\left( 11 \right),\) it is given by

\[\begin{split} \begin{matrix}\left\lbrack v\left( 6 \right)v\left( 7.5 \right)v\left( 9 \right)v\left( 11 \right) \right\rbrack \end{matrix} &= \left\lbrack 0.290472 \ \ 19.6905\ \ 1.08571 \right\rbrack\begin{bmatrix} 6^{2} & 7.5^{2} & 9^{2} & 11^{2} \\ 6 & 7.5 & 9 & 11 \\ 1 & 1 & 1 & 1 \end{bmatrix}\\ &= \left\lbrack 0.290472 \ \ 19.6905 \ \ 1.08571 \right\rbrack\begin{bmatrix} 36 & 56.25 & 81 & 121 \\ 6 & 7.5 & 9 & 11 \\ 1 & 1 & 1 & 1 \end{bmatrix}\\ &= \left\lbrack 129.686 \ \ 165.104\ \ 201.828 \ \ 252.828 \right\rbrack \end{split} \nonumber \]

\[v(6) = 129.686\ m/s \nonumber \]

\[v(7.5) = 165.1\ 04\ m/s \nonumber \]

\[v(9) = 201.828\ m/s \nonumber \]

\[v(11) = 252.828\ m/s \nonumber \]

Can we use Naive Gauss elimination methods to find the determinant of a square matrix?

One of the more efficient ways to find the determinant of a square matrix is by taking advantage of the following two theorems on a determinant of matrices coupled with Naive Gauss elimination.

Theorem \(\PageIndex{1}\)

Let \(\lbrack A\rbrack\) be a \(n \times n\) matrix. Then, if \(\lbrack B\rbrack\) is a \(n \times n\) matrix that results from adding or subtracting a multiple of one row to another row, then \(det(A) = det(B)\) (The same is true for column operations also).

Theorem \(\PageIndex{2}\)

Let \(\lbrack A\rbrack\) be a \(n \times n\) matrix that is upper triangular, lower triangular or diagonal, then

\[\begin{split} \det(A) &= a_{11} \times a_{22} \times ... \times a_{ii} \times ... \times a_{nn}\\ &= \prod_{i = 1}^{n}a_{ii} \end{split} \nonumber \]

This implies that if we apply the forward elimination steps of the Naive Gauss elimination method, the determinant of the matrix stays the same according to Theorem \(\PageIndex{1}\). Then since at the end of the forward elimination steps, the resulting matrix is upper triangular, the determinant will be given by Theorem \(\PageIndex{2}\).

Find the determinant of

\[\lbrack A\rbrack = \begin{bmatrix} 25 & 5 & 1 \\ 64 & 8 & 1 \\ 144 & 12 & 1 \\ \end{bmatrix} \nonumber \]

Remember in Example 1, we conducted the steps of forward elimination of unknowns using the Naive Gauss elimination method on \(\lbrack A\rbrack\) to give

\[\left\lbrack B \right\rbrack = \begin{bmatrix} 25 & 5 & 1 \\ 0 & - 4.8 & - 1.56 \\ 0 & 0 & 0.7 \\ \end{bmatrix} \nonumber \]

According to Theorem 2

\[\begin{split} det(A) &= det(B)\\ &= 25 \times ( - 4.8) \times 0.7\\ &= - 84.00 \end{split} \nonumber \]

What if I cannot find the determinant of the matrix using the Naive Gauss elimination method, for example, if I get division by zero problems during the Naive Gauss elimination method?

Well, you can apply Gaussian elimination with partial pivoting. However, the determinant of the resulting upper triangular matrix may differ by a sign. The following theorem applies in addition to the previous two to find the determinant of a square matrix.

Theorem \(\PageIndex{3}\)

Let \(\lbrack A\rbrack\) be a \(n \times n\) matrix. Then, if \(\lbrack B\rbrack\) is a matrix that results from switching one row with another row, then \(det(B) = - det(A)\) .

\[\lbrack A\rbrack = \begin{bmatrix} 10 & - 7 & 0 \\ - 3 & 2.099 & 6 \\ 5 & - 1 & 5 \\ \end{bmatrix} \nonumber \]

The end of the forward elimination steps of Gaussian elimination with partial pivoting, we would obtain

\[\lbrack B\rbrack = \begin{bmatrix} 10 & - 7 & 0 \\ 0 & 2.5 & 5 \\ 0 & 0 & 6.002 \\ \end{bmatrix} \nonumber \]

\[\begin{split} \det\left( B \right) &= 10 \times 2.5 \times 6.002\\ &= 150.05 \end{split} \nonumber \]

Since rows were switched once during the forward elimination steps of Gaussian elimination with partial pivoting,

\[\begin{split} \det\left( A \right) &= - det(B)\\ &= - 150.05 \end{split} \nonumber \]

\[\det(A) = \frac{1}{\det\left( A^{- 1} \right)} \nonumber \]

\[\lbrack A\rbrack\lbrack A\rbrack^{- 1} = \lbrack I\rbrack \nonumber \]

\[\det(AA^{- 1}) = det(I) \nonumber \]

\[\det(A)\det(A^{-1}) = 1 \nonumber \]

\[\det(A) = \frac{1}{\det(A^{-1})} \nonumber \]

If \({\lbrack A\rbrack}\) is a \({n \times n}\) matrix and \({\det(A) \neq 0}\) , what other statements are equivalent to it?

