Solving Equations By Multiplication And Division

Last Updated on February 14, 2024.

Questions about Solving equations by multiplication & division are included in the GED Math test.

Make sure you’ll understand the steps that will lead to the correct answers.

This lesson is provided by Onsego GED Prep.

Video Transcription

We have stated that equations that produce identical solutions are equivalent. Furthermore, we’ve seen that adding identical numbers to both sides of the equations produces equivalent equations.

Similarly, if we subtract the same numbers from both sides of the equations, we also come up with equivalent equations.

We also may make similar statements when looking at multiplication & division.

So let’s look at: multiplying both equation sides by an identical quantity.

Well, let’s see of multiplying both equation sides by the same quantity will indeed not change the solution set.

So let’s see if when \(a = b\), if multiplying both the equation sides by \(c\) will produce an equivalent equation:
\(a \cdot c = b \cdot c\) provided \( c\)  isn’t equal to \(0\).

We can make a similar statement about division.

Dividing both equation sides by identical quantities will not change the solution set.

So let’s see if when \(a = b\), if dividing both of the equation sides by \(c\) will produce an equivalent equation:
\(a \cdot c = b \cdot c\) provided \( c\)  isn’t equal to \(0\).

Multiplication & Division as Inverse Operations:

Well, the inverse of multiplication is division. When we start with a number \(“x”\), and then multiply it by a number \(“a”\), then if we divide the result by our number \(“a”\) will bring us back to our original number \(“x”\).

In symbols: \(\frac{ax}{a=x}\)

The inverse of division is multiplication. When we begin with the number \(“x”\) and divide it by a number \(“a”\), multiplying the result by our number \(“a”\) will bring us back to our original number \(“x”\).

In symbols: \(a\,(\frac{x}{a})=x\)

Check out this example:

Solve this equation for \(x \div 3x = 24\).

The solution: To undo the multiplying by \(3\) effects, we divide both equation sides by \(3\):

\(3x = 24\) is our original equation.

\(3x \div 3 = 24 \div 3\) \(x = 8\)

So we’ve divided both sides of our equation by \(3\). On the left side, dividing by \(3\) “undoes” the multiplying by \(3-effect\) and returns us back to \(x\).

On the right side, \(\frac{2}{3}=8\).

The solution: To check this, substitute the solution \(8\) into our original equation:
\(3x = 24\) is the original equation.
\(3(8) = 24\) (we substituted \(8\) for \(x\)).
\(24 = 24\) (we’ve just simplified both sides).

The fact that the last line of this check is a true statement that guarantees that, in fact, \(8\) (eight) is a solution of the original \(3x = 24\).

One more example:

Solve the following equation for \(x:\)  \(x\div 7 = 12\).

The solution: to undo the dividing by \(7\) effects, we need to multiply both equation sides by \(7\).

The original equation: \(\frac{x}{7} = 12\)

We multiply both equation sides by \(7\):

\(7 \cdot \frac{x}{7} = 7 \cdot 12\)

On the left side, multiplying by \(7\) “undoes” the dividing by \(7\) effects and brings us back to \(x\).

On the right side,\(\,7 \cdot 12 = 84\).

The solution.

To check all, just substitute our solution 84 into the initial equation. So substitute \(84\) for \(x\).

\(\frac{84}{7} = 12\) \(12 = 12\)

The fact that this last line of the check is a true statement is guaranteeing that \(84\) is a solution of \(\frac{x}{7} = 12\).