how to solve algebraic equations with a fraction

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Inverse operations

Solve equations with fractions

Here you will learn about how to solve equations with fractions, including solving equations with one or more operations. You will also learn about solving equations with fractions where the unknown is the denominator of a fraction.

Students will first learn how to solve equations with fractions in 7th grade as part of their work with expressions and equations and expand that knowledge in 8th grade.

What are equations with fractions?

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

The numerator (top number) in a fraction is divided by the denominator (bottom number).

To solve equations with fractions, you will use the “balancing method” to apply the inverse operation to both sides of the equation in order to work out the value of the unknown variable.

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,

\begin{aligned} \cfrac{2x+3}{5} \, &= 7\\ \colorbox{#cec8ef}{$\times \, 5$} \; & \;\; \colorbox{#cec8ef}{$\times \, 5$} \\\\ 2x+3&=35 \\ \colorbox{#cec8ef}{$-\,3$} \; & \;\; \colorbox{#cec8ef}{$- \, 3$} \\\\ 2x & = 32 \\ \colorbox{#cec8ef}{$\div \, 2$} & \; \; \; \colorbox{#cec8ef}{$\div \, 2$}\\\\ x & = 16 \end{aligned}

Common Core State Standards

How does this relate to 7th grade and 8th grade math?

• Grade 7: Expressions and Equations (7.EE.A.1) Apply properties of operations as strategies to add, subtract, factor, and expand linear expressions with rational coefficients.
• Grade 8: Expressions and Equations (8.EE.C.7) Solve linear equations in one variable.
• Grade 8: Expressions and Equations (8.EE.C.7b) Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms.

How to solve equations with fractions

In order to solve equations with fractions:

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.

[FREE] Solve Equations with Fractions Worksheet (Grade 6 to 8)

Use this worksheet to check your grade 6 to 8 students’ understanding of solving equations with fractions. 15 questions with answers to identify areas of strength and support!

Solve equations with fractions examples

Example 1: equations with one operation.

Solve for x \text{: } \cfrac{x}{5}=4 .

The unknown is x.

Looking at the left hand side of the equation, the x is divided by 5.

\cfrac{x}{5}

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 ”.

You will multiply both sides of the equation by 5.

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

The final answer is x=20.

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

\cfrac{20}{5}=20\div5=4

Example 2: equations with one operation

Solve for x \text{: } \cfrac{x}{3}=8 .

Looking at the left hand side of the equation, the x is divided by 3.

\cfrac{x}{3}

The inverse of “dividing by 3 ” is “multiplying by 3 ”.

You will multiply both sides of the equation by 3.

The final answer is x=24.

\cfrac{24}{3}=24\div3=8

Example 3: equations with two operations

Solve for x \text{: } \cfrac{x \, + \, 1}{2}=7 .

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

\cfrac{x \, + \, 1}{2}

First, clear the fraction by multiplying both sides of the equation by 2.

Then, subtract 1 from both sides.

The final answer is x=13.

\cfrac{13 \, +1 \, }{2}=\cfrac{14}{2}=14\div2=7

Example 4: equations with two operations

Solve for x \text{: } \cfrac{x}{4}-2=3 .

Looking at the left hand side of the equation, x is divided by 4 and then 2 is subtracted.

\cfrac{x}{4}-2

First, add 2 to both sides of the equation.

Then, multiply both sides of the equation by 4.

\cfrac{20}{4}-2=20\div4-2=5-2=3

Example 5: equations with three operations

Solve for x \text{: } \cfrac{3x}{5}+1=7 .

Looking at the left hand side of the equation, x is multiplied by 3, then divided by 5 , and then 1 is added.

\cfrac{3x}{5}+1

First, subtract 1 from both sides of the equation.

Then, multiply both sides of the equation by 5.

Finally, divide both sides by 3.

The final answer is x=10.

\cfrac{3 \, \times \, 10}{5}+1=\cfrac{30}{5}+1=6+1=7

Example 6: equations with three operations

Solve for x \text{: } \cfrac{2x-1}{7}=3 .

