Solving Equations by Simplifying Like Terms

Simplify the like terms in an equation involves combining the elements that belong to the same group. In other words: in all first-degree equations with one unknown, there are elements that belong to the group of unknowns (variables) and elements that belong to the group of numbers. The goal is to unite all the elements of each of the mentioned groups into respective sides to thus arrive at the result of the equation.

For example

A1 - Solving Equations by Simplifying Like Terms

$X+2X=5+1$

In this equation, we can clearly see that the elements $X$ and $2X$ belong to the group of unknowns, and therefore, we can combine them.

Conversely, the elements $5$ and $1$ belong to the group of numbers and thus can also be combined.

$3X=6$

$X=2$

The result of the equation is $2$ .

Detailed explanation

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Test yourself on solution of an equation using addition of like terms!

$$ 7x+4x+5x=0 $$

$x=\text{?}$

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Below, we provide you with some examples where we apply this method.

Example 1

$6X-1=5X+5$

In this equation, we can clearly see that the elements $6X$ and $5X$ belong to the group of variables, and therefore, we can combine them.

Conversely, the elements $(-1)$ and $5$ belong to the group of numbers, and thus they can also be combined.

$6X-5X=5+1$

$X=6$

The result of the equation is $6$ .

Exercises on Equations by Simplifying Like Terms

Exercise 2

Assignment

$7a+8b+4a+9b=\text{?}$

Solution

We arrange the corresponding elements

$7a+4a+8b+9b=\text{?}$

We add accordingly

$11a+8b+9b=\text{?}$

$11a+17b$

Answer

$11a+17b$

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Test your knowledge

Question 1

$$ 7m+3m-40m=0 $$

$m=\text{?}$

Question 2

$$ 2a+3a+45a=0 $$

$a=\text{?}$

Question 3

$$ 8-b=6 $$

Exercise 3

Assignment:

$3x+4x+7+2=\text{?}$

Solution

We add the corresponding elements

$7x+7+2=\text{?}$

$\text{7x+9}$

Answer

$\text{7x+9}$

Exercise 4

Assignment

$18x-7+4x-9-8x=\text{?}$

Solution

We input the corresponding elements

$18x+4x-8x-7-9=\text{?}$

We solve accordingly

$22x-8x-7-9=\text{?}$

$14x-7-9=\text{?}$

Answer

$14x-16$

Do you know what the answer is?

Question 1

$$ -16+a=-17 $$

Question 2

$$ 2+4y-2y=4 $$

Question 3

$$ x+x=8 $$

Exercise 5

Assignment

$7.3\cdot4a+2.3+8a=\text{?}$

Solution

First, we solve the multiplication exercise

$29.2a+2.3+8a=$

We arrange the terms accordingly

$29.2a+8a+2.3=$

We add the terms accordingly

$37.2a+2.3$

Answer

$37.2a+2.3$

Exercise 6

Assignment

$\frac{3}{8}a+\frac{14}{9}b+1\frac{1}{9}b+\frac{6}{8}a=\text{?}$

Solution

We input the corresponding elements

$\frac{3}{8}a+\frac{6}{8}a+\frac{14}{9}b+1\frac{1}{9}b=$

We add accordingly

$\frac{3+6}{8}a+\frac{14}{9}b+1\frac{1}{9}b=$

We convert the mixed number into an improper fraction

$\frac{3+6}{8}a+\frac{14}{9}b+\frac{10}{9}b=$

We add accordingly

$\frac{9}{8}a+\frac{14+10}{9}b=$

$\frac{9}{8}a+\frac{24}{9}b=$

We convert the improper fractions into mixed numbers

$1\frac{1}{8}a+\frac{24}{9}b=$

$1\frac{1}{8}a+2\frac{6}{9}b$

Answer

$1\frac{1}{8}a+2\frac{6}{9}b$

Check your understanding

Question 1

Find the value of the parameter X

$$ x+3-8x=4+3-x $$

Question 2

Find the value of the parameter X

$$ -7+3x-8x=9+3-5x $$

Question 3

Solve for X:

$$ -5x+20-3x=40+2-6x $$

examples with solutions for solution of an equation using addition of like terms

Exercise #1

Solve for x:

$8(-2-x)=16$

Video Solution

Step-by-Step Solution

First, we divide both sections by 8:

$\frac{8(-2-x)}{8}=\frac{16}{8}$

Keep in mind that the 8 in the fraction of the left section is reduced, so the equation we get is:

$-2-x=2$

We move the minus 2 to the right section and maintain the plus and minus signs accordingly:

$-x=2+2$

$-x=4$

We divide both sides by minus 1 and maintain the plus and minus signs accordingly when we divide:

$\frac{-x}{-1}=\frac{4}{-1}$

$x=-4$

Answer

-4

Exercise #2

Solve for x:

$5+x=3$

Video Solution

Step-by-Step Solution

We will rearrange the equation so that x remains on the left side and we will move similar elements to the right side.

Remember that when we move a positive number, it will become a negative number, so we will get:

$x=3-5$

$x=-2$

Answer

-2

Exercise #3

Solve for x:

$-9-x=3+2x$

Video Solution

Step-by-Step Solution

To solve the equation, we will move similar elements to one side.

On the right side, we place the elements with X, while in the left side we place the elements without X.

Remember that when we move sides, the plus and minus signs change accordingly, so we get:

$-9-3=2x+x$

We calculate both sides: $-12=3x$

Finally, divide both sides by 3:

$-\frac{12}{3}=\frac{3x}{3}$

$-4=x$

Answer

-4

Exercise #4

Solve for x:

$-\frac{1}{2}+\frac{1}{3}x=\frac{1}{5}+x$

Video Solution

Step-by-Step Solution

We will move the elements with the X to the left side and the elements without the X to the right side, changing the plus and minus signs accordingly.

First, we move the minus X to the left section:

$-\frac{1}{2}+\frac{1}{3}x+x=\frac{1}{5}$

Now we move the minus 1/2 to the right section:

$\frac{1}{3}x+x=\frac{1}{5}+\frac{1}{2}$

We will find a common denominator for the fractions on the right side and reduce accordingly. Convert the mixed fraction on the left side into a simple fraction:

$1\frac{1}{3}x=\frac{2+5}{10}$

$\frac{4}{3}x=\frac{7}{10}$

Multiply by $\frac{3}{4}$ to reduce the left side:

$x=\frac{7}{10}\times\frac{3}{4}=\frac{7\times3}{10\times4}=\frac{21}{40}$

Answer

$\frac{21}{40}$

Exercise #5

Solve for x:

$-3(x+1)+5x-4=-3+5(x-1)$

Video Solution

Step-by-Step Solution

First, we will expand the parentheses on both sides:

$-3x-3+5x-4=-3+5x-5$

Enter the like terms in both sections. Let's start with the left section:

$-3x+5x=2x$

$-3-4=-7$

Calculate the like terms on the right side:

$-3-5=-8$

Now, we obtain the equation:

$2x-7=-8+5x$

To the right side we will move the members without the X, while to the left side we move those with the X, keeping the plus and minus signs as appropriate:

$2x-5x=-8+7$

$-3x=-1$

Finally, we divide both sides by -3:

$\frac{-1}{-3}=\frac{-3x}{-3}$

$\frac{1}{3}=x$

Answer

$\frac{1}{3}$

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