how to solve equations fractions

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Course: Algebra 1   >   Unit 2

• Why we do the same thing to both sides: Variable on both sides
• Intro to equations with variables on both sides
• Equations with variables on both sides: 20-7x=6x-6
• Equations with variables on both sides

Equation with variables on both sides: fractions

• Equations with variables on both sides: decimals & fractions
• Equation with the variable in the denominator

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Video transcript

Module 5: Multi-Step Linear Equations

Solving equations by clearing fractions, learning outcomes.

• Use the least common denominator to eliminate fractions from a linear equation before solving it
• Solve equations with fractions that require several steps

You may feel overwhelmed when you see fractions in an equation, so we are going to show a method to solve equations with fractions where you use the common denominator to eliminate the fractions from an equation. The result of this operation will be a new equation, equivalent to the first, but with no fractions.

Pay attention to the fact that each term in the equation gets multiplied by the least common denominator. That’s what makes it equal to the original!

Solve: [latex]\Large\frac{1}{8}\normalsize x+\Large\frac{1}{2}=\Large\frac{1}{4}[/latex]

In the last example, the least common denominator was [latex]8[/latex]. Now it’s your turn to find an LCD, and clear the fractions before you solve these linear equations.

Notice that once we cleared the equation of fractions, the equation was like those we solved earlier in this chapter. We changed the problem to one we already knew how to solve!

Solve equations by clearing the Denominators

• Find the least common denominator of all the fractions in the equation.
• Multiply both sides of the equation by that LCD. This clears the fractions.
• Isolate the variable terms on one side, and the constant terms on the other side.
• Simplify both sides.
• Use the multiplication or division property to make the coefficient on the variable equal to [latex]1[/latex].

Here’s an example where you have three variable terms. After you clear fractions with the LCD, you will simplify the three variable terms, then isolate the variable.

Solve: [latex]7=\Large\frac{1}{2}\normalsize x+\Large\frac{3}{4}\normalsize x-\Large\frac{2}{3}\normalsize x[/latex]

Show Solution

Solution: We want to clear the fractions by multiplying both sides of the equation by the LCD of all the fractions in the equation.

Now here’s a similar problem for you to try. Clear the fractions, simplify, then solve.

One of the most common mistakes when you clear fractions is forgetting to multiply BOTH sides of the equation by the LCD. If your answer doesn’t check, make sure you have multiplied both sides of the equation by the LCD.

In the next example, we’ll have variables and fractions on both sides of the equation. After you clear the fractions using the LCD, you will see that this equation is similar to ones with variables on both sides that we solved previously. Remember to choose a variable side and a constant side to help you organize your work.

Solve: [latex]x+\Large\frac{1}{3}=\Large\frac{1}{6}\normalsize x-\Large\frac{1}{2}[/latex]

Now you can try solving an equation with fractions that has variables on both sides of the equal sign. The answer may be a fraction.

In the following video we show another example of how to solve an equation that contains fractions and variables on both sides of the equal sign.

In the next example, we start with an equation where the variable term is locked up in some parentheses and multiplied by a fraction. You can clear the fraction, or if you use the distributive property it will eliminate the fraction.  Can you see why?

Solve: [latex]1=\Large\frac{1}{2}\normalsize\left(4x+2\right)[/latex]

Now you can try solving an equation that has the variable term in parentheses that are multiplied by a fraction.

• Question ID 142514, 142542. Authored by : Lumen Learning. License : CC BY: Attribution . License Terms : IMathAS Community License, CC-BY + GPL
• Solve a Linear Equation with Parentheses and a Fraction 2/3(9x-12)=8+2x. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/1dmEoG7DkN4 . License : CC BY: Attribution
• Ex 1: Solve an Equation with Fractions with Variable Terms on Both Sides. Authored by : James Sousa (Mathispower4u.com). Located at : https://youtu.be/G5R9jySFMpw . License : CC BY: Attribution
• Question ID 71948. Authored by : Alyson Day. License : CC BY: Attribution . License Terms : IMathAS Community License, CC-BY + GPL
• Prealgebra. Provided by : OpenStax. License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]

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Solving Equations with Fractions

I know fractions are difficult, but with these easy step-by step instructions you'll be solving equations with fractions in no time.

Do you start to get nervous when you see fractions? Do you have to stop and review all the rules for adding, subtracting, multiplying and dividing fractions?

If so, you are just like almost every other math student out there! But... I am going to make your life so much easier when it comes to solving equations with fractions!

Our first step when solving these equations is to get rid of the fractions because they are not easy to work with!

Let see what happens with a typical two-step equation with the distributive property.

In this problem, we would typically distribute the 3/4 throughout the parenthesis and then solve. Let's see what happens:

Yuck! That just made this problem worse! Now we have two fractions to contend with and that means subtracting fractions and multiplying fractions.

So... let's stop here and say,

We DO NOT want to do this! DO NOT distribute fractions.

We are going to learn how to get rid of the fractions and make this much more simple!

