Solving Equations by Multiplying or Dividing Both Sides by the Same Number

πŸ†Practice solving an equation by multiplication/ division

Multiplying or Dividing Both Sides of the Equation

Sometimes when solving equations, we may encounter variables with coefficients, which we need to remove to isolate the variable and find its value.
Exactly for those cases, and many more, we have the ability to multiply or divide both sides of the equation by the same number to maintain balance and solve for the variable.

With this method, we can multiply or divide both sides of the equation by the same element without thereby altering the overall value of the equation. This means that the final result of the equation will not be affected because we have multiplied or divided both sides by the same element or number.Β 

In order to so we need to follow these two steps:
  1. Identify the Coefficient: Determine if multiplication or division is needed to isolate the variable.
  2. Apply Operation to Both Sides: Multiply or divide by the coefficient’s reciprocal.
Solving Equations by Multiplying or Dividing Both Sides by the Same Number

It's important to remember that when we multiply or divide both sides of an equation, the equation's balance should remain unchanged. This means we can always reverse the operation to return to the original equation. If reversing leads to a different result, it indicates that an error was made in the calculations.

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Test yourself on solving an equation by multiplication/ division!

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\( 4x:30=2 \)

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

Example 1

3X=24 3X=24

We solve the equation and find the numerical value of X X by dividing both sides of the equation by the number 3 3 .

In this way, we neutralize and isolate the X X on the left side of the equation, while on the right side we obtain the result of the equation.

3X=24 3X=24 / :3 :3

X=8 X=8

The result of the equation is 8 8 .


Example 2

X2=5 \frac{X}{2}=5

We solve the equation and find the numerical value of X by multiplying both sides of the equation by the number 2. This way, we neutralize and isolate X on the left side of the equation, while on the right side we obtain the result of the equation.

X2=5 \frac{X}{2}=5 Β / Γ—2 \times2

X=10 X=10

The result of the equation is 10 10 .


Examples and exercises with solutions for solving equations by multiplying or dividing both sides by the same number

Exercise #1

Solve the equation

20:4x=5 20:4x=5

Video Solution

Step-by-Step Solution

To solve the exercise, we first rewrite the entire division as a fraction:

204x=5 \frac{20}{4x}=5

Actually, we didn't have to do this step, but it's more convenient for the rest of the process.

To get rid of the fraction, we multiply both sides of the equation by the denominator, 4X.

20=5*4X

20=20X

Now we can reduce both sides of the equation by 20 and we will arrive at the result of:

X=1

Answer

x=1 x=1

Exercise #2

Solve for X:

3x=18 3x=18

Video Solution

Step-by-Step Solution

We use the formula:

aβ‹…x=b a\cdot x=b

x=ba x=\frac{b}{a}

Note that the coefficient of X is 3.

Therefore, we will divide both sides by 3:

3x3=183 \frac{3x}{3}=\frac{18}{3}

Then divide accordingly:

x=6 x=6

Answer

6 6

Exercise #3

Solve for X:

x4=3 \frac{x}{4}=3

Video Solution

Step-by-Step Solution

We use the formula:

aβ‹…x=b a\cdot x=b

x=ba x=\frac{b}{a}

We multiply the numerator by X and write the exercise as follows:

x4=3 \frac{x}{4}=3

We multiply by 4 to get rid of the fraction's denominator:

4Γ—x4=3Γ—4 4\times\frac{x}{4}=3\times4

Then, we remove the common factor from the left side and perform the multiplication on right side to obtain:

x=12 x=12

Answer

12 12

Exercise #4

Solve for X:

5x=38 5x=\frac{3}{8}

Video Solution

Step-by-Step Solution

ax=cb ax=\frac{c}{b}

x=cbβ‹…a x=\frac{c}{b\cdot a}

Answer

340 \frac{3}{40}

Exercise #5

βˆ’y5=βˆ’25 \frac{-y}{5}=-25

Video Solution

Step-by-Step Solution

We begin by multiplying the simple fraction by y:

βˆ’15Γ—y=βˆ’25 \frac{-1}{5}\times y=-25

We then reduce both terms by βˆ’15 -\frac{1}{5}

y=βˆ’25βˆ’15 y=\frac{-25}{-\frac{1}{5}}

Finally we multiply the fraction by negative 5:

y=βˆ’25Γ—(βˆ’5)=125 y=-25\times(-5)=125

Answer

y=125 y=125

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