Question

solve the system by graphing.
7
\

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

a.
(3, -1)
b.
(1, 3)
c.
(-1, 3)
d.
(3, 1)
use a matrix to solve the system.
8.
\

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

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

Explanation:

Problem 7: Solve the system by graphing

Step1: Rewrite equations in slope - intercept form

For the first equation \(-3x - 2y=-7\), solve for \(y\):
\(-2y = 3x - 7\)
\(y=-\frac{3}{2}x+\frac{7}{2}\) (slope \(m =-\frac{3}{2}\), y - intercept \(b=\frac{7}{2}=3.5\))

For the second equation \(3x - y = 10\), solve for \(y\):
\(y = 3x-10\) (slope \(m = 3\), y - intercept \(b=- 10\))

Step2: Analyze the graphs

We can also test the intersection points given in the options.

Test option b: \((3,-1)\)
For \(y =-\frac{3}{2}x+\frac{7}{2}\), when \(x = 3\), \(y=-\frac{9}{2}+\frac{7}{2}=\frac{-9 + 7}{2}=-1\)
For \(y = 3x-10\), when \(x = 3\), \(y=9 - 10=-1\)
So \((3,-1)\) satisfies both equations.

Now check the graphs: The graph in option a (the first graph) has the intersection point \((3,-1)\) (from the coordinates and the equations we derived, the slopes and intercepts match the lines in graph a as well, but the key is the intersection point \((3,-1)\) which satisfies both equations).

Step1: Write the system in matrix form \(AX = B\)

The coefficient matrix \(A=

$$\begin{bmatrix}2&6\\5&-1\end{bmatrix}$$

\), the variable matrix \(X=

$$\begin{bmatrix}x\\y\end{bmatrix}$$

\), and the constant matrix \(B=

$$\begin{bmatrix}38\\15\end{bmatrix}$$

\)

First, find the inverse of \(A\). The determinant of \(A\) is \(\det(A)=(2\times(-1))-(6\times5)=-2 - 30=-32\)

The inverse of \(A\) is \(A^{-1}=\frac{1}{\det(A)}

$$\begin{bmatrix}-1&-6\\-5&2\end{bmatrix}$$

=\frac{1}{-32}

$$\begin{bmatrix}-1&-6\\-5&2\end{bmatrix}$$

=

$$\begin{bmatrix}\frac{1}{32}&\frac{6}{32}\\\frac{5}{32}&-\frac{2}{32}\end{bmatrix}$$

=

$$\begin{bmatrix}\frac{1}{32}&\frac{3}{16}\\\frac{5}{32}&-\frac{1}{16}\end{bmatrix}$$

\)

Step2: Multiply \(A^{-1}\) with \(B\)

\(X = A^{-1}B=

$$\begin{bmatrix}\frac{1}{32}&\frac{3}{16}\\\frac{5}{32}&-\frac{1}{16}\end{bmatrix}$$

$$\begin{bmatrix}38\\15\end{bmatrix}$$

\)

First element (x - coordinate):
\(\frac{1}{32}\times38+\frac{3}{16}\times15=\frac{38}{32}+\frac{45}{16}=\frac{38 + 90}{32}=\frac{128}{32}=4\)

Second element (y - coordinate):
\(\frac{5}{32}\times38-\frac{1}{16}\times15=\frac{190}{32}-\frac{30}{32}=\frac{190 - 30}{32}=\frac{160}{32}=5\)

We can also test the options:

Test option a: \((4,5)\)
For \(2x + 6y\): \(2\times4+6\times5 = 8 + 30=38\)
For \(5x-y\): \(5\times4-5=20 - 5 = 15\)
So \((4,5)\) satisfies both equations.

Answer:

a. The graph with intersection point \((3, - 1)\) (the first graph)

Problem 8: Use a matrix to solve the system \(

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

\)