So... what do we do? We are going to get rid of just the denominator in the fraction, so we will be left with the numerator, or just an integer!

I know, easier said than done! It's really not hard, but before I get into it, I want to go over one algebra definition.

We need to discuss the word term.

In Algebra, each term within an equation is separated by a plus (+) sign, minus (-) sign or an equals sign (=). Variable or quantities that are multiplied or divided are considered the same term.

That last example is the most important to remember. If a quantity is in parentheses, it it considered one term!

Let's look at a few examples of how to solve these crazy looking problems!

Example 1 - Equations with Fractions

Take a look at this example on video if you are feeling overwhelmed.

Hopefully you were able to follow that example. I know it's tough, but if you can get rid of the fraction, it will make these problems so much easier. Keep going, you'll get the hang of it!

In the next example, you will see two fractions.  Since they have the same denominator, we will multiply by the denominator and get rid of both fractions.

Example 2 - Equations with Fractions with the Same Denominator

Did you notice how multiplying by 2 (the denominator of both fractions) allowed us to get rid of the fractions?  This is the best way to deal with equations that contain fractions.

In the next example, you will see what happens when you have 2 fractions that have different denominators. 

We still want to get rid of the fractions all in one step. Therefore, we need to multiply all terms by the least common multiple.  Remember how to find the LCM?  If not, check out the LCM lesson here .

Example 3 - Equations with Two Fractions with Different Denominators

Yes, the equations are getting harder, but if you take it step-by-step, you will arrive at the correct solution. Keep at it - I know you'll get it!

• Solving Equations
• Equations with Fractions

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Equations With Fractions

Here we will learn about equations with fractions, including solving equations with fractions where the unknown is the denominator of a fraction.

There are also equations with fractions worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

What are equations with fractions?

Equations with fractions involve solving equations where the unknown variable is part of the numerator and/or the denominator of the fraction.  

To solve equations with fractions we need to work out what the value of the unknown variable. We solve equations by using the “balancing method” by applying the inverse operation to both sides of the equation.

The inverse operation of addition is subtraction .

The inverse operation of subtraction is addition .

The inverse operation of multiplication is division .

The inverse operation of division is multiplication .

For example,

How to solve equations with fractions

In order to solve equations with fraction:

• Identify the operations that are being applied to the unknown variable.
• Apply the inverse operations, one at a time, to both sides of the equation.
• Write the final answer, checking that it is correct.

Equations with fractions worksheet

Get your free Equations with fractions worksheet of 20+ questions and answers. Includes reasoning and applied questions.

Related lessons on solving equations

Equations with fractions  is part of our series of lessons to support revision on  solving equations . You may find it helpful to start with the main solving equations lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

• Solving equations
• Quadratic equations  
• Linear equations
• Forming and solving equations
• Iteration maths

Equations with fractions examples

Example 1: equations with one operation.

Identify the operations that are being applied to the unknown variable .

The unknown variable is x .  Looking at the left hand side of the equation, the  x is divided by 5 (the denominator of the fraction).

2 Apply the inverse operations, one at a time, to both sides of the equation .

The inverse of “dividing by 5 ” is “multiplying by 5 ”.  We need to multiply both sides of the equation by 5 .

3 Write the final answer, checking that it is correct .

The final answer is:

We can check the answer by substituting the answer back into the original equation.

Example 2: equations with one operation

The unknown variable is x . Looking at the left hand side of the equation, the x is divided by 3 (the denominator of the fraction).

Apply the inverse operations, one at a time, to both sides of the equation .

The inverse of “dividing by 3 ” is “multiplying by 3 ”.  We need to multiply both sides of the equation by 3 .

Write the final answer, checking that it is correct .

Example 3: equations with two operations

Solve: 

The unknown variable is x . Looking at the left hand side of the equation, 1 is added to x and then divided by 2 (the denominator of the fraction).

We need to do the inverse operations in the reverse order. First we need to multiply both sides of the equation by 2 . Then we need to subtract 1 from both sides.

Example 4: equations with two operations

The unknown variable is x .  Looking at the left hand side of the equation, x is divided by 4 (the denominator of the fraction) and then 2 is subtracted.

We need to do the inverse operations in the reverse order. First we need to add 2 to both sides of the equation. Then we need to multiply both sides of the equation by 4 .

Example 5: equations with three operations

The unknown variable is x .  Looking at the left hand side of the equation, x is multiplied by 3 , then divided by 4 (the denominator of the fraction) and then 1 is added.

We need to do the inverse operations in the reverse order. First we need to subtract 1 to both sides of the equation. Then we need to multiply both sides of the equation by 5 and finally divide both sides by 3 .

Example 6: equations with three operations

The unknown variable is x . Looking at the left hand side of the equation, x is multiplied by 2 , then 1 is subtracted. Then we divide by 7 (the denominator).

We need to do the inverse operations in the reverse order. First we need to multiply both sides of the equation by 7 .  Then we need to add 1 to both sides and finally divide both sides by 2 .

