Question

which of these strategies would eliminate a variable in the system of equations?\

$$\begin{cases}2x - 6y = 6\\\\6x - 4y = 2\\end{cases}$$

\choose 2 answers:\
a multiply the top equation by -3, then add the equations.\
b multiply the bottom equation by 3, then subtract the bottom equation from the top equation.\
c multiply the bottom equation by $-\dfrac{3}{2}$, then add the

Explanation:

Step1: Test Strategy A

Multiply top eq by $-3$:
$-3(2x - 6y) = -3(6)$
$\implies -6x + 18y = -18$
Add to bottom eq:
$(-6x + 18y) + (6x - 4y) = -18 + 2$
$\implies 14y = -16$ (x eliminated)

Step2: Test Strategy B

Multiply bottom eq by 3:
$3(6x - 4y) = 3(2)$
$\implies 18x - 12y = 6$
Subtract from top eq:
$(2x - 6y) - (18x - 12y) = 6 - 6$
$\implies -16x + 6y = 0$ (no variable eliminated)

Step3: Test Strategy C

Multiply bottom eq by $-\frac{3}{2}$:
$-\frac{3}{2}(6x - 4y) = -\frac{3}{2}(2)$
$\implies -9x + 6y = -3$
Add to top eq:
$(2x - 6y) + (-9x + 6y) = 6 + (-3)$
$\implies -7x = 3$ (y eliminated)

Answer:

A. Multiply the top equation by $-3$, then add the equations.
C. Multiply the bottom equation by $-\frac{3}{2}$, then add the