Looking at the left hand side of the equation, x is multiplied by 2, then 1 is subtracted, and the last operation is divided by 7 (the denominator).

\cfrac{2x-1}{7}

First, multiply both sides of the equation by 7.

Next, add 1 to both sides.

The final answer is x=11.

\cfrac{2 \, \times \, 11-1}{7}=\cfrac{22-1}{7}=\cfrac{21}{7}=3

Example 7: equations with the unknown as the denominator

Solve for x \text{: } \cfrac{24}{x}=6 .

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

\cfrac{24}{x}

You need to multiply both sides of the equation by x.

Then, you can divide both sides by 6.

The final answer is x=4.

\cfrac{24}{4}=24\div4=6

Example 8: equations with the unknown as the denominator

Solve for x \text{: } \cfrac{18}{x}-6=3 .

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

\cfrac{18}{x}-6

First, add 6 to both sides of the equation.

Then, multiply both sides of the equation by x.

Finally, divide both sides by 9.

The final answer is x=2.

\cfrac{18}{2}-6=9-6=3

Teaching tips for solving equations with fractions

• When students first start working through practice problems and word problems, provide step-by-step instructions to assist them with solving linear equations.
• Introduce solving equations with fractions with one-step problems, then two-step problems, before introducing multi-step problems.
• Students will need lots of practice with solving linear equations. These standards provide the foundation for work with future linear equations in Algebra I and II.
• Provide opportunities for students to explain their thinking through writing. Ensure that they are using key vocabulary, such as, absolute value, coefficient, equation, common factors, inequalities, simplify, etc.

Easy mistakes to make

• The solution to an equation can be any type of number The unknowns do not have to be integers (whole numbers and their negative opposites). The solutions can be fractions or decimals. They can also be positive or negative numbers.
• The unknown of an equation can be on either side of the equation The unknown, 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.

• Lowest common denominator (LCD) It is common to get confused between solving equations involving fractions and adding and subtracting fractions. When adding and subtracting, you need to work out the lowest/least common denominator (sometimes called the least common multiple or LCM). When you solve equations involving fractions, multiply both sides of the equation by the denominator of the fraction.

Related math equations lessons

• Math equations
• Rearranging equations
• How to find the equation of a line
• Substitution
• Linear equations
• Writing linear equations
• Solving equations
• Identity math
• One step equations

Practice solve equations with fractions questions

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

You will multiply both sides of the equation by 6, because the inverse of “dividing by 6 ” is “multiplying by 6 ”.

The final answer is x = 18.

\cfrac{18}{6}=18 \div 6=3

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

Then subtract 4 from both sides.

The final answer is x = 10.

\cfrac{10 \, + \, 4}{2}=\cfrac{14}{2}=14 \div 2=7

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

First, add 5 to both sides of the equation.

Then multiply both sides of the equation by 8.

The final answer is x = 48.

\cfrac{48}{8}-5=48 \div 8-5=1

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

First, multiply both sides of the equation by 4.

Next, subtract 2 from both sides.

The final answer is x = 2.

\cfrac{3 \, \times \, 2+2}{4}=\cfrac{6 \, + \, 2}{4}=\cfrac{8}{4}=8 \div 4=2

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

Then multiply both sides of the equation by 7.

Finally, divide both sides by 4.

The final answer is x = 14.

\cfrac{4 \, \times \, 14}{7}-2=\cfrac{56}{7}-2=56 \div 7-2=6

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

Then you divide both sides by 7.

The final answer is x = 6.

\cfrac{42}{6}=42 \div 6=7

Solve equations with fractions FAQs

Yes, you still follow the order of operations when solving equations with fractions. You will start with any operations in the numerator and follow PEMDAS (parenthesis, exponents, multiply/divide, add/subtract), followed by any operations in the denominator. Then you will solve the rest of the equation as usual.

The next lessons are

• Inequalities
• Types of graphs
• Math formulas
• Coordinate plane
• Number patterns
• Algebraic expressions

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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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1.26: Solving Fractional Equations

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A fractional equation is an equation involving fractions which has the unknown in the denominator of one or more of its terms.

