To solve the given system of equations by eliminating one variable through addition, we need to manipulate the equations such that the coefficients of either [tex]\( x \)[/tex] or [tex]\( y \)[/tex] will cancel out when the equations are added together.
Consider the given system of equations:
[tex]\[
\begin{array}{l}
4x - 2y = 7 \quad \text{(Equation 1)} \\
3x - 3y = 15 \quad \text{(Equation 2)}
\end{array}
\][/tex]
We want to manipulate these equations such that when we add them, either the [tex]\( x \)[/tex] terms or the [tex]\( y \)[/tex] terms cancel out.
Let's analyze option C: multiply the top equation by [tex]\(-3\)[/tex] and the bottom equation by [tex]\(2\)[/tex].
1. Multiply Equation 1 by [tex]\(-3\)[/tex]:
[tex]\[
-3 \times (4x - 2y) = -3 \times 7
\][/tex]
This gives us:
[tex]\[
-12x + 6y = -21 \quad \text{(Equation 3)}
\][/tex]
2. Multiply Equation 2 by [tex]\(2\)[/tex]:
[tex]\[
2 \times (3x - 3y) = 2 \times 15
\][/tex]
This gives us:
[tex]\[
6x - 6y = 30 \quad \text{(Equation 4)}
\][/tex]
Now, we add Equation 3 and Equation 4:
[tex]\[
(-12x + 6y) + (6x - 6y) = -21 + 30
\][/tex]
Simplifying the left-hand side:
[tex]\[
-12x + 6x + 6y - 6y = -21 + 30
\][/tex]
[tex]\[
-6x = 9
\][/tex]
As we can see, multiplying the top equation by [tex]\(-3\)[/tex] and the bottom equation by [tex]\(2\)[/tex] successfully eliminates the [tex]\( y \)[/tex] variable when the equations are added.
Therefore, the correct answer is:
[tex]\[
\boxed{\text{C. Multiply the top equation by -3 and the bottom equation by 2.}}
\][/tex]