  • \(\lbrack A\rbrack\) is invertible.
  • \(\lbrack A\rbrack^{- 1}\) exists.
  • \(\lbrack A\rbrack\ \lbrack X\rbrack = \lbrack C\rbrack\) has a unique solution.
  • \(\lbrack A\rbrack\ \lbrack X\rbrack = \lbrack 0\rbrack\) solution is \(\lbrack X\rbrack = \lbrack{0}\rbrack\) .
  • \(\lbrack A\rbrack\ \lbrack A\rbrack^{- 1} = \lbrack I\rbrack = \lbrack A\rbrack^{- 1}\ \lbrack A\rbrack\) .

Are there any pitfalls of the Naïve Gauss elimination method?

Yes, there are two pitfalls of the Naïve Gauss elimination method.

Division by zero: It is possible for division by zero to occur during the beginning of the \(n - 1\) steps of forward elimination.

For example

\[5x_{2} + 6x_{3} = 11 \nonumber \]

\[4x_{1} + 5x_{2} + 7x_{3} = 16 \nonumber \]

\[9x_{1} + 2x_{2} + 3x_{3} = 15 \nonumber \]

will result in division by zero in the first step of forward elimination as the coefficient of \(x_{1}\) in the first equation is zero as is evident when we write the equations in matrix form.

\[\begin{bmatrix} 0 & 5 & 6 \\ 4 & 5 & 7 \\ 9 & 2 & 3 \\ \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{bmatrix} = \begin{bmatrix} 11 \\ 16 \\ 15 \\ \end{bmatrix} \nonumber \]

But what about the equations below: Is division by zero a problem?

\[5x_{1} + 6x_{2} + 7x_{3} = 18 \nonumber \]

\[10x_{1} + 12x_{2} + 3x_{3} = 25 \nonumber \]

\[20x_{1} + 17x_{2} + 19x_{3} = 56 \nonumber \]

Written in matrix form,

\[\begin{bmatrix} 5 & 6 & 7 \\ 10 & 12 & 3 \\ 20 & 17 & 19 \\ \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{bmatrix} = \begin{bmatrix} 18 \\25 \\ 56 \\ \end{bmatrix} \nonumber \]

there is no issue of division by zero in the first step of forward elimination. The pivot element is the coefficient of \(x_{1}\) in the first equation, 5, and that is a non-zero number. However, at the end of the first step of forward elimination, we get the following equations in matrix form

\[\begin{bmatrix} 5 & 6 & 7 \\ 0 & 0 & - 11 \\ 0 & - 7 & - 9 \\ \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{bmatrix} = \begin{bmatrix} 18 \\ - 11 \\ - 16 \\ \end{bmatrix} \nonumber \]

Now at the beginning of the \(2^{nd}\) step of forward elimination, the coefficient of \(x_{2}\) in Equation 2 would be used as the pivot element. That element is zero and hence would create the division by zero problem.

So it is important to consider that the possibility of division by zero can occur at the beginning of any step of forward elimination.

Round-off error

The Naïve Gauss elimination method is prone to round-off errors. This is true when there are large numbers of equations as errors propagate. Also, if there is subtraction of numbers from each other, it may create large errors. See the example below.

Remember Example 2 where we used Naïve Gauss elimination to solve

\[20x_{1} + 15x_{2} + 10x_{3} = 45 \nonumber \]

\[- 3x_{1} - 2.249x_{2} + 7x_{3} = 1.751 \nonumber \]

\[5x_{1} + x_{2} + 3x_{3} = 9 \nonumber \]

using six significant digits with chopping in your calculations? Repeat the problem, but now use five significant digits with chopping in your calculations.

Writing in the matrix form

\[\begin{bmatrix} 20 & 15 & 10 \\ - 3 & - 2.249 & 7 \\ 5 & 1 & 3 \\ \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{bmatrix} = \begin{bmatrix} 45 \\ 1.751 \\ 9 \\ \end{bmatrix} \nonumber \]

Divide Row 1 by 20 and then multiply it by –3, that is, multiply Row 1 by \(- 3/20 = - 0.15\) .

\[\begin{pmatrix} \begin{bmatrix} 20 & 15 & 10 \\ \end{bmatrix} & \left\lbrack 45 \right\rbrack \\ \end{pmatrix} \times - 0.15\ \text{gives Row 1 as} \nonumber \]

\[\begin{matrix} \left\lbrack - 3 \right.\ & - 2.25 & \left. \ - 1.5 \right\rbrack \\ \end{matrix} \ \ \ \ \ \ \ \ \left\lbrack - 6.75 \right\rbrack \nonumber \]

Subtract the result from Row 2

\[\frac{\begin{matrix} \ \ \lbrack\begin{matrix} -3 & \ \ \ -2.249 & \ \ \ \ \ 7 \\ \end{matrix}\ \ \ \ \ \ | & 1.751\rbrack \\ - \begin{matrix} \lbrack -3 & \ \ \ \ -2.25 \ \ -1.5 \ \ \\ \end{matrix}\ \ \ | & -6.75\rbrack \\ \end{matrix}}{\ \ \ \ \begin{matrix} \begin{matrix} 0 & \ \ \ \ \ 0.001 & \ \ \ \ 8.5 \\ \end{matrix} & \ \ \ \ \ 8.501 \end{matrix}} \nonumber \]

\[\begin{bmatrix} 20 & 15 & 10 \\ 0 & 0.001 & 8.5 \\ 5 & 1 & 3 \\ \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{bmatrix} = \begin{bmatrix} 45 \\ 8.501 \\ 9 \\ \end{bmatrix} \nonumber \]

