Which is the best reason why step 1 is a good first step in the solution shown? Math item stem image
CLEAR CHECK

Multiplying by 3 isolates x on one side of the equation.

3 is a factor of 12.

The same thing was done to both sides of the equation.

Multiplying by 3 is easier than subtracting 12.

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Step-by-step explanation:

Let's think about how we can solve for ttt in the following equation:

\qquad 6t = 546t=546, t, equals, 54

We want to get ttt by itself on the left hand side of the equation. So, what can we do to undo multiplying by 6?

We should divide by 6 because the inverse operation of multiplication is division!

Here's how dividing by 6 on each side looks:

\begin{aligned} 6t &= 54 \dfrac{6t}{\blueD{6}} &= \dfrac{54}{\blueD{ 6}}\small\gray{\text{Divide each side by six.}} t &= \greenD{9}\small\gray{\text{Simplify.}} \end{aligned}6t66tt=54=654          Divide each side by six.=9          Simplify.

Let's check our work.

It's always a good idea to check our solution in the original equation to make sure we didn't make any mistakes:

\qquad \begin{aligned} 6t &= 54 \\ 6 \cdot \greenD9 &\stackrel{\large?}{=} 54\\ 54 &= 54 \end{aligned}6t6⋅954=54=?54=54

Yes, t = \greenD{9}t=9t, equals, start color #1fab54, 9, end color #1fab54 is a solution!

Solving a division equation using inverse operations

Now, let's try to solve a slightly different type of equation:

\qquad \dfrac x5 = 75x=7start fraction, x, divided by, 5, end fraction, equals, 7

We want to get xxx by itself on the left hand side of the equation. So, what can we do to cancel out dividing by 5?

We can multiply by 5 because the inverse operation of division is multiplication!

Here's how multiplying by 5 on each side looks:

\begin{aligned} \dfrac x5 &= 7 \dfrac x5 \cdot \blueD{5} &= 7 \cdot \blueD{5}\small\gray{\text{Multiply each side by five.}} x &= \greenD{35}\small\gray{\text{Simplify.}} \end{aligned}5x5x⋅5x=7=7⋅5          Multiply each side by five.=35          Simplify.

Let's check our work.

\qquad \begin{aligned} \dfrac x5 &= 7 \dfrac{\greenD{35}}{5} &\stackrel{\large?}{=} 7 7 &= 7 \end{aligned}5x5357

ok then what happened.