Example 24.1

The following are examples of fractional equations:

a) \(\frac{3}{x}=\frac{9}{20}\)

b) \(\frac{x-2}{x+2}=\frac{3}{5}\)

c) \(\frac{3}{x-3}=\frac{4}{x-5}\)

d) \(\frac{3}{4}-\frac{1}{8 x}=0\)

e) \(\frac{x}{6}-\frac{2}{3 x}=\frac{2}{3}\)

The Cross-Product property can be used to solve fractional equations.

Cross-Product Property

If \(\frac{A}{B}=\frac{C}{D}\) then \(A \cdot D=B \cdot C\).

Using this property we can transform fractional equations into non-fractional ones. We must take care when applying this property and use it only when there is a single fraction on each side of the equation. So, fractional equations can be divided into two categories.

I. Single Fractions on Each Side of the Equation

Equations a), b) and c) in Example 24.1 fall into this category. We solve these equations here.

a) Solve \(\frac{3}{x}=\frac{9}{20}\)

\[\begin{array}{ll} \text{Cross-Product} & 3 \cdot 20=9 \cdot x \\ \text{Linear Equation} & 60=9 x \\ \text{Divide by 9 both sides} & \frac{60}{9}=x \end{array}\nonumber\]

The solution is \(x=\frac{60}{9}=\frac{20}{3}\).

\[\begin{array}{ll} \text{Cross-Product} & 5 \cdot(x-2)=3 \cdot(x+2) \\ \text{Remove parentheses} & 5 x-10=3 x+6 \\ \text{Linear Equation: isolate the variable} & 5 x-3 x=10+6 \\ & 2 x=16 \\ \text{Divide by 2 both sides} & \frac{2 x}{2}=\frac{16}{2}\end{array}\nonumber\]

the solution is \(x=8\).

\[\begin{array}{ll} \text{Cross-Product} & 3 \cdot(x-5)=4 \cdot(x-3) \\ \text{Remove parentheses} & 3 x-15=4 x-12 \\ \text{Linear Equation: isolate the variable} & 3 x-4 x=15-12 \\ & -x=3 \\ \text{Divide by 2 both sides} & \frac{-x}{-1}=\frac{3}{-1}\end{array}\nonumber\]

The solution is \(x=-3\)

Note: If you have a fractional equation and one of the terms is not a fraction, you can always account for that by putting 1 in the denominator. For example:

\[\frac{3}{x}=15\nonumber\]

We re-write the equation so that all terms are fractions.

\[\frac{3}{x}=\frac{15}{1}\nonumber\]

\[\begin{array}{ll} \text{Cross-Product} & 3 \cdot 1=15 \cdot x \\ \text{Linear Equation: isolate the variable} & 3=15 x \\ \text{Divide by 15 both sides} & \frac{3}{15}=\frac{15 x}{15} \end{array}\nonumber\]

The solution is \(x=\frac{3}{15}=\frac{3 \cdot 1}{3 \cdot 5}=\frac{1}{5}\).

II. Multiple Fractions on Either Side of the Equation

Equations d) and e) in Example 24.1 fall into this category. We solve these equations here.

We use the technique for combining rational expressions we learned in Chapter 23 to reduce our problem to a problem with a single fraction on each side of the equation.

d) Solve \(\frac{3}{4}-\frac{1}{8 x}=0\)

First we realize that there are two fractions on the LHS of the equation and thus we cannot use the Cross-Product property immediately. To combine the LHS into a single fraction we do the following:

\[\begin{array}{ll} \text{Find the LCM of the denominators} & 8 x \\ \text{Rewrite each fraction using the LCM} & \frac{3 \cdot 2 x}{8 x}-\frac{1}{8 x}=0 \\ \text{Combine into one fraction} & \frac{6 x-1}{8 x}=0 \\ \text{Re-write the equation so that all terms are fractions} & \frac{6 x-1}{8 x}=\frac{0}{1} \\ \text{Cross-Product} & (6 x-1) \cdot 1=8 x \cdot 0 \\ \text{Remove parentheses} & 6 x-1=0 \\\text{Linear Equation: isolate the variable} & 6 x=1 \\ \text{Divide by 6 both sides} & \frac{6 x}{6}=\frac{1}{6} \end{array}\nonumber\]