Example 7: equations with the unknown as the denominator

The unknown variable is x . Looking at the left hand side of the equation, x is the denominator. 24 is divided by x .

We need to multiply both sides of the equation by x . Then we can divide both sides by 6 .

Example 8: equations with the unknown as the denominator

The unknown variable is x . Looking at the left hand side of the equation, x is the denominator. 18 is divided by x and then 6 is subtracted.

First we add 6 to both sides of the equation. Then we need to multiply both sides of the equation by x .  Then we can divide both sides by 9 .

Common misconceptions

• Types of number

The solution to an equation can be different types of number. The unknown does not have to be an integer (whole numbers), it can also be a fraction or a decimal and can be positive or negative.

• The side of the equation that th unknown is on 

The unknown variable, represented by a letter, is often on the left hand side of the equations however it doesn’t have to be. It could also be on the right hand side of an equation.

• Multiplying both sides of an equation

When multiplying each side of the equation of a number, it is a common mistake to forget to multiply every term. E.g. Solve: \frac{x}{2}+3=9

Here we have not multiplied the +3 by 2 resulting in the incorrect answer:

Here we have correctly multiplied each term by the denominator:

• Lowest common denominator (LCD)

It is common to get confused between solving equations involving fractions and adding and subtracting fractions. When adding and subtracting we need to work out the lowest/least common denominator (sometimes called the lowest common multiple or lcm) whereas when we solve equations involving fractions we need to multiply both sides of the equation by the denominator of the fraction.

Practice equations with fractions questions

1. Solve: \frac{x}{6}=3

2. Solve: \frac{x+4}{2}=7

3. Solve: \frac{x}{8}-5=1

4. Solve: \frac{3x+2}{4}=2

5. Solve: \frac{4x}{7}-2=6

6. Solve: \frac{42}{x}=7

Equations with fractions GCSE questions

1. Solve \frac{x}{5}=3

for the correct answer

2. Solve \frac{x-3}{7}=2

for the correct first step

3. Solve \frac{5a+6}{2}=23

5a+6=46 for the correct first step

5a=40 for the correct second step

x=8 for the correct answer

Learning checklist

You have now learned how to:

• Solve equations when there are fractions
• Solve fractions where the unknown is the denominator

The next lessons are

• Factorising
• Rearranging equations
• Simultaneous equations

Still stuck?

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8.4 Solve Equations with Fraction or Decimal Coefficients

Learning objectives.

Solve equations with fraction coefficients

• Solve equations with decimal coefficients

Be Prepared 8.4

Before you get started, take this readiness quiz.

• Multiply: 8 · 3 8 . 8 · 3 8 . If you missed this problem, review Example 4.28
• Find the LCD of 5 6 and 1 4 . 5 6 and 1 4 . If you missed this problem, review Example 4.63
• Multiply: 4.78 4.78 by 100 . 100 . If you missed this problem, review Example 5.18

Solve Equations with Fraction Coefficients

Let’s use the General Strategy for Solving Linear Equations introduced earlier to solve the equation 1 8 x + 1 2 = 1 4 . 1 8 x + 1 2 = 1 4 .

This method worked fine, but many students don’t feel very confident when they see all those fractions. So we are going to show an alternate method to solve equations with fractions. This alternate method eliminates the fractions.

We will apply the Multiplication Property of Equality and multiply both sides of an equation by the least common denominator of all the fractions in the equation. The result of this operation will be a new equation, equivalent to the first, but with no fractions. This process is called clearing the equation of fractions . Let’s solve the same equation again, but this time use the method that clears the fractions.

Example 8.37

Solve: 1 8 x + 1 2 = 1 4 . 1 8 x + 1 2 = 1 4 .

Try It 8.73

Solve: 1 4 x + 1 2 = 5 8 . 1 4 x + 1 2 = 5 8 .

Try It 8.74

Solve: 1 6 y − 1 3 = 1 6 . 1 6 y − 1 3 = 1 6 .

Notice in Example 8.37 that once we cleared the equation of fractions, the equation was like those we solved earlier in this chapter. We changed the problem to one we already knew how to solve! We then used the General Strategy for Solving Linear Equations.

Solve equations with fraction coefficients by clearing the fractions.

• Step 1. Find the least common denominator of all the fractions in the equation.
• Step 2. Multiply both sides of the equation by that LCD. This clears the fractions.
• Step 3. Solve using the General Strategy for Solving Linear Equations.

Example 8.38

Solve: 7 = 1 2 x + 3 4 x − 2 3 x . 7 = 1 2 x + 3 4 x − 2 3 x .

We want to clear the fractions by multiplying both sides of the equation by the LCD of all the fractions in the equation.

Try It 8.75

Solve: 6 = 1 2 v + 2 5 v − 3 4 v . 6 = 1 2 v + 2 5 v − 3 4 v .

Try It 8.76

Solve: −1 = 1 2 u + 1 4 u − 2 3 u . −1 = 1 2 u + 1 4 u − 2 3 u .