Divide Row 1 by 20 and then multiply it by \(5\) , that is, multiply Row 1 by \(5/20 = 0.25\) .

\[\begin{pmatrix} \begin{bmatrix} 20 & 15 & 10 \\ \end{bmatrix} & \left\lbrack 45 \right\rbrack \\ \end{pmatrix} \times 0.25\ \text{gives Row 1 as} \nonumber \]

\[\begin{matrix} \left\lbrack 5 \right.\ & 3.75 & \left. \ 2.5 \right\rbrack \\ \end{matrix} \ \ \ \ \ \ \ \ \left\lbrack 11.25 \right\rbrack \nonumber \]

Subtract the result from Row 3

\[\frac{\begin{matrix} \ \ \lbrack\begin{matrix} 5 & \ \ \ \ \ \ 1 & \ \ \ \ \ \ 3 \\ \end{matrix}\ \ \ \ \ \ \ | & 9\rbrack \\ - \begin{matrix} \lbrack 5 & \ \ \ \ 3.75 \ \ \ \ \ 2.5 \ \ \\ \end{matrix}\ \ \ | & 11.25\rbrack \\ \end{matrix}}{\ \ \ \begin{matrix} \begin{matrix} 0 &\ -2.75 \ \ \ \ \ \ 0.5 \\ \end{matrix} & \ -2.25 \end{matrix}} \nonumber \]

\[\begin{bmatrix} 20 & 15 & 10 \\ 0 & 0.001 & 8.5 \\ 0 & - 2.75 & 0.5 \\ \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{bmatrix} = \begin{bmatrix} 45 \\ 8.501 \\ - 2.25 \\ \end{bmatrix} \nonumber \]

Now for the second step of forward elimination, we will use Row 2 as the pivot equation and eliminate Row 3: Column 2.

Divide Row 2 by \(0.001\) and then multiply it by \(-2.75\) , that is, multiply Row 2 by \(- 2.75/0.001 = - 2750\) .

\[\begin{pmatrix} \begin{bmatrix} 0 & 0.001 & 8.5 \\ \end{bmatrix} & \left\lbrack 8.501 \right\rbrack \\ \end{pmatrix} \times - 2750\ \text{gives Row 2 as} \nonumber \]

\[\begin{matrix} \left\lbrack 0 \right.\ & - 2.75 & \left. \ - 23375 \right\rbrack \\ \end{matrix}\ \ \ \ \ \ \ \ \left\lbrack - 23377.75 \right\rbrack \nonumber \]

Rewriting within 5 significant digits with chopping

\[\begin{matrix} \left\lbrack 0 \right.\ & - 2.75 & \left. \ - 23375 \right\rbrack \\ \end{matrix} \ \ \ \ \ \ \ \ \left\lbrack - 23377 \right\rbrack \nonumber \]

\[\frac{\begin{matrix} \ \ \lbrack\begin{matrix} 0 & \ \ \ \ \ \ -2.75 & \ \ \ \ \ \ 0.5 \\ \end{matrix}\ \ \ \ \ \ \ | & -2.25\rbrack \\ - \begin{matrix} \lbrack 0 & \ \ \ \ -2.75 \ \ \ \ \ -23375 \ \ \\ \end{matrix}\ \ \ | & -23377\rbrack \\ \end{matrix}}{\begin{matrix} \begin{matrix} 0 &\ \ \ \ \ \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ \ \ \ 23375 \\ \end{matrix} & \ \ \ \ \ \ \ \ 23374 \end{matrix}} \nonumber \]

Rewriting within 6 significant digits with chopping

\[\begin{matrix} \left\lbrack 0 \right.\ & 0 & \left. \ 23375 \right\rbrack \\ \end{matrix} \ \ \ \ \ \ \ \ \left\lbrack - 23374 \right\rbrack \nonumber \]

\[\begin{bmatrix} 20 & 15 & 10 \\ 0 & 0.001 & 8.5 \\ 0 & 0 & 23375 \\ \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{bmatrix} = \begin{bmatrix} 45 \\ 8.501 \\ 23374 \\ \end{bmatrix} \nonumber \]

This is the end of the forward elimination steps.

We can now solve the above equations by back substitution. From the third equation,

\[\begin{split} 23375x_{3} &= 23374\\ x_{3} &= \frac{23374}{23375}\\ &= 0.99995 \end{split} \nonumber \]

Substituting the value of \(x_{3}\) in the second equation

\[0.001x_{2} + 8.5x_{3} = 8.501 \nonumber \]

\[\begin{split} x_{2} &= \frac{8.501 - 8.5x_{3}}{0.001}\\ &= \frac{8.501 - 8.5 \times 0.99995}{0.001}\\ &= \frac{8.501 - 8.499575}{0.001}\\ &= \frac{8.501 - 8.4995}{0.001}\\ &= \frac{0.0015}{0.001}\\ &= 1.5 \end{split} \nonumber \]

