the following system $$ \begin{array}{l}8 x+3 y=2 \ -2 x+7 y=-16 \ x=\square \ y=\square\end{array} $$

Question

UpStudy AI · Accepted Answer

To solve the system of equations:

$$
\begin{cases}
8x + 3y = 2 \quad \text{(1)} \
-2x + 7y = -16 \quad \text{(2)}
\end{cases}
$$

**Step 1: Eliminate $$ x $$**

Multiply equation (2) by 4 to make the coefficient of $$ x $$ equal in magnitude to that in equation (1):

$$
-2x \times 4 + 7y \times 4 = -16 \times 4 \
-8x + 28y = -64 \quad \text{(3)}
$$

Now, add equation (1) and equation (3):

$$
(8x + 3y) + (-8x + 28y) = 2 + (-64) \
31y = -62 \
y = \frac{-62}{31} \
y = -2
$$

**Step 2: Solve for $$ x $$**

Substitute $$ y = -2 $$ back into equation (1):

$$
8x + 3(-2) = 2 \
8x - 6 = 2 \
8x = 8 \
x = 1
$$

**Solution:**

$$
\begin{array}{l}
8x + 3y = 2 \
-2x + 7y = -16 \
x = 1 \
y = -2
\end{array}
$$
$$ x = 1 $$ and $$ y = -2 $$.

the following system \( \begin{array}{l}8 x+3 y=2 \\ -2 x+7 y=-16 \\ x=\square \\ y=\square\end{array} \)