In the next example, we’ll have variables and fractions on both sides of the equation.

Example 8.39

Solve: x + 1 3 = 1 6 x − 1 2 . x + 1 3 = 1 6 x − 1 2 .

Try It 8.77

Solve: a + 3 4 = 3 8 a − 1 2 . a + 3 4 = 3 8 a − 1 2 .

Try It 8.78

Solve: c + 3 4 = 1 2 c − 1 4 . c + 3 4 = 1 2 c − 1 4 .

In Example 8.40 , we’ll start by using the Distributive Property. This step will clear the fractions right away!

Example 8.40

Solve: 1 = 1 2 ( 4 x + 2 ) . 1 = 1 2 ( 4 x + 2 ) .

Try It 8.79

Solve: −11 = 1 2 ( 6 p + 2 ) . −11 = 1 2 ( 6 p + 2 ) .

Try It 8.80

Solve: 8 = 1 3 ( 9 q + 6 ) . 8 = 1 3 ( 9 q + 6 ) .

Many times, there will still be fractions, even after distributing.

Example 8.41

Solve: 1 2 ( y − 5 ) = 1 4 ( y − 1 ) . 1 2 ( y − 5 ) = 1 4 ( y − 1 ) .

Try It 8.81

Solve: 1 5 ( n + 3 ) = 1 4 ( n + 2 ) . 1 5 ( n + 3 ) = 1 4 ( n + 2 ) .

Try It 8.82

Solve: 1 2 ( m − 3 ) = 1 4 ( m − 7 ) . 1 2 ( m − 3 ) = 1 4 ( m − 7 ) .

Solve Equations with Decimal Coefficients

Some equations have decimals in them. This kind of equation will occur when we solve problems dealing with money and percent. But decimals are really another way to represent fractions. For example, 0.3 = 3 10 0.3 = 3 10 and 0.17 = 17 100 . 0.17 = 17 100 . So, when we have an equation with decimals, we can use the same process we used to clear fractions—multiply both sides of the equation by the least common denominator .

Example 8.42

Solve: 0.8 x − 5 = 7 . 0.8 x − 5 = 7 .

The only decimal in the equation is 0.8 . 0.8 . Since 0.8 = 8 10 , 0.8 = 8 10 , the LCD is 10 . 10 . We can multiply both sides by 10 10 to clear the decimal.

Try It 8.83

Solve: 0.6 x − 1 = 11 . 0.6 x − 1 = 11 .

Try It 8.84

Solve: 1.2 x − 3 = 9 . 1.2 x − 3 = 9 .

Example 8.43

Solve: 0.06 x + 0.02 = 0.25 x − 1.5 . 0.06 x + 0.02 = 0.25 x − 1.5 .

Look at the decimals and think of the equivalent fractions.

0.06 = 6 100 , 0.02 = 2 100 , 0.25 = 25 100 , 1.5 = 1 5 10 0.06 = 6 100 , 0.02 = 2 100 , 0.25 = 25 100 , 1.5 = 1 5 10

Notice, the LCD is 100 . 100 .

By multiplying by the LCD we will clear the decimals.

Try It 8.85

Solve: 0.14 h + 0.12 = 0.35 h − 2.4 . 0.14 h + 0.12 = 0.35 h − 2.4 .

Try It 8.86

Solve: 0.65 k − 0.1 = 0.4 k − 0.35 . 0.65 k − 0.1 = 0.4 k − 0.35 .

The next example uses an equation that is typical of the ones we will see in the money applications in the next chapter. Notice that we will distribute the decimal first before we clear all decimals in the equation.

Example 8.44

Solve: 0.25 x + 0.05 ( x + 3 ) = 2.85 . 0.25 x + 0.05 ( x + 3 ) = 2.85 .

Try It 8.87

Solve: 0.25 n + 0.05 ( n + 5 ) = 2.95 . 0.25 n + 0.05 ( n + 5 ) = 2.95 .

Try It 8.88

Solve: 0.10 d + 0.05 ( d − 5 ) = 2.15 . 0.10 d + 0.05 ( d − 5 ) = 2.15 .

ACCESS ADDITIONAL ONLINE RESOURCES

• Solve an Equation with Fractions with Variable Terms on Both Sides
• Ex 1: Solve an Equation with Fractions with Variable Terms on Both Sides
• Ex 2: Solve an Equation with Fractions with Variable Terms on Both Sides
• Solving Multiple Step Equations Involving Decimals
• Ex: Solve a Linear Equation With Decimals and Variables on Both Sides
• Ex: Solve an Equation with Decimals and Parentheses

Section 8.4 Exercises

Practice makes perfect.

In the following exercises, solve the equation by clearing the fractions.