Substituting the value of \(x_{3}\) and \(x_{2}\) in the first equation,

\[\begin{split} x_{1} &= \frac{45 - 15x_{2} - 10x_{3}}{20}\\ &= \frac{45 - 15 \times 1.5 - 10 \times 0.99995}{20}\\ &= \frac{45 - 22.5 - 9.9995}{20}\\ &= \frac{22.5 - 9.9995}{20}\\ &= \frac{12.5005}{20}\\ &= \frac{12.500}{20}\\ &= 0.625 \end{split} \nonumber \]

Hence the solution is

\[\begin{split} \left\lbrack X \right\rbrack &= \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{bmatrix}\\ &= \begin{bmatrix} 0.625 \\ 1.5 \\ 0.99995 \\ \end{bmatrix} \end{split} \nonumber \]

Compare this with the exact solution of

\[\left\lbrack X \right\rbrack = \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix} \nonumber \]

What are some techniques for improving the Naïve Gauss elimination method?

As seen in Example 3 , round off errors were large when five significant digits were used as opposed to six significant digits. One method of decreasing the round-off error would be to use more significant digits, that is, use double or quad precision for representing the numbers. However, this would not avoid possible division by zero errors in the Naïve Gauss elimination method. To avoid division by zero as well as reduce (not eliminate) round-off error, Gaussian elimination with partial pivoting is the method of choice.

How does Gaussian elimination with partial pivoting differ from Naïve Gauss elimination?

The two methods are the same, except in the beginning of each step of forward elimination, a row switching is done based on the following criterion. If there are \(n\) equations, then there are \(n - 1\) forward elimination steps. At the beginning of the \(k^{th}\) step of forward elimination, one finds the maximum of

\[\left| a_{kk} \right|,\left| a_{k + 1,k} \right|,\ldots\ldots,\left| a_{nk} \right| \nonumber \]

Then if the maximum of these values is \(\left| a_{pk} \right|\) in the \(p^{th}\) row, \(k \leq p \leq n\) , then switch rows \(p\) and \(k\) .

The other steps of forward elimination are the same as the Naïve Gauss elimination method. The back substitution steps stay exactly the same as the Naïve Gauss elimination method.

In the previous two examples, we used Naive Gauss elimination to solve

using five and six significant digits with chopping in the calculations. Using five significant digits with chopping, the solution found was

This is different from the exact solution of

\[\begin{split} \left\lbrack X \right\rbrack &= \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{bmatrix}\\ &= \begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix} \end{split} \nonumber \]

Find the solution using Gaussian elimination with partial pivoting using five significant digits with chopping in your calculations.

Now for the first step of forward elimination, the absolute value of the first column elements below Row 1 is

\[\left| 20 \right|,\left| - 3 \right|,\left| 5 \right| \nonumber \]

\[20,\ 3,\ 5 \nonumber \]

So, the largest absolute value is in the Row \(1\) . So as per Gaussian elimination with partial pivoting, the switch is between Row \(1\) and Row \(1\) to give

Divide Row \(1\) by \(20\) and then multiply it by \(-3\) , that is, multiply Row \(1\) by \(\displaystyle - 3/20 = - 0.15\) .

\[\begin{matrix} \left\lbrack - 3 \right.\ & - 2.25 & \left. \ - 1.5 \right\rbrack \\ \end{matrix}\ \ \ \ \ \ \ \ \left\lbrack - 6.75 \right\rbrack \nonumber \]

Divide Row \(1\) by \(20\) and then multiply it by \(5\) , that is, multiply Row \(1\) by \(5/20 = 0.25\) .

\[\begin{matrix} \left\lbrack 5 \right.\ & 3.75 & \left. \ 2.5 \right\rbrack \\ \end{matrix}\ \ \ \ \ \ \ \ \left\lbrack 11.25 \right\rbrack \nonumber \]

\[\displaystyle \frac{\begin{matrix} \ \ \lbrack\begin{matrix} 5 & \ \ \ \ \ \ 1 & \ \ \ \ \ \ 3 \\ \end{matrix}\ \ \ \ \ \ \ | & 9\rbrack \\ - \begin{matrix} \lbrack 5 & \ \ \ \ 3.75 \ \ \ \ \ 2.5 \ \ \\ \end{matrix}\ \ \ | & 11.25\rbrack \\ \end{matrix}}{\ \ \ \begin{matrix} \begin{matrix} 0 &\ -2.75 \ \ \ \ \ \ 0.5 \\ \end{matrix} & \ -2.25 \end{matrix}} \nonumber \]

This is the end of the first step of forward elimination.

Now for the second step of forward elimination, the absolute value of the second column elements below Row 1 is

\[\left| 0.001 \right|,\left| - 2.75 \right| \nonumber \]

\[0.001,\ 2.75 \nonumber \]

So, the largest absolute value is in Row \(3\) . So, Row \(2\) is switched with Row \(3\) to give

\[\begin{bmatrix} 20 & 15 & 10 \\ 0 & - 2.75 & 0.5 \\ 0 & 0.001 & 8.5 \\ \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{bmatrix} = \begin{bmatrix} 45 \\ - 2.25 \\ 8.501 \\ \end{bmatrix} \nonumber \]

Divide Row \(2\) by \(-2.75\) and then multiply it by \(0.001\) , that is, multiply Row \(2\) by \(0.001/ - 2.75 = - 0.00036363\) .

