Question

3b. solve the system of equations.\

$$\begin{cases}-x - 3y = 10\\\\8x + 5y = -4\\end{cases}$$

\bigcirc no solution (parallel lines)\bigcirc (2, -4)\bigcirc (-4, 2)\bigcirc (-2, -4)\bigcirc infinitely many solutions (same line)\bigcirc (4, -2)

Explanation:

Step1: Isolate $x$ from first equation

From $-x - 3y = 10$, rearrange to get:
$x = -3y - 10$

Step2: Substitute $x$ into second equation

Substitute $x = -3y - 10$ into $8x + 5y = -4$:
$8(-3y - 10) + 5y = -4$
Expand and simplify:
$-24y - 80 + 5y = -4$
$-19y - 80 = -4$

Step3: Solve for $y$

Isolate $y$:
$-19y = -4 + 80$
$-19y = 76$
$y = \frac{76}{-19} = -4$

Step4: Solve for $x$

Substitute $y=-4$ into $x = -3y -10$:
$x = -3(-4) -10 = 12 -10 = 2$

Answer:

(2, -4)