1 4 x − 1 2 = − 3 4 1 4 x − 1 2 = − 3 4

3 4 x − 1 2 = 1 4 3 4 x − 1 2 = 1 4

5 6 y − 2 3 = − 3 2 5 6 y − 2 3 = − 3 2

5 6 y − 1 3 = − 7 6 5 6 y − 1 3 = − 7 6

1 2 a + 3 8 = 3 4 1 2 a + 3 8 = 3 4

5 8 b + 1 2 = − 3 4 5 8 b + 1 2 = − 3 4

2 = 1 3 x − 1 2 x + 2 3 x 2 = 1 3 x − 1 2 x + 2 3 x

2 = 3 5 x − 1 3 x + 2 5 x 2 = 3 5 x − 1 3 x + 2 5 x

1 4 m − 4 5 m + 1 2 m = −1 1 4 m − 4 5 m + 1 2 m = −1

5 6 n − 1 4 n − 1 2 n = −2 5 6 n − 1 4 n − 1 2 n = −2

x + 1 2 = 2 3 x − 1 2 x + 1 2 = 2 3 x − 1 2

x + 3 4 = 1 2 x − 5 4 x + 3 4 = 1 2 x − 5 4

1 3 w + 5 4 = w − 1 4 1 3 w + 5 4 = w − 1 4

3 2 z + 1 3 = z − 2 3 3 2 z + 1 3 = z − 2 3

1 2 x − 1 4 = 1 12 x + 1 6 1 2 x − 1 4 = 1 12 x + 1 6

1 2 a − 1 4 = 1 6 a + 1 12 1 2 a − 1 4 = 1 6 a + 1 12

1 3 b + 1 5 = 2 5 b − 3 5 1 3 b + 1 5 = 2 5 b − 3 5

1 3 x + 2 5 = 1 5 x − 2 5 1 3 x + 2 5 = 1 5 x − 2 5

1 = 1 6 ( 12 x − 6 ) 1 = 1 6 ( 12 x − 6 )

1 = 1 5 ( 15 x − 10 ) 1 = 1 5 ( 15 x − 10 )

1 4 ( p − 7 ) = 1 3 ( p + 5 ) 1 4 ( p − 7 ) = 1 3 ( p + 5 )

1 5 ( q + 3 ) = 1 2 ( q − 3 ) 1 5 ( q + 3 ) = 1 2 ( q − 3 )

1 2 ( x + 4 ) = 3 4 1 2 ( x + 4 ) = 3 4

1 3 ( x + 5 ) = 5 6 1 3 ( x + 5 ) = 5 6

In the following exercises, solve the equation by clearing the decimals.

0.6 y + 3 = 9 0.6 y + 3 = 9

0.4 y − 4 = 2 0.4 y − 4 = 2

3.6 j − 2 = 5.2 3.6 j − 2 = 5.2

2.1 k + 3 = 7.2 2.1 k + 3 = 7.2

0.4 x + 0.6 = 0.5 x − 1.2 0.4 x + 0.6 = 0.5 x − 1.2

0.7 x + 0.4 = 0.6 x + 2.4 0.7 x + 0.4 = 0.6 x + 2.4

0.23 x + 1.47 = 0.37 x − 1.05 0.23 x + 1.47 = 0.37 x − 1.05

0.48 x + 1.56 = 0.58 x − 0.64 0.48 x + 1.56 = 0.58 x − 0.64

0.9 x − 1.25 = 0.75 x + 1.75 0.9 x − 1.25 = 0.75 x + 1.75

1.2 x − 0.91 = 0.8 x + 2.29 1.2 x − 0.91 = 0.8 x + 2.29

0.05 n + 0.10 ( n + 8 ) = 2.15 0.05 n + 0.10 ( n + 8 ) = 2.15

0.05 n + 0.10 ( n + 7 ) = 3.55 0.05 n + 0.10 ( n + 7 ) = 3.55

0.10 d + 0.25 ( d + 5 ) = 4.05 0.10 d + 0.25 ( d + 5 ) = 4.05

0.10 d + 0.25 ( d + 7 ) = 5.25 0.10 d + 0.25 ( d + 7 ) = 5.25

0.05 ( q − 5 ) + 0.25 q = 3.05 0.05 ( q − 5 ) + 0.25 q = 3.05

0.05 ( q − 8 ) + 0.25 q = 4.10 0.05 ( q − 8 ) + 0.25 q = 4.10

Everyday Math

Coins Taylor has $2.00 $2.00 in dimes and pennies. The number of pennies is 2 2 more than the number of dimes. Solve the equation 0.10 d + 0.01 ( d + 2 ) = 2 0.10 d + 0.01 ( d + 2 ) = 2 for d , d , the number of dimes.

Stamps Travis bought $9.45 $9.45 worth of 49-cent 49-cent stamps and 21-cent 21-cent stamps. The number of 21-cent 21-cent stamps was 5 5 less than the number of 49-cent 49-cent stamps. Solve the equation 0.49 s + 0.21 ( s − 5 ) = 9.45 0.49 s + 0.21 ( s − 5 ) = 9.45 for s , s , to find the number of 49-cent 49-cent stamps Travis bought.

