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