Question

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name: period: solve the systems of equations by graphing.1. $y=\frac{2}{3}x - 2$ m= b= $y=2x + 2$ m= b= solution: (__, )2. $y=-\frac{3}{2}x + 5$ m= b= $y=-\frac{2}{3}x + 5$ m= b= solution: (, )3. $y=-\frac{1}{2}x + 2$ m= b= $y=-3x - 3$ m= b= solution: (, )4. $y=-3$ m= b= $y=x - 3$ m= b= solution: (, __)

Explanation:

Step1: Solve system 1 algebraically

Set $\frac{2}{3}x - 2 = 2x + 2$
$\frac{2}{3}x - 2x = 2 + 2$
$\frac{2}{3}x - \frac{6}{3}x = 4$
$-\frac{4}{3}x = 4$
$x = 4 \times (-\frac{3}{4}) = -3$
Substitute $x=-3$ into $y=2x+2$:
$y=2(-3)+2=-6+2=-4$

Step2: Analyze system 2

$y=-\frac{3}{2}x+5$ and $y=-\frac{2}{3}x+5$ are distinct parallel lines? No, slopes are $-\frac{3}{2}$ and $-\frac{2}{3}$ (not equal). Set equal:
$-\frac{3}{2}x+5 = -\frac{2}{3}x+5$
$-\frac{3}{2}x + \frac{2}{3}x = 0$
$-\frac{9}{6}x + \frac{4}{6}x = 0$
$-\frac{5}{6}x=0 \implies x=0$
Substitute $x=0$: $y=5$

Step3: Solve system 3 algebraically

Set $-\frac{1}{2}x + 2 = -3x - 3$
$-\frac{1}{2}x + 3x = -3 - 2$
$\frac{5}{2}x = -5$
$x = -5 \times \frac{2}{5} = -2$
Substitute $x=-2$ into $y=-\frac{1}{2}x+2$:
$y=-\frac{1}{2}(-2)+2=1+2=3$

Step4: Solve system 4 algebraically

Set $-3 = x - 3$
$x = -3 + 3 = 0$
$y=-3$

Answer:

1. Solution: $(-3, -4)$
2. Solution: $(0, 5)$
3. Solution: $(-2, 3)$
4. Solution: $(0, -3)$