Solving Equations by Adding or Subtracting the Same Number from Both Sides

๐Ÿ†Practice solution of an equation by adding/subtracting two sides

This method allows us to add or subtract the same element from both sides of the equation without changing the final result, that is, the outcome of the equation will not be affected by the fact that we have added or subtracted the same element from both sides.

Let's see what the logic of this method is:

Josรฉ and Isabel, for example, are twin siblings who receive their weekly allowance for the first time.

Josรฉ and Isabel receive 10 10 euros each, so at this moment they have exactly 10 10 euros per person.

After a month, each has received another 2 2 euros, so now each has 12 12 euros.

We see that adding 2 2 euros to the amount each of them had has not affected the equivalence between them: both still have the same amount of money.

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Test yourself on solution of an equation by adding/subtracting two sides!

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Solve for X:

\( 1.5-x=\text{2}.8 \)

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

Example 1

X+5+2=3 X+5+2=3

If we are asked what the value of the expression X+5 X+5 is, we can leave it on the left side of the equation if we subtract the number 2 2 from both sides.

X+5+2=3 X+5+2=3 ย  / โˆ’2 -2

X+5=1 X+5=1

Here we see that the expression X+5 X+5 is equivalent to 1 1 .


Example 2

X+7โˆ’4=10 X+7-4=10

If we are asked what the value of the expression X+7 X+7 is, we can leave it on the left side of the equation if we add the number 4 4 to both sides of the equation.

X+7โˆ’4=10 X+7-4=10 / +4 +4

X+7=14 X+7=14

Here we see that the expression X+7 X+7 is equivalent to 14 14 .



Examples and exercises with solutions for solving equations by adding or subtracting the same number from both sides

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Solve the equation and find Y:

20ร—y+8ร—2โˆ’7=14 20\times y+8\times2-7=14

examples.explanation_title

First, we will put parentheses around the two multiplication exercises:

(20ร—y)+(8ร—2)โˆ’7=14 (20\times y)+(8\times2)-7=14

We solve the exercises within the parentheses:

20y+16โˆ’7=14 20y+16-7=14

We simplify:

20y+9=14 20y+9=14

We move the sections:

20y=14โˆ’9 20y=14-9

20y=5 20y=5

We divide by 20:

y=520 y=\frac{5}{20}

y=55ร—4 y=\frac{5}{5\times4}

We simplify:

y=14 y=\frac{1}{4}

examples.solution_title

4

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92โˆ’xร—2โˆ’24โ€‰โฃ:4=64 92-x\times2-24\colon4=64 Calculate X.

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First, we solve the multiplication and division exercises, we will put them in parentheses to avoid confusion:

92โˆ’(xร—2)โˆ’(24โ€‰โฃ:4)=64 92-(x\times2)-(24\colon4)=64

92โˆ’2xโˆ’6=64 92-2x-6=64

Reduce:

86โˆ’2x=64 86-2x=64

Move the sides:

โˆ’2x=64โˆ’86 -2x=64-86

โˆ’2x=โˆ’22 -2x=-22

Divide by negative 2:

x=โˆ’22โˆ’2 x=\frac{-22}{-2}

x=11 x=11

examples.solution_title

11

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Solve for X:

5xโˆ’8=10x+22 5x-8=10x+22

examples.explanation_title

First, we arrange the two sections so that the right side contains the values with the coefficient x and the left side the numbers without the x

Let's remember to maintain the plus and minus signs accordingly when we move terms between the sections.

First, we move a5x 5x to the right section and then the 22 to the left side. We obtain the following equation:

โˆ’8โˆ’22=10xโˆ’5x -8-22=10x-5x

We subtract both sides accordingly and obtain the following equation:

โˆ’30=5x -30=5x

We divide both sections by 5 and obtain:

โˆ’6=x -6=x

examples.solution_title

โˆ’6 -6

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What is the missing number?

23โˆ’12ร—(โˆ’6)+?โ€‰โฃ:7=102 23-12\times(-6)+?\colon7=102

examples.explanation_title

First, we solve the multiplication exercise:

12ร—(โˆ’6)=โˆ’72 12\times(-6)=-72

Now we get:

23โˆ’(โˆ’72)+xโ€‰โฃ:7=102 23-(-72)+x\colon7=102

Let's pay attention to the minus signs, remember that a negative times a negative equals a positive.

We multiply them one by one to be able to open the parentheses:

23+72+xโ€‰โฃ:7=102 23+72+x\colon7=102

We reduce:

95+x:7=102 95+x:7=102

We move the sections:

x:7=102โˆ’95 x:7=102-95

x:7=7 x:7=7

x7=7 \frac{x}{7}=7

Multiply by 7:

x=7ร—7=49 x=7\times7=49

examples.solution_title

49

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Solve for X:

18xโˆ’23x+15x=1 \frac{1}{8}x-\frac{2}{3}x+\frac{1}{5}x=1

examples.explanation_title

The common denominator of 8, 3, and 5 is 120.

Now we multiply each numerator by the corresponding number to reach 120 and thus cancel the fractions and obtain the following equation:

(1ร—xร—15)โˆ’(2ร—xร—40)+(1ร—xร—24)=1ร—120 (1\times x\times15)-(2\times x\times40)+(1\times x\times24)=1\times120

We multiply the exercises in parentheses accordingly:

15xโˆ’80x+24x=120 15x-80x+24x=120

We will solve the left side (from left to right) and will obtain:

(15xโˆ’80x)+24x=120 (15x-80x)+24x=120

โˆ’65x+24x=120 -65x+24x=120

โˆ’41x=120 -41x=120

We reduce both sides by โˆ’41 -41

โˆ’41xโˆ’41=120โˆ’41 \frac{-41x}{-41}=\frac{120}{-41}

We find that x is equalx=โˆ’12041 x=-\frac{120}{41}

examples.solution_title

โˆ’12041 -\frac{120}{41}

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