Writing Exercises

Explain how to find the least common denominator of 3 8 , 1 6 , and 2 3 . 3 8 , 1 6 , and 2 3 .

If an equation has several fractions, how does multiplying both sides by the LCD make it easier to solve?

If an equation has fractions only on one side, why do you have to multiply both sides of the equation by the LCD?

In the equation 0.35 x + 2.1 = 3.85 , 0.35 x + 2.1 = 3.85 , what is the LCD? How do you know?

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ Overall, after looking at the checklist, do you think you are well-prepared for the next Chapter? Why or why not?

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How to Solve Linear Equations with Fractions – A Step-by-Step Guide

To solve linear equations with fractions , I first clear the fractions by finding the least common denominator (LCD) and multiplying each term of the equation by this number.

This crucial step transforms the equation into a more straightforward format without fractions , which simplifies the process of isolating the variable.

Solving these equations involves the familiar steps of combining like terms and isolating the variable to one side of the equation . The goal is to determine the value of the variable that makes the equation true.

My approach ensures that the process of solving becomes a smooth transition from an equation laden with fractions to one that’s more manageable, akin to standard linear equations.

Keeping the balance between both sides of the equation is paramount as I manipulate it to find a solution. Engaging with this technique not only enhances my algebraic agility but also prepares me for tackling a broader range of mathematical problems.

Stay tuned as I unwrap this process step-by-step, ensuring you’ll confidently handle these equations on your own.

Strategies for Solving Linear Equations With Fractions

When I encounter a linear equation with fractions , I like to start by finding the least common denominator ( LCD ) of all the fractions . This strategy involves multiple steps which I will outline straightforwardly.

First, to clear the equation of fractions , I multiply every term by the LCD . For example, if I have the equation $\frac{2}{3}x + \frac{1}{4} = \frac{3}{8}x + \frac{5}{12}$, I would multiply each term by 24, the LCD of 3, 4, 8, and 12. This simplifies the equation to $16x + 6 = 9x + 10$.

Here’s how I simplify further:

Combine like terms. I bring all variables to one side and constants to the other, creating an equivalent equation . From our equation , it would look like this: $16x – 9x = 10 – 6$, which simplifies to $7x = 4$.

Solve for the variable using inverse operations . In this case, I divide both sides by 7 to isolate x: $x = \frac{4}{7}$.

In some cases, there could be variables on both sides of the equation . My approach here is to first make sure all variables and fractions are cleared before combining like terms.

Throughout this process, I check my work by ensuring each multiplication and division step maintains the balance of the equation . The left-hand side should always be equal to the right-hand side after every operation.

Let me summarize the steps in the following table for easy reference:

It’s essential to keep in mind negative numbers and to apply the multiplication property of negative ones when fractions with negative numerators or denominators appear. This helps to maintain accuracy throughout the solving equations process.

Solving linear equations with fractions doesn’t have to be intimidating. I’ve walked you through a methodical process that can make these problems much more manageable.

Remember to start by finding the least common denominator (LCD) to clear the fractions and simplify the equation into a more familiar form. By multiplying each term by the LCD, our equation looks like any other linear equation without fractions.

Once the equation is clear of fractions, you can proceed with the steps we’re all used to: isolate the variable, combine like terms, and perform operations to solve for the unknown.

It’s crucial to maintain balance by performing the same operations on both sides of the equation. If our initial problem was $\frac{3}{4}x + 2 = \frac{3}{8}x – 4$ , after clearing the fractions and simplifying we would end up with an equation like $6x + 16 = 3x – 32$ , which is straightforward to solve for $x$ .

In my experience, practicing these steps can significantly boost your confidence in handling equations with fractional or decimal coefficients.

Always double-check your work by substituting the solution back into the original equation to ensure it holds. This not only verifies your answer but also reinforces your understanding of the solving process .

By mastering these concepts, you’re not just learning to solve another type of equation; you’re enhancing your overall mathematical skill set, which will be incredibly useful in more advanced math and real-world problem-solving.

Keep practicing, stay curious, and don’t hesitate to revisit previous sections if you need a refresher on the methods used to tackle these linear equations .

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How To Solve Equations With Fractions

What is an equation.

An equation is a statement used in mathematics to show that two items are equal. In fractions, an equation is used to find the value of a fraction when one or more of its parts are unknown.

If you’re stuck on a math problem that involves fractions, don’t worry! In this article, we’ll show you how to solve equations with fractions step by step.

Step 1: Find the least common denominator

Firstly we need to find the least common denominator (LCD) of the fractions found in the equation, which is the smallest number that can be a common denominator for both of the fractions. For this equation the LCD is 12 as this is the lowest common multiple of 4 and 6.

Step 2: Multiply the least common denominator

Multiply the LCD to both sides of the equation.

Step 3: Simplify the equation

Simplify both sides of the equation and make sure we’re only working with whole numbers.

Step 4: Simplify until there's one term on both sides

Move all terms with the variable on one side and further simplify both sides of the Equations so we have one term on both sides.