\[\begin{pmatrix} \begin{bmatrix} 0 & - 2.75 & 0.5 \\ \end{bmatrix} & \left\lbrack - 2.25 \right\rbrack \\ \end{pmatrix} \times - 0.00036363\ \text{gives Row 2 as} \nonumber \]

\[\begin{matrix} \left\lbrack 0 \right.\ & 0.00099998 & \left. \ - 0.00018182 \right\rbrack \\ \end{matrix}\ \ \ \ \ \ \ \ \left\lbrack 0.00081816 \right\rbrack \nonumber \]

\[\displaystyle \frac{\begin{matrix} \ \ \ \lbrack \begin{matrix} 0 & \ \ \ \ \ \ \ \ \ \ \ 0.001 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 8.5 \\ \end{matrix} \ \ \ \ \ \ \ \ \ \ \ \ \ \ | & 8.501\rbrack \\ - \begin{matrix} \lbrack 0 & \ \ \ \ 0.00099998 \ \ \ \ \ -0.00018182 \ \ \\ \end{matrix}\ \ \ | & 0.00081816\rbrack \\ \end{matrix}}{\ \ \begin{matrix} \ \begin{matrix} 0 &\ \ \ \ \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 8.50018182 \\ \end{matrix} & \ \ \ \ \ \ \ \ \ 8.50018184 \end{matrix}} \nonumber \]

Rewriting within \(5\) significant digits with chopping

\[\begin{matrix} \left\lbrack 0 \right.\ & 0 & \left. \ 8.5001 \right\rbrack \\ \end{matrix}\ \ \ \ \ \ \ \ \left\lbrack 8.5001 \right\rbrack \nonumber \]

\[\begin{bmatrix} 20 & 15 & 10 \\ 0 & - 2.75 & 0.5 \\ 0 & 0 & 8.5001 \\ \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{bmatrix} = \begin{bmatrix} 45 \\ - 2.25 \\ 8.5001 \\ \end{bmatrix} \nonumber \]

\[8.5001x_{3} = 8.5001 \nonumber \]

\[\begin{split} x_{3} &= \frac{8.5001}{8.5001}\\ &= 1 \end{split} \nonumber \]

Substituting the value of \(x_{3}\) in Row \(2\)

\[- 2.75x_{2} + 0.5x_{3} = - 2.25 \nonumber \]

\[\begin{split} x_{2} &= \frac{- 2.25 - 0.5x_{2}}{- 2.75}\\ &= \frac{- 2.25 - 0.5 \times 1}{- 2.75}\\ &= \frac{- 2.25 - 0.5}{- 2.75}\\ &= \frac{- 2.75}{- 2.75}\\ &= 1\end{split} \nonumber \]

Substituting the value of \(x_{3}\) and \(x_{2}\) in Row 1

\[\begin{split} x_{1} &= \frac{45 - 15x_{2} - 10x_{3}}{20}\\ &= \frac{45 - 15 \times 1 - 10 \times 1}{20}\\ &= \frac{45 - 15 - 10}{20}\\ &= \frac{30 - 10}{20}\\ &= \frac{20}{20}\\ &= 1 \end{split} \nonumber \]

Gaussian Elimination Method for Solving Simultaneous Linear Equations Quiz

The goal of forward elimination steps in the Naïve Gauss elimination method is to reduce the coefficient matrix to a (an) _____________ matrix.

(A) diagonal

(B) identity

(C) lower triangular

(D) upper triangular

Division by zero during forward elimination steps in Naïve Gaussian elimination of the set of equations \(\left\lbrack A \right\rbrack\left\lbrack X \right\rbrack = \left\lbrack C \right\rbrack\) implies the coefficient matrix \(\left\lbrack A \right\rbrack\)

(A) is invertible

(B) is nonsingular

(C) may be singular or nonsingular

(D) is singular

Using a computer with four significant digits with chopping, the Naïve Gauss elimination solution to

\[\begin{matrix} 0.0030x_{1} + 55.23x_{2} = 58.12 \\ 6.239x_{1} - 7.123x_{2} = 47.23 \\ \end{matrix} \nonumber \] is

(A) \(x_{1} = 26.66;\ x_{2} = 1.051\)

(B) \(x_{1} = 8.769;\ x_{2} = 1.051\)

(C) \(x_{1} = 8.800;\ x_{2} = 1.000\)

(D) \(x_{1} = 8.771;\ x_{2} = 1.052\)

Using a computer with four significant digits with chopping, the Gaussian elimination with partial pivoting solution to

At the end of the forward elimination steps of the Naïve Gauss elimination method on the following equations

\[\begin{bmatrix} {4.2857 \times 1}{0}^7 & {- 9.2307 \times 1}{0}^5 & {0} & {0} \\ {4.2857 \times 1}{0}^7 & {- 5.4619 \times 1}{0}^5 & {- 4.2857 \times 1}{0}^7 & {5.4619 \times 1}{0}^5 \\ {- 6.5} & {- 0.15384} & {6.5} & {0.15384} \\ {0} & {0} & {4.2857 \times 1}{0}^7 & {- 3.6057 \times 1}{0}^5 \\ \end{bmatrix}{\ \ }\begin{bmatrix} {c}_1 \\ {c}_2 \\ {c}_3 \\ {c}_4 \\ \end{bmatrix}{=}\begin{bmatrix} {- 7.887 \times 1}{0}^3 \\ {0} \\ {0.007} \\ {0} \\ \end{bmatrix} \nonumber \]

