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Algebraic expressions Combining like terms Expanding expressions PolynomialsHere you will learn about solving equations, including linear and quadratic algebraic equations, and how to solve them.

Students will first learn about solving equations in grade 8 as a part of expressions and equations, and again in high school as a part of reasoning with equations and inequalities.

Every week, we teach lessons on solving equations to students in schools and districts across the US as part of our online one-on-one math tutoring programs. On this page we’ve broken down everything we’ve learnt about teaching this topic effectively.

Solving equations is a step-by-step process to find the value of the variable. A variable is the unknown part of an equation, either on the left or right side of the equals sign. Sometimes, you need to solve multi-step equations which contain algebraic expressions.

To do this, you must use the order of operations, which is a systematic approach to equation solving. When you use the order of operations, you first solve any part of an equation located within parentheses. An equation is a mathematical expression that contains an equals sign.

For example,

\begin{aligned}y+6&=11\\\\ 3(x-3)&=12\\\\ \cfrac{2x+2}{4}&=\cfrac{x-3}{3}\\\\ 2x^{2}+3&x-2=0\end{aligned}

There are two sides to an equation, with the left side being equal to the right side. Equations will often involve algebra and contain unknowns, or variables, which you often represent with letters such as x or y.

You can solve simple equations and more complicated equations to work out the value of these unknowns. They could involve fractions, decimals or integers.

How does this relate to 8 th grade and high school math?

**Grade 8 – Expressions and Equations (8.EE.C.7)**Solve linear equations in one variable.

**High school – Reasoning with Equations and Inequalities (HSA.REI.B.3)**

Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.

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

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

DOWNLOAD FREEIn order to solve equations, you need to work out the value of the unknown variable by adding, subtracting, multiplying or dividing both sides of the equation by the same value.

**Combine like terms**.**Simplify the equation by using the opposite operation to both sides.****Isolate the variable on one side of the equation.**

Solve for x.

5q-4q=9**Combine like terms.**

Combine the q terms on the left side of the equation. To do this, subtract 4q from both sides.

(5 q-4 q)=9-4 qThe goal is to simplify the equation by combining like terms. Subtracting 4q from both sides helps achieve this.

After you combine like terms, you are left with q=9-4q.

2**Simplify the equation by using the opposite operation on both sides.**

Add 4q to both sides to isolate q to one side of the equation.

q+4q=9-4q+4qThe objective is to have all the q terms on one side. Adding 4q to both sides accomplishes this.

After you move the variable to one side of the equation, you are left with 5q=9.

3**Isolate the variable on one side of the equation.**

Divide both sides of the equation by 5 to solve for q.

5q \div 5 = 9 \div 5Dividing by 5 allows you to isolate q to one side of the equation in order to find the solution. After dividing both sides of the equation by 5, you are left with q=1 \cfrac{4}{5} \, .

Solve for x.

7v=8–8v**Combine like terms.**

Combine the v terms on the same side of the equation. To do this, add 8v to both sides.

7v+8v=8-8v+8v

After combining like terms, you are left with the equation 15v=8.

**Simplify the equation by using the opposite operation on both sides and isolate the variable to one side.**

Divide both sides of the equation by 15 to solve for v. This step will isolate v to one side of the equation and allow you to solve.

15v \div 15=8 \div 15

The final solution to the equation 7v=8-8v is \cfrac{8}{15} \, .

Solve for x.

3(c-5)-4=2**Combine like terms by using the distributive property.**

The 3 outside the parentheses needs to be multiplied by both terms inside the parentheses. This is called the distributive property.

\begin{aligned}& 3 \times c=3 c \\\\ & 3 \times(-5)=-15 \\\\ &3 c-15-4=2\end{aligned}

Once the 3 is distributed on the left side, rewrite the equation and combine like terms. In this case, the like terms are the constants on the left, –15 and –4. Subtract –4 from –15 to get –19.

3c-19=2

**Simplify the equation by using the opposite operation on both sides.**

The goal is to isolate the variable, c, on one side of the equation. By adding 19 to both sides, you move the constant term to the other side.

\begin{aligned}& 3 c-19+19=2+19 \\\\
& 3 c=21\end{aligned}

**Isolate the variable to one side of the equation.**

To solve for c, you want to get c by itself.