Step 5: Divide the coefficient on both sides

Once the variable is isolated on one side, divide the coefficient on both sides to solve for the unknown variable.

Examples of how to solve equations with fractions

Q1) Find the value of x in:

First, let’s find the least common denominator (LCD) of the fractions:

6=2×3 15=3×5 LCD:2×3×5=30

Multiply 30 on both sides of the equation. Make sure to simplify after distributing 30.

Move 6x on on the left-hand side of the equation to isolate the term with the variable.

20x-6x=60 14x=60

Divide 14 on both sides of the equation to solve for x.

Q2) An unknown fraction is added to 1 and we divide the sum by 3. The result is equal to 3/4. What is the value of the unknown fraction?

We can let n be the unknown number. We can set up the equation to solve the problem.

First, look for the least common denominator (LCD). Since 3 and 4 don’t share any common factors, we find their product.

Multiplying 12 to both sides of the equation, we have:

Move 4 on the right-hand side of the equation.

4n=9-4 4n=5

Divide both sides by 4 to isolate n.

Hence, the unknown fraction is equal to 5/4.

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Solving Systems Of Equations With Fractions Or Decimals

In these lessons, we help Algebra 1 students learn to solve systems of linear equations with fractions.

Related Pages Solving Systems of Equations or Simultaneous Equations using algebra Solving Simultaneous Equations Using Matrices More Lessons On Matrices Algebra Lessons

How To Solve Systems Of Equations With Fractions?

When a system includes an equation with fractions as coefficients: Step 1. Eliminate the fractions by multiplying each side of the equation by a common denominator. Step 2: Solve the resulting system using the addition method, elimination method, or the substitution method.

The following diagrams show how to solve systems of equations using the Substitution Method and the Elimination Method.

Systems of Equations with Fractions

Students learn to solve systems of linear equations that involve fractions. Students also learn to solve linear systems of equations by the method of their choice using the following rules: if one of the variables cancels out when the equations are added together, then use addition, and if a variable is already isolated in one of the equations, then use substitution.

Example: Solve the following systems of equations y = 1/3 x - 1 3x - 2y = 4

Solving a System Involving Fractions with Substitution

How to solve a system of 2 equations with 2 variables involving fractions using substitution?

Example: y = -3/4 x + 12 y = 1/3 x - 2

Solving a System Involving Fractions with Elimination

How to solve a linear system by the process of elimination where the equations contain fractions?

Example: x/2 + y/8 = 4 x/3 - y/2 = -2

Solving systems of linear equations involving decimals and fractions

• Scale the system so that integer coefficients exist.
• Use elimination to solve for the solutions, if they exist.

Examples: System 1 0.05x - 0.03y = 0.21 0.07x + 0.02y = 0.16

System 2 x/4 + y/6 = 1 x - y = 3

System 3 (x + 3)/4 + (y - 1)/3 = 1 2x - y = 1

Solving System Of Linear Equations by Adding (with decimals)

Examples: 3.5x + 2.5y = 17 -1.5x - 7.5y = -33

Solving Systems of Linear Equations with Fractions

Solve a system of linear equations using multiplication with addition or subtraction.

Examples: x/2 + y/3 = -4 x/5 + y/5 = -2

We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.

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3.6: Solving Equations with Fractions or Decimals

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Learning Objectives

By the end of this section, you will be able to:

• Solve equations with fraction coefficients
• Solve equations with decimal coefficients

Solve Equations with Fraction Coefficients

Let’s use the general strategy for solving linear equations introduced earlier to solve the equation, \(\frac{1}{8}x+\frac{1}{2}=\frac{1}{4}\).

This method worked fine, but many students do not feel very confident when they see all those fractions. So, we are going to show an alternate method to solve equations with fractions. This alternate method eliminates the fractions.

We will apply the Multiplication Property of Equality and multiply both sides of an equation by the least common denominator of all the fractions in the equation. The result of this operation will be a new equation, equivalent to the first, but without fractions. This process is called “clearing” the equation of fractions.

Let’s solve a similar equation, but this time use the method that eliminates the fractions.

Exercise \(\PageIndex{1}\): How to Solve Equations with Fraction Coefficients

Solve: \(\frac{1}{6}y - \frac{1}{3} = \frac{5}{6}\)

Exercise \(\PageIndex{2}\)

Solve: \(\frac{1}{4}x + \frac{1}{2} = \frac{5}{8}\)

\(x= \frac{1}{2}\)

Exercise \(\PageIndex{3}\)

Solve: \(\frac{1}{8}x + \frac{1}{2} = \frac{1}{4}\)

Notice in Exercise \(\PageIndex{1}\), once we cleared the equation of fractions, the equation was like those we solved earlier in this chapter. We changed the problem to one we already knew how to solve! We then used the General Strategy for Solving Linear Equations.

STRATEGY TO SOLVE EQUATIONS WITH FRACTION COEFFICIENTS.