the resulting equations in matrix form are given by

\[\begin{bmatrix} {4.2857 \times 1}{0}^7 & {- 9.2307 \times 1}{0}^5 & {0} & {0} \\ {0} & {3.7688 \times 1}{0}^5 & {- 4.2857 \times 1}{0}^7 & {5.4619 \times 1}{0}^5 \\ {0} & {0} & {- 26.9140} & {0.579684} \\ {0} & {0} & {0} & {5.62500 \times 1}{0}^5 \\ \end{bmatrix}{\ \ }\begin{bmatrix} {c}_1 \\ {c}_2 \\ {c}_3 \\ {c}_4 \\ \end{bmatrix}{=}\begin{bmatrix} {- 7.887 \times 1}{0}^3 \\ {7.887 \times 1}{0}^3 \\ {1.19530 \times 1}{0}^-2 \\ {1.90336 \times 1}{0}^4 \\ \end{bmatrix} \nonumber \]

The determinant of the original coefficient matrix is

(A) \({0.00}\)

(B) \({4.2857 \times 1}{0}^7\)

(C) \({5.486 \times 1}{0}^19\)

(D) \({- 2.445 \times 1}{0}^20\)

The following data is given for the velocity of the rocket as a function of time. To find the velocity at \(t = 21s\) , you are asked to use a quadratic polynomial, \(v(t) = at^{2} + {bt} + c\) to approximate the velocity profile.

The correct set of equations that will find a , b and c are

(A) \(\begin{bmatrix} {176} & {14} & {1} \\ {225} & {15} & {1} \\ {400} & {20} & {1} \\ \end{bmatrix}\begin{bmatrix} {a} \\ {b} \\ {c} \\ \end{bmatrix}{=}\begin{bmatrix} {227.04} \\ {362.78} \\ {517.35} \\ \end{bmatrix}\)

(B) \(\begin{bmatrix} {225} & {15} & {1} \\ {400} & {20} & {1} \\ {900} & {30} & {1} \\ \end{bmatrix}\begin{bmatrix} {a} \\ {b} \\ {c} \\ \end{bmatrix}{=}\begin{bmatrix} {362.78} \\ {517.35} \\ {602.97} \\ \end{bmatrix}\)

(C) \(\begin{bmatrix} {0} & {0} & {1} \\ {225} & {15} & {1} \\ {400} & {20} & {1} \\ \end{bmatrix}\begin{bmatrix} {a} \\ {b} \\ {c} \\ \end{bmatrix}{=}\begin{bmatrix} {0} \\ {362.78} \\ {517.35} \\ \end{bmatrix}\)

(D) \(\begin{bmatrix} 400 & 20 & 1 \\ 900 & 30 & 1 \\ 1225 & 35 & 1 \\ \end{bmatrix}\begin{bmatrix} a \\ b \\ c \\ \end{bmatrix} = \begin{bmatrix} 517.35 \\ 602.97 \\ 901.67 \\ \end{bmatrix}\)

Gaussian Elimination Method for Solving Simultaneous Linear Equations Exercise

Use Naïve Gauss elimination to solve

\[{4x_{1} + x_{2} - x_{3} = - 2} \nonumber \] \[{5x_{1} + x_{2} + 2x_{3} = 4} \nonumber \] \[{6x_{1} + x_{2} + x_{3} = 6} \nonumber \]

Assume that you are using a computer with four significant digits with chopping. Use Naïve Gauss elimination method to solve

\[[A]= \begin{bmatrix} 10 & - 7 & 0 \\ - 3 & 2.099 & 6 \\ 5 & - 1 & 5 \\ \end{bmatrix} \nonumber \]

Find the determinant of [A] using forward elimination step of naïve Gauss elimination method.

At the end of forward elimination steps using naïve Gauss elimination method on the coefficient matrix

\[[A]= \begin{bmatrix} 25 & c & 1 \\ 64 & a & 1 \\ 144 & b & 1 \\ \end{bmatrix} \nonumber \]

[A] reduces to

\[[B]= \begin{bmatrix} 25 & 5 & 1 \\ 0 & - 4.8 & - 1.56 \\ 0 & 0 & 0.7 \\ \end{bmatrix} \nonumber \]

What is the determinant of \([A]\) ?

Using Gaussian elimination with partial pivoting to solve

Assume that you are using a computer with four significant digits with chopping, use Gaussian elimination with partial pivoting to solve

How to Solve Simultaneous Equations

How to Solve Simultaneous Equations

It has been said that “If your child wants to score AL1 for PSLE Math, he must know how to solve Simultaneous Equations.”

Someone also told me that the ability to solve Simultaneous Equations is what differentiates the ‘AL1’ students and ‘A’ students. To a certain extent, I think it is true. Most of the challenging problem sums in PSLE Math has 2 unknowns. To solve for 2 unknowns, you need to create 2 equations and solve them at the same time. That is we call them Simultaneous Equations.

Solving Simultaneous Equations is an important skill to solve challenging questions in PSLE Math papers.

There are 2 popular methods to solve Simultaneous Equations: The elimination Method and the Substitution Method

Elimination Method

We use the elimination method to eliminate one of the variables.