Dividing both sides by 3 accomplishes this.

On the left side, \cfrac{3c}{3} simplifies to c, and on the right, \cfrac{21}{3} simplifies to 7.

The final solution is c=7.

As an additional step, you can plug 7 back into the original equation to check your work.

Solve for x.

2x+5=15**Combine like terms by simplifying.**

Using steps to solve, you know that the goal is to isolate x to one side of the equation. In order to do this, you must begin by subtracting from both sides of the equation.

\begin{aligned} & 2x+5=15 \\\\ & 2x+5-5=15-5 \\\\ & 2x=10 \end{aligned}

**Continue to simplify the equation by using the opposite operation on both sides.**

Continuing with steps to solve, you must divide both sides of the equation by 2 to isolate x to one side.

\begin{aligned} & 2x \div 2=10 \div 2 \\\\
& x= 5 \end{aligned}

**Isolate the variable to one side of the equation and check your work.**

Plugging in 5 for x in the original equation and making sure both sides are equal is an easy way to check your work. If the equation is not equal, you must check your steps.

\begin{aligned}& 2(5)+5=15 \\\\
& 10+5=15 \\\\
& 15=15\end{aligned}

Solve the following equation by factoring.

2x^2+3x-20=0**Combine like terms by factoring the equation by grouping.**

Multiply the coefficient of the quadratic term by the constant term.

2 x (-20) = -40

Look for two numbers that multiply to give you –40 and add up to the coefficient of 3. In this case, the numbers are 8 and –5 because 8 x -5=–40, and 8+–5=3.

Split the middle term using those two numbers, 8 and –5. Rewrite the middle term using the numbers 8 and –5.

2x^2+8x-5x-20=0

Group the terms in pairs and factor out the common factors.

2x^2+8x-5x-20=2x(x + 4)-5(x+4)=0

Now, you’ve factored the equation and are left with the following simpler equations 2x-5 and x+4.

**Simplify the equation by using the opposite operation on both sides.**

This step relies on understanding the zero product property, which states that if two numbers multiply to give zero, then at least one of those numbers must equal zero.

Let’s relate this back to the factored equation (2x-5)(x+4)=0

Because of this property, either (2x-5)=0 or (x+4)=0

**Isolate the variable for each equation and solve.**

When solving these simpler equations, remember that you must apply each step to both sides of the equation to maintain balance.

\begin{aligned}& 2 x-5=0 \\\\ & 2 x-5+5=0+5 \\\\ & 2 x=5 \\\\ & 2 x \div 2=5 \div 2 \\\\ & x=\cfrac{5}{2}
\end{aligned}

\begin{aligned}& x+4=0 \\\\ & x+4-4=0-4 \\\\ & x=-4\end{aligned}

The solution to this equation is x=\cfrac{5}{2} and x=-4.

Solve the following quadratic equation.

x^{2}+2x-5=0**Combine like terms by factoring the quadratic equation when terms are isolated to one side.**

To factorize a quadratic expression like this, you need to find two numbers that multiply to give -5 (the constant term) and add to give +2 (the coefficient of the x term).

The two numbers that satisfy this are -1 and +5.

So you can split the middle term 2x into -1x+5x: x^2-1x+5x-5-1x+5x

Now you can take out common factors x(x-1)+5(x-1).

And since you have a common factor of (x-1), you can simplify to (x+5)(x-1).

The numbers -1 and 5 allow you to split the middle term into two terms that give you common factors, allowing you to simplify into the form (x+5)(x-1).

**Simplify the equation by using the opposite operation on both sides.**

This step relies on understanding the zero product property, which states that if two numbers multiply to give zero, then at least one of those numbers must equal zero.

Let’s relate this back to the factored equation (x+5)(x-1)=0.

Because of this property, either (x+5)=0 or (x-1)=0.

**Isolate the variable for each equation and solve.**

Now, you can solve the simple equations resulting from the zero product property.

\begin{aligned}& x+5=0 \\\\ & x+5-5=0-5 \\\\ & x=-5 \\\\\\ & x-1=0 \\\\ & x-1+1=0+1 \\\\ & x=1\end{aligned}

The solutions to this quadratic equation are x=1 and x=-5.