The solution is \(x=\frac{1}{6}\).

e) Solve \(\frac{x}{6}+\frac{2}{3 x}=\frac{2}{3}\)

\[\begin{array}{ll} \text{Find the LCM of the denominators of LHS} & 6x \\ \text{Rewrite each fraction on LHS using their LCM} & \frac{x \cdot x}{6 x}+\frac{2 \cdot 2}{6 x}=\frac{2}{3} \\ \frac{x^{2}+4}{6 x}=\frac{2}{3} \text{Combine into one fraction} & \left(x^{2}+4\right) \cdot 3=6 x \cdot 2 \\ \text{Cross-Product} & 3 x^{2}+12=12 x \\ \text{Remove parentheses} & 3 x^{2}-12 x+12=0 \\ \text{Quadratic Equation: Standard form} & 3 x^{2}-12 x+12=0 \\\text{Quadratic Equation: Factor} & 3 \cdot x^{2}-3 \cdot 4 x+3 \cdot 4=0 \\ & 3\left(x^{2}-4 x+4\right)=0 \\ & 3(x-2)(x-2)=0 \\ \text{Divide by 3 both sides} & \frac{3(x-2)(x-2)}{3}=\frac{0}{3} \\ & (x-2)(x-2)=0 \\ \text{Quadratic Equation: Zero-Product Property} & (x-2)=0 \text { or }(x-2)=0 \end{array}\nonumber\]

Since both factors are the same, then \(x-2=0\) gives \(x=2\). The solution is \(x=2\)

Note: There is another method to solve equations that have multiple fractions on either side. It uses the LCM of all denominators in the equation. We demonstrate it here to solve the following equation: \(\frac{3}{2}-\frac{9}{2 x}=\frac{3}{5}\)

\[\begin{array} \text{Find the LCM of all denominators in the equation} & 10x \\ \text{Multiply every fraction (both LHS and RHS) by the LCM} & 10 x \cdot \frac{3}{2}-10 x \cdot \frac{9}{2 x}=10 x \cdot \frac{3}{5} \\ & \frac{10 x \cdot 3}{2}-\frac{10 x \cdot 9}{2 x}=\frac{10 x \cdot 3}{5} \\ \text{Simplify every fraction} & \frac{5 x \cdot 3}{1}-\frac{5 \cdot 9}{1}=\frac{2 x \cdot 3}{1} \\ \text{See how all denominatiors are now 1, thus can be disregarded} & 5 x \cdot 3-5 \cdot 9=2 x \cdot 3 \\ \text{Solve like you would any other equation} & 15 x-45=6 x \\ \text{Linear equation: islolate the variable} & 15 x-6 x=45 \\ & 9 x=45 \\ & x=\frac{45}{9} \\ & x=5 \end{array} \nonumber\]

The solution is \(x=5\)

Exit Problem

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

Solving Equations

What is an equation.

An equation says that two things are equal. It will have an equals sign "=" like this:

That equations says:

what is on the left (x − 2)  equals  what is on the right (4)

So an equation is like a statement " this equals that "

What is a Solution?

A Solution is a value we can put in place of a variable (such as x ) that makes the equation true .

Example: x − 2 = 4

When we put 6 in place of x we get:

which is true

So x = 6 is a solution.

How about other values for x ?

• For x=5 we get "5−2=4" which is not true , so x=5 is not a solution .
• For x=9 we get "9−2=4" which is not true , so x=9 is not a solution .

In this case x = 6 is the only solution.

You might like to practice solving some animated equations .

More Than One Solution

There can be more than one solution.

Example: (x−3)(x−2) = 0

When x is 3 we get:

(3−3)(3−2) = 0 × 1 = 0

And when x is 2 we get:

(2−3)(2−2) = (−1) × 0 = 0

which is also true

So the solutions are:

x = 3 , or x = 2

When we gather all solutions together it is called a Solution Set

The above solution set is: {2, 3}

Solutions Everywhere!

