Question

solve each system by graphing.
1)
$5x + 3y = 12$
$-x + 3y = -6$
2)
$2x + y = 2$
$-x + y = -4$
3)
$3x + y = -3$
$y = 3$
4)
$y = \frac{1}{3}x + 2$
$y = -\frac{1}{2}x + 2$
5)
$y = -x + 4$
$y = 2$
6)
$y = -2x + 2$
$y = \frac{1}{3}x - 5$
7)
$y = \frac{3}{2}x + 3$
$y = -3$
8)
$-6x + 4y = -12$
$y = -3$

Explanation:

Problem 1: Solve \(

$$\begin{cases} 5x + 3y = 12 \\ -x + 3y = -6 \end{cases}$$

\) by Graphing

Step 1: Rewrite equations in slope - intercept form (\(y = mx + b\))

For \(5x+3y = 12\):
Subtract \(5x\) from both sides: \(3y=-5x + 12\)
Divide by 3: \(y=-\frac{5}{3}x + 4\)
For \(-x + 3y=-6\):
Add \(x\) to both sides: \(3y=x - 6\)
Divide by 3: \(y=\frac{1}{3}x-2\)

Step 2: Find intersection by graphing (or solving)

Subtract the second equation from the first equation:
\((5x + 3y)-(-x + 3y)=12-(-6)\)
\(5x+3y + x-3y=18\)
\(6x = 18\)
\(x = 3\)
Substitute \(x = 3\) into \(y=\frac{1}{3}x-2\): \(y=\frac{1}{3}(3)-2=1 - 2=-1\)

Step 1: Rewrite in slope - intercept form

For \(2x + y = 2\): \(y=-2x + 2\)
For \(-x + y=-4\): \(y=x - 4\)

Step 2: Find intersection

Subtract the second equation from the first:
\((2x + y)-(-x + y)=2-(-4)\)
\(2x + y+x - y=6\)
\(3x = 6\)
\(x = 2\)
Substitute \(x = 2\) into \(y=x - 4\): \(y=2-4=-2\)

Step 1: Substitute \(y = 3\) into the first equation

\(3x+3=-3\)

Step 2: Solve for \(x\)

\(3x=-6\)
\(x=-2\)

Answer:

\((3,-1)\)

Problem 2: Solve \(

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

\) by Graphing