The elimination method is used when the values of the coefficients of one variable in both equations are the same. [Coefficient = Number in front of the variable]

Example of the Elimination Method

3x + 7y = 38  – (1) 3x + 6y = 33  – (2)

In this example, we notice that the coefficients of x are the same in both equations ( 3x = 3x ).

Hence, we use equation 1 minus equation 2 to eliminate  x to find the value of  y.

7y − 6y = 38 − 33

Next, we use the value of  y to substitute into either of the equations to find the value of  x .

3x + 7 x 5 = 38

3x + 35 = 38

Substitution Method

In this method, we first rearrange one equation to express one variable in terms of the other variable.

Next, we substitute this expression into the other equation to obtain an equation in only one variable.

Example of the Substitution Method

4x + y = 11   – (1) 7x + 2y = 20  – (2)

From the first equation, we can rearrange the equation to make y in terms of x and label this as Equation 3.

y = 11 − 4x  – (3)

Next, we substitute Equation 3 into Equation 2 to obtain one equation in  x only.

7x + 2(11 − 4x) = 20

7x + 22 − 8x = 20

x = 22 − 20

Now, we can substitute the value of  x into Equation 3 to find the value of  y .

y = 11 – 4 x 2

Can My Child Use Model Drawing to Solve Simultaneous Equations?

Yes, your child can also use Model Drawing to solve Simultaneous Equations!

Model drawing is great for visual learners. As primary school students’ minds are still not trained to handle complex Algebra yet, model drawing maybe more appealing for them.

Nonetheless, Algebra is a must-know skill as you will be learning how to use Algebra to solve more complicated equations in Secondary Math . That is why I strongly encourage primary school students to pick up Algebra well.

I shall go through how to solve Simultaneous Equations using Models and Algebra. Watch the 2 videos below to learn how.

Ahmad and Bernard had a total of  $240. Ahmad spent 2/3 of his money and Bernard spent 3/5 of his money. As a result, Bernard had $8 more than Ahmad. How much did they spend altogether ?

Using Model to Solve Simultaneous Equations

Using Algebra to Solve Simultaneous Equations

After watching the videos, which method do you prefer?

Many students also ask me, “Which method is better?”

Either one is fine as long as you know how to apply it correctly. Marks will be awarded as long as your workings are logical. In fact, you may have other methods instead of Model and Algebra. 

These methods are covered more in-depth in my online course. More details can be found below. 

8 Must Know Methods to Solve Problem Sums - Online Course

Don't go for exams with knowing these 8 methods they are crucial.

8 Must Know Methods for PSLE Math

Receive all the recorded lessons, practice questions, and guidance from Mr Jimmy!

If you’re looking for more resources to give your child a better chance at succeeding in their primary school exams, don’t hesitate to check out our PSLE math tuition classes .

We provide personalised guidance and support so your child can feel confident and secure during their exam preparation. With the right resources and strategy, anything is achievable!

Enroll your child at Jimmy Maths today!

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solving problems using simultaneous equations

IMAGES

  1. Simultaneous Linear Equations

    solving problems using simultaneous equations

  2. Solving Simultaneous Equations Graphically

    solving problems using simultaneous equations

  3. SOLVING PROBLEMS USING SIMULTANEOUS EQUATIONS

    solving problems using simultaneous equations

  4. Method of Elimination Steps to Solve Simultaneous Equations

    solving problems using simultaneous equations

  5. Solving Simultaneous Equations with Different Coefficients

    solving problems using simultaneous equations

  6. Solve Simultaneous Equations Algebraically

    solving problems using simultaneous equations

VIDEO

  1. SOLVING SIMULTANEOUS EQUATIONS USING ELIMINATION METHOD

  2. Solving Simultaneous Equations using the Substitution Method

  3. Solving Simultaneous Equations

  4. SOLVING SIMULTANEOUS EQUATIONS 1

  5. S2 Ex7C Solving Simultaneous Equations by Algebraic Methods 04B: Elimination

  6. Simultaneous equation (Additional Mathematics)

COMMENTS

  1. Solving Simultaneous Equations: Worksheets with Answers

    Mathster is a fantastic resource for creating online and paper-based assessments and homeworks. They have kindly allowed me to create 3 editable versions of each worksheet, complete with answers. Worksheet Name. 1. 2. 3. Simultaneous Equations - Elimination Method. 1. 2.

  2. Simultaneous Equations

    There are three different approaches to solve the simultaneous equations such as substitution, elimination, and augmented matrix method. Among these three methods, the two simplest methods will effectively solve the simultaneous equations to get accurate solutions. Here we are going to discuss these two important methods, namely, Elimination Method

  3. Simultaneous Equations

    We can solve simultaneous equations using different methods such as substitution method, elimination method, and graphically. In this article, we will explore the concept of simultaneous equations and learn how to solve them using different methods of solving.

  4. Simultaneous Equations

    When solving simultaneous equations you will need different methods depending on what sort of simultaneous equations you are dealing with. There are two sorts of simultaneous equations you will need to solve: linear simultaneous equations quadratic simultaneous equations

  5. Solving simultaneous equations

    The most common method for solving simultaneous equations is the elimination method which means one of the unknowns will be removed from each equation. The remaining unknown can then be...

  6. Simultaneous Equations

    This algebra video tutorial shows you how to solve simultaneous equations using the substitution method, the elimination method, graphical method, systems of...

