Solution of an Equation using Addition of Like Terms - Examples, Exercises and Solutions

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 X+2X=5+1

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

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

3X=6 3X=6

X=2 X=2

The result of the equation is 2 2 .


Suggested Topics to Practice in Advance

  1. Solving Equations by Adding or Subtracting the Same Number from Both Sides
  2. Solving Equations by Multiplying or Dividing Both Sides by the Same Number

Practice Solution of an Equation using Addition of Like Terms

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

Exercise #1

Solve for x:

8(2x)=16 8(-2-x)=16

Video Solution

Step-by-Step Solution

First, we divide both sections by 8:

8(2x)8=168 \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:

2x=2 -2-x=2

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

x=2+2 -x=2+2

x=4 -x=4

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

x1=41 \frac{-x}{-1}=\frac{4}{-1}

x=4 x=-4

Answer

-4

Exercise #2

Solve for x:

5+x=3 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=35 x=3-5

x=2 x=-2

Answer

-2

Exercise #3

Solve for x:

9x=3+2x -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:

93=2x+x -9-3=2x+x

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

Finally, divide both sides by 3:

123=3x3 -\frac{12}{3}=\frac{3x}{3}

4=x -4=x

Answer

-4

Exercise #4

Solve for x:

12+13x=15+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:

12+13x+x=15 -\frac{1}{2}+\frac{1}{3}x+x=\frac{1}{5}

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

13x+x=15+12 \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:

113x=2+510 1\frac{1}{3}x=\frac{2+5}{10}

43x=710 \frac{4}{3}x=\frac{7}{10}

Multiply by34 \frac{3}{4} to reduce the left side:

x=710×34=7×310×4=2140 x=\frac{7}{10}\times\frac{3}{4}=\frac{7\times3}{10\times4}=\frac{21}{40}

Answer

2140 \frac{21}{40}

Exercise #5

Solve for x:

3(x+1)+5x4=3+5(x1) -3(x+1)+5x-4=-3+5(x-1)

Video Solution

Step-by-Step Solution

First, we will expand the parentheses on both sides:

3x3+5x4=3+5x5 -3x-3+5x-4=-3+5x-5

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

3x+5x=2x -3x+5x=2x

34=7 -3-4=-7

Calculate the like terms on the right side:

35=8 -3-5=-8

Now, we obtain the equation:

2x7=8+5x 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:

2x5x=8+7 2x-5x=-8+7

3x=1 -3x=-1

Finally, we divide both sides by -3:

13=3x3 \frac{-1}{-3}=\frac{-3x}{-3}

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

Answer

13 \frac{1}{3}

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

Exercise #1

Solve for x:

8+x3(x2)=5(2+x)4+3x -8+x-3(x-2)=5(2+x)-4+3x

Video Solution

Step-by-Step Solution

First, we will expand the parentheses on both sides by multiplying their contents by the number outside:

8+x3×x+(3)×(2)=5×2+5×x4+3x -8+x-3\times x+(-3)\times(-2)=5\times2+5\times x-4+3x

8+x3x+6=10+5x4+3x -8+x-3x+6=10+5x-4+3x

Now we collect like terms on both sides:

22x=6+8x -2-2x=6+8x

We move 8x to the left and -2 to the right side, remembering to leave the plus and minus signs unchanged accordingly:

2x8x=6+2 -2x-8x=6+2

We add the terms together:

10x=8 -10x=8

Finally, we divide both sides by negative 10:

10x10=810 \frac{-10x}{-10}=\frac{8}{-10}

x=810 x=-\frac{8}{10}

Answer

810 -\frac{8}{10}

Exercise #2

Solve for x:

15x+14x+120x15=31025+210x -\frac{1}{5}x+\frac{1}{4}x+\frac{1}{20}x-\frac{1}{5}=\frac{3}{10}-\frac{2}{5}+\frac{2}{10}x

Video Solution

Step-by-Step Solution

  • Move similar terms to one side.

  • Create common denominators using the least common multiple of the different fractions.

  • Reduction of fractions.

Answer

-1

Exercise #3

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

x=? x=\text{?}

Video Solution

Answer

0 0

Exercise #4

7m+3m40m=0 7m+3m-40m=0

m=? m=\text{?}

Video Solution

Answer

0

Exercise #5

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

a=? a=\text{?}

Video Solution

Answer

0 0

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

Exercise #1

8b=6 8-b=6

Video Solution

Answer

2 2

Exercise #2

16+a=17 -16+a=-17

Video Solution

Answer

1 -1

Exercise #3

2+4y2y=4 2+4y-2y=4

Video Solution

Answer

1 1

Exercise #4

x+x=8 x+x=8

Video Solution

Answer

4

Exercise #5

Find the value of the parameter X

x+38x=4+3x x+3-8x=4+3-x

Video Solution

Answer

23 -\frac{2}{3}

Topics learned in later sections

  1. Solving Equations Using the Distributive Property
  2. First-degree equations with one unknown
  3. Solution of an equation