Some equations are true for all allowed values and are then called Identities

Example: sin(−θ) = −sin(θ) is one of the Trigonometric Identities

Let's try θ = 30°:

sin(−30°) = −0.5 and

−sin(30°) = −0.5

So it is true for θ = 30°

Let's try θ = 90°:

sin(−90°) = −1 and

−sin(90°) = −1

So it is also true for θ = 90°

Is it true for all values of θ ? Try some values for yourself!

How to Solve an Equation

There is no "one perfect way" to solve all equations.

A Useful Goal

But we often get success when our goal is to end up with:

x = something

In other words, we want to move everything except "x" (or whatever name the variable has) over to the right hand side.

Example: Solve 3x−6 = 9

Now we have x = something ,

and a short calculation reveals that x = 5

Like a Puzzle

In fact, solving an equation is just like solving a puzzle. And like puzzles, there are things we can (and cannot) do.

Here are some things we can do:

• Add or Subtract the same value from both sides
• Clear out any fractions by Multiplying every term by the bottom parts
• Divide every term by the same nonzero value
• Combine Like Terms
• Expanding (the opposite of factoring) may also help
• Recognizing a pattern, such as the difference of squares
• Sometimes we can apply a function to both sides (e.g. square both sides)

Example: Solve √(x/2) = 3

And the more "tricks" and techniques you learn the better you will get.

Special Equations

There are special ways of solving some types of equations. Learn how to ...

• solve Quadratic Equations
• solve Radical Equations
• solve Equations with Sine, Cosine and Tangent

Check Your Solutions

You should always check that your "solution" really is a solution.

How To Check

Take the solution(s) and put them in the original equation to see if they really work.

Example: solve for x:

2x x − 3 + 3 = 6 x − 3     (x≠3)

We have said x≠3 to avoid a division by zero.

Let's multiply through by (x − 3) :

2x + 3(x−3) = 6

Bring the 6 to the left:

2x + 3(x−3) − 6 = 0

Expand and solve:

2x + 3x − 9 − 6 = 0

5x − 15 = 0

5(x − 3) = 0

Which can be solved by having x=3

Let us check x=3 using the original question:

2 × 3 3 − 3 + 3  =   6 3 − 3

Hang On: 3 − 3 = 0 That means dividing by Zero!

And anyway, we said at the top that x≠3 , so ...

x = 3 does not actually work, and so:

There is No Solution!

That was interesting ... we thought we had found a solution, but when we looked back at the question we found it wasn't allowed!

This gives us a moral lesson:

"Solving" only gives us possible solutions, they need to be checked!

• Note down where an expression is not defined (due to a division by zero, the square root of a negative number, or some other reason)
• Show all the steps , so it can be checked later (by you or someone else)

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How to Solve an Algebraic Expression

Last Updated: April 6, 2024 Fact Checked

This article was co-authored by David Jia . David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. There are 10 references cited in this article, which can be found at the bottom of the page. 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 491,500 times.

An algebraic expression is a mathematical phrase that contains numbers and/or variables. Though it cannot be solved because it does not contain an equals sign (=), it can be simplified. You can, however, solve algebraic equations , which contain algebraic expressions separated by an equals sign. If you want to know how to master this mathematical concept, then see Step 1 to get started.

Understanding the Basics

• Algebraic expression : 4x + 2
• Algebraic equation : 4x + 2 = 100

• 3x 2 + 5 + 4x 3 - x 2 + 2x 3 + 9 =
• 3x 2 - x 2 + 4x 3 + 2x 3 + 5 + 9 =
• 2x 2 + 6x 3 + 14

• You can see that each coefficient can be divisible by 3. Just "factor out" the number 3 by dividing each term by 3 to get your simplified equation.
• 3x/3 + 15/3 = 9x/3 + 30/3 =
• x + 5 = 3x + 10

• (3 + 5) 2 x 10 + 4
• First, follow P, the operation in the parentheses:
• = (8) 2 x 10 + 4
• Then, follow E, the operation of the exponent:
• = 64 x 10 + 4
• Next, do multiplication:
• And last, do addition:

• 5x + 15 = 65 =
• 5x/5 + 15/5 = 65/5 =
• x + 3 = 13 =

Joseph Meyer

To solve an equation for a variable like "x," you need to manipulate the equation to isolate x. Use techniques like the distributive property, combining like terms, factoring, adding or subtracting the same number, and multiplying or dividing by the same non-zero number to isolate "x" and find the answer.