  7. SOLVING PROBLEMS USING SIMULTANEOUS EQUATIONS

    0:00 / 8:58 Introduction SOLVING PROBLEMS USING SIMULTANEOUS EQUATIONS Classnotes with Nicole 5.21K subscribers Subscribe Subscribed 15 Save 2.7K views 3 years ago 📘 Year 9-10 Maths 📘 In this...

  8. Solving Systems of Equations (Simultaneous Equations)

    If you have two different equations with the same two unknowns in each, you can solve for both unknowns. There are three common methods for solving: addition/subtraction, substitution, and graphing. Addition/subtraction method This method is also known as the elimination method. To use the addition/subtraction method, do the following:

  9. Linear simultaneous equations

    Familiar Attempted Not started Quiz Unit test Geometrical representation Learn Solutions to systems of equations: dependent vs. independent Number of solutions to a system of equations Solution of simultaneous equations by use of graphs Learn Systems of equations with graphing: 5x+3y=7 & 3x-2y=8 Practice

  10. Simultaneous equations

    Solve the simultaneous equations: \ (y = 2x\) \ (x + y = 6\) One way to solve them is by using the substitution method. Begin by labelling the equations (1) and (2): \ (y = 2x\) (1) \...

  11. Simultaneous Equations Practice Questions

    . Click here for Answers . Simultaneous Equations Answers - Corbettmaths Watch on The Corbettmaths Practice Questions on Simultaneous Equations

  12. Simultaneous Linear Equations

    Simultaneous linear equations in two variables involve two unknown quantities to represent real-life problems. It helps in establishing a relationship between quantities, prices, speed, time, distance, etc results in a better understanding of the problems. We all use simultaneous linear equations in our daily life without knowing it.

  13. Simultaneous Equations Calculator

    Simultaneous equations can be used to solve a wide range of problems in finance, science, engineering, and other fields. They are often used to find the values of variables that make multiple equations or expressions true at the same time. Show more Related Symbolab blog posts

  14. Simultaneous Equations

    Problems on Simultaneous Equations Introduction Mathematics plays a vital role in our life; without mathematics many situations go wrong. There is no problem in physical science that can be solved without converting it into mathematics. Whenever a problem is converted to mathematics it gives an equation in some variable form.

  15. 4.4 Solving simultaneous equations

    We can solve simultaneous equations algebraically using substitution and elimination methods. We will also show that a system of simultaneous equations can be solved graphically. Siyavula's open Mathematics Grade 10 textbook, chapter 4 on Equations and inequalities covering 4.4 Solving simultaneous equations.

  16. Word Problems on Simultaneous Linear Equations

    Solving the solution of two variables of system equation that leads for the word problems on simultaneous linear equations is the ordered pair (x, y) which satisfies both the linear equations. Problems of different problems with the help of linear simultaneous equations:

  17. Word problems that lead to simultaneous equations. Examples

    H ERE ARE SOME EXAMPLES of problems that lead to simultaneous equations. Example 1. Andre has more money than Bob. If Andre gave Bob $20, they would have the same amount. While if Bob gave Andre $22, Andre would then have twice as much as Bob. How much does each one actually have? Solution . Let x be the amount of money that Andre has.

  18. Solving Problems Using Simultaneous Equations

    Solving Problems Using Simultaneous Equations ChavezMathClub 1.77K subscribers 12K views 10 years ago Here are some examples of how you can use simultaneous equations to solve word...

  19. SIMULTANEOUS EQUATIONS

    Solving Simultaneous Equations (Different x Coefficients) Fill in the Blanks ( Editable Word | PDF | Answers) Solving Simultaneous Equations Sort It Out ( Editable Word | PDF | Answers) Linear Simultaneous Equations Crack the Code ( Editable Word | PDF | Answers) Linear Simultaneous Equations Worded Problems Practice Strips ( Editable Word ...

  20. How to Solve Simultaneous Equations Using Elimination Method

    5. Solve to find the first unknown variable from the resulting (rather shortened) equation. Divide both sides by the coefficient of the left side. Take 5 to the other side.It will look like this:x = 25/5. [5] 6. 25 divided by 5 makes 5 so we have now found the value of "x" which is 5. 7. Find the value of "y".

  21. 6: Gaussian Elimination Method for Solving Simultaneous Linear Equations

    One of the most popular techniques for solving simultaneous linear equations is the Gaussian elimination method. The approach is designed to solve a general set of n equations and n unknowns. a11x1 + a12x2 + a13x3 + … + a1nxn = b1 a21x1 + a22x2 + a23x3 + … + a2nxn = b2 ⋮ ⋮ an1x1 + an2x2 + an3x3 + … + annxn = bn.

  22. How to Solve Simultaneous Equations

    Example of the Substitution Method. 4x + y = 11 - (1) 7x + 2y = 20 - (2) From the first equation, we can rearrange the equation to make y in terms of x and label this as Equation 3. y = 11 − 4x - (3) Next, we substitute Equation 3 into Equation 2 to obtain one equation in x only. 7x + 2 (11 − 4x) = 20. 7x + 22 − 8x = 20.

  23. Word Problems on Simultaneous Linear Equations

    As a result, we can solve the problem. Simultaneous equations are ones that need simultaneous solutions. Word problems, also known as applied problems, are situations in which unknowns are described using words. For success with word (or applied) difficulties, practice and mastery of certain basic translation processes are essential. Solving ...