- Use physical manipulatives like balance scales as a visual aid. Show how you need to keep both sides of the equation balanced, like a scale. Add or subtract the same thing from both sides to keep it balanced when solving. Use this method to practice various types of equations.

- Emphasize the importance of undoing steps to isolate the variable. If you are solving for x and 3 is added to x, subtracting 3 undoes that step and isolates the variable x.

- Relate equations to real-world, relevant examples for students. For example, word problems about tickets for sports games, cell phone plans, pizza parties, etc. can make the concepts click better.

- Allow time for peer teaching and collaborative problem solving. Having students explain concepts to each other, work through examples on whiteboards, etc. reinforces the process and allows peers to ask clarifying questions. This type of scaffolding would be beneficial for all students, especially English-Language Learners. Provide supervision and feedback during the peer interactions.

**Forgetting to distribute or combine like terms**One common mistake is neglecting to distribute a number across parentheses or combine like terms before isolating the variable. This error can lead to an incorrect simplified form of the equation.

**Misapplying the distributive property**

Incorrectly distributing a number across terms inside parentheses can result in errors. Students may forget to multiply each term within the parentheses by the distributing number, leading to an inaccurate equation.

**Failing to perform the same operation on both sides**

It’s crucial to perform the same operation on both sides of the equation to maintain balance. Forgetting this can result in an imbalanced equation and incorrect solutions.

**Making calculation errors**

Simple arithmetic mistakes, such as addition, subtraction, multiplication, or division errors, can occur during the solution process. Checking calculations is essential to avoid errors that may propagate through the steps.

**Ignoring fractions or misapplying operations**

When fractions are involved, students may forget to multiply or divide by the common denominator to eliminate them. Misapplying operations on fractions can lead to incorrect solutions or complications in the final answer.

- Math equations
- Rearranging equations
- How to find the equation of a line
- Solve equations with fractions
- Linear equations
- Writing linear equations
- Substitution
- Identity math
- One step equation

1. Solve 4x-2=14.

x=3

x=4

x=14

x=3.5

4x-2=14

Add 2 to both sides.

4x=16

Divide both sides by 4.

x=4

2. Solve 3x-8=x+6.

x=7

x=4

x=8

x=6

3x-8=x+6

Add 8 to both sides.

3x=x+14

Subtract x from both sides.

2x=14

Divide both sides by 2.

x=7

3. Solve 3(x+3)=2(x-2).

x=2

x=–3

x=–13

x=13

3(x+3)=2(x-2)

Expanding the parentheses.

3x+9=2x-4

Subtract 9 from both sides.

3x=2x-13

Subtract 2x from both sides.

x=-13

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

x=2

x=–3

x=–13

x=13

\cfrac{2 x+2}{3}=\cfrac{x-3}{2}

Multiply by 6 (the lowest common denominator) and simplify.

2(2x+2)=3(x-3)

Expand the parentheses.

4x+4=3x-9

Subtract 4 from both sides.

4x=3x-13

Subtract 3x from both sides.

x=-13

5. Solve \cfrac{3 x^{2}}{2}=24.

x=±4

x=±2

x=±3

x=±6

\cfrac{3 x^{2}}{2}=24

Multiply both sides by 2.

3 x^{2}=48

Divide both sides by 3.

x^{2}=16

Square root both sides.

x=\pm 4

6. Solve by factoring:

x^{2}-13 x+30=0.

x=-3, \; x=10

x=-3, \; x=-10

x=3, \; x=10

x=3, \; x=-10

x^{2}-13 x+30=0

Use factoring to find simpler equations.

(x-3)(x-10)=0

Set each set of parentheses equal to zero and solve.

x=3 or x=10

The first step in solving a simple linear equation is to simplify both sides by combining like terms. This involves adding or subtracting terms to isolate the variable on one side of the equation.

Performing the same operation on both sides of the equation maintains the equality. This ensures that any change made to one side is also made to the other, keeping the equation balanced and preserving the solutions.

To handle variables on both sides of the equation, start by combining like terms on each side. Then, move all terms involving the variable to one side by adding or subtracting, and simplify to isolate the variable. Finally, perform any necessary operations to solve for the variable.

To deal with fractions in an equation, aim to eliminate them by multiplying both sides of the equation by the least common denominator. This helps simplify the equation and make it easier to isolate the variable. Afterward, proceed with the regular steps of solving the equation.

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