• Find the least common denominator of all the fractions in the equation.
• Multiply both sides of the equation by that LCD. This clears the fractions.
• Solve using the General Strategy for Solving Linear Equations.

Exercise \(\PageIndex{4}\)

Solve: \(6 = \frac{1}{2}v + \frac{2}{5}v - \frac{3}{4}v\)

We want to clear the fractions by multiplying both sides of the equation by the LCD of all the fractions in the equation.

Exercise \(\PageIndex{5}\)

Solve: \(7 = \frac{1}{2}x + \frac{3}{4}x - \frac{2}{3}x\)

Exercise \(\PageIndex{6}\)

Solve: \(-1 = \frac{1}{2}u + \frac{1}{4}u - \frac{2}{3}u\)

\(u = -12\)

In the next example, we again have variables on both sides of the equation.

Exercise \(\PageIndex{7}\)

Solve: \(a + \frac{3}{4} = \frac{3}{8}a - \frac{1}{2}\)

Exercise \(\PageIndex{8}\)

Solve: \(x + \frac{1}{3} = \frac{1}{6}x - \frac{1}{2}\)

Exercise \(\PageIndex{9}\)

Solve: \(c + \frac{3}{4} = \frac{1}{2}c - \frac{1}{4}\)

In the next example, we start by using the Distributive Property. This step clears the fractions right away.

Exercise \(\PageIndex{10}\)

Solve: \(-5 = \frac{1}{4}(8x + 4)\)

Exercise \(\PageIndex{11}\)

Solve: \(-11 = \frac{1}{2}(6p + 2)\)

Exercise \(\PageIndex{12}\)

Solve: \(8 = \frac{1}{3}(9q + 6)\)

In the next example, even after distributing, we still have fractions to clear.

Exercise \(\PageIndex{13}\)

Solve: \(\frac{1}{2}(y - 5) = \frac{1}{4}(y - 1)\)

Exercise \(\PageIndex{14}\)

Solve: \(\frac{1}{5}(n + 3) = \frac{1}{4}(n + 2)\)

Exercise \(\PageIndex{15}\)

Solve: \(\frac{1}{2}(m - 3) = \frac{1}{4}(m - 7)\)

Exercise \(\PageIndex{16}\)

Solve: \(\frac{5x - 3}{4} = \frac{x}{2}\)

Exercise \(\PageIndex{17}\)

Solve: \(\frac{4y - 7}{3} = \frac{y}{6}\)

Exercise \(\PageIndex{18}\)

Solve: \(\frac{-2z - 5}{4} = \frac{z}{8}\)

Exercise \(\PageIndex{19}\)

Solve: \(\frac{a}{6} + 2 = \frac{a}{4} + 3\)

Exercise \(\PageIndex{20}\)

Solve: \(\frac{b}{10} + 2 = \frac{b}{4} + 5\)

\(b = -20\)

Exercise \(\PageIndex{21}\)

Solve: \(\frac{c}{6} + 3 = \frac{c}{3} + 4\)

Exercise \(\PageIndex{22}\)

Solve: \(\frac{4q + 3}{2}+ 6 = \frac{3q + 5}{4}\)

Exercise \(\PageIndex{23}\)

Solve: \(\frac{3r + 5}{6}+ 1 = \frac{4r + 3}{3}\)

Exercise \(\PageIndex{24}\)

Solve: \(\frac{2s + 3}{2}+ 1 = \frac{3s + 2}{4}\)

Solve Equations with Decimal Coefficients

Some equations have decimals in them. This kind of equation will occur when we solve problems dealing with money or percentages. But decimals can also be expressed as fractions. For example, \(0.3 = \frac{3}{10}\) and \(0.17 = \frac{17}{100}\). So, with an equation with decimals, we can use the same method we used to clear fractions—multiply both sides of the equation by the least common denominator.

Exercise \(\PageIndex{25}\)

Solve: \(0.06x + 0.02 = 0.25x - 1.5\)

Look at the decimals and think of the equivalent fractions.

\(0.06 = \frac { 6 } { 100 } \quad 0.02 = \frac { 2 } { 100 } \quad 0.25 = \frac { 25 } { 100 } \quad 1.5 = 1 \frac { 5 } { 10 }\)

Notice, the LCD is 100.

By multiplying by the LCD, we will clear the decimals from the equation.

Exercise \(\PageIndex{26}\)

Solve: \(0.14h + 0.12 = 0.35h - 2.4\)

Exercise \(\PageIndex{27}\)

Solve: \(0.65k - 0.1 = 0.4k - 0.35\)

The next example uses an equation that is typical of the money applications in the next chapter. Notice that we distribute the decimal before we clear all the decimals.

Exercise \(\PageIndex{28}\)

Solve: \(0.25x + 0.05(x + 3) = 2.85\)

Exercise \(\PageIndex{29}\)

Solve: \(0.25n + 0.05(n + 5) = 2.95\)

Exercise \(\PageIndex{30}\)

Solve: \(0.10d + 0.05(d -5) = 2.15\)

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