Question

solve the system by graphing.

1. \

$$\begin{cases} -3x - 2y = -7 \\\\ 3x - y = 10 \\end{cases}$$

a.
graph a
(3, -1)
b.
graph b
(1, 3)
c.
graph c
(-1, 3)
d.
graph d
(3, 1)
use a matrix to solve the system.

1. \

$$\begin{cases} 2x + 6y = 38 \\\\ 5x - y = 15 \\end{cases}$$

a. (4, 5)
b. (-4, -5)
c. (5, 4)
d. (-4, 5)

Explanation:

Step1: Rewrite equations in slope - intercept form

For the first equation \(-3x - 2y=-7\), we solve for \(y\):
\(-2y = 3x - 7\)
\(y=-\frac{3}{2}x+\frac{7}{2}\)

For the second equation \(3x - y = 10\), we solve for \(y\):
\(-y=-3x + 10\)
\(y = 3x-10\)

Step2: Analyze the graphs

We can also test the intersection points given in the options. Let's test the point \((3, - 1)\) in both equations.

For the first equation \(-3x-2y=-7\):
Left - hand side (LHS)=\(-3(3)-2(-1)=-9 + 2=-7\), which is equal to the right - hand side (RHS) \(-7\).

For the second equation \(3x - y = 10\):
LHS \(=3(3)-(-1)=9 + 1 = 10\), which is equal to the RHS \(10\).

Now let's check the other options:

• For the point \((-1,3)\):

In the first equation, LHS \(-3(-1)-2(3)=3 - 6=-3
eq - 7\).

• For the point \((1,3)\):

In the first equation, LHS \(-3(1)-2(3)=-3-6=-9
eq - 7\).

• For the point \((3,1)\):

In the second equation, LHS \(3(3)-1 = 9 - 1=8
eq10\).

Answer:

a. \((3,-1)\)