Solve an Algebraic Equation

• 4x + 16 = 25 -3x =
• 4x = 25 -16 - 3x
• 4x + 3x = 25 -16 =
• 7x/7 = 9/7 =

• First, subtract 12 from both sides.
• 2x 2 + 12 -12 = 44 -12 =
• Next, divide both sides by 2.
• 2x 2 /2 = 32/2 =
• Solve by taking the square root of both sides, since that will turn x 2 into x.
• √x 2 = √16 =
• State both answers:x = 4, -4

• First, cross multiply to get rid of the fraction. You have to multiply the numerator of one fraction by the denominator of the other.
• (x + 3) x 3 = 2 x 6 =
• Now, combine like terms. Combine the constant terms, 9 and 12, by subtracting 9 from both sides.
• 3x + 9 - 9 = 12 - 9 =
• Isolate the variable, x, by dividing both sides by 3 and you've got your answer.
• 3x/3 = 3/3 =

• First, move everything that isn't under the radical sign to the other side of the equation:
• √(2x+9) = 5
• Then, square both sides to remove the radical:
• (√(2x+9)) 2 = 5 2 =
• Now, solve the equation as you normally would by combining the constants and isolating the variable:
• 2x = 25 - 9 =

• |4x +2| - 6 = 8 =
• |4x +2| = 8 + 6 =
• |4x +2| = 14 =
• 4x + 2 = 14 =
• Now, solve again by flipping the sign of the term on the other side of the equation after you've isolated the absolute value:
• 4x + 2 = -14
• 4x = -14 -2
• 4x/4 = -16/4 =
• Now, just state both answers: x = -4, 3

Community Q&A

• The degree of a polynomial is the highest power within the terms. Thanks Helpful 9 Not Helpful 1
• Once you're done, replace the variable with the answer, and solve the sum to see if it makes sense. If it does, then, congratulations! You just solved an algebraic equation! Thanks Helpful 7 Not Helpful 3
• To cross-check your answer, visit wolfram-alpha.com. They give the answer and often the two steps. Thanks Helpful 8 Not Helpful 5

You Might Also Like

• ↑ https://www.math4texas.org/Page/527
• ↑ https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-expressions-and-variables/cc-6th-combining-like-terms/v/combining-like-terms-2
• ↑ https://www.mathsisfun.com/algebra/factoring.html
• ↑ https://www.mathsisfun.com/operation-order-pemdas.html
• ↑ https://sciencing.com/tips-for-solving-algebraic-equations-13712207.html
• ↑ https://www.mathsisfun.com/algebra/equations-solving.html
• ↑ https://tutorial.math.lamar.edu/Classes/Alg/SolveExpEqns.aspx
• ↑ https://www.mathsisfun.com/algebra/fractions-algebra.html
• ↑ https://math.libretexts.org/Courses/Coastline_College/Math_C045%3A_Beginning_and_Intermediate_Algebra_(Chau_Duc_Tran)/10%3A_Roots_and_Radicals/10.07%3A_Solve_Radical_Equations
• ↑ https://www.mathplanet.com/education/algebra-1/linear-inequalitites/solving-absolute-value-equations-and-inequalities

About This Article

If you want to solve an algebraic expression, first understand that expressions, unlike equations, are mathematical phrase that can contain numbers and/or variables but cannot be solved. For example, 4x + 2 is an expression. To reduce the expression, combine like terms, for example everything with the same variable. After you've done that, factor numbers by finding the lowest common denominator. Then, use the order of operations, which is known by the acronym PEMDAS, to reduce or solve the problem. To learn how to solve algebraic equations, keep scrolling! Did this summary help you? Yes No

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