Question

the tables of ordered pairs represent some points on the graphs of two different lines. write a system of equations that could represent the two lines.
5.
line q

x	-9	-3	2
y	0	18	33

line v

x	-4	0	10
y	10	0	3

equation 1:
equation 2:
6.

x	y₁	y₂
-4	-10	-1
-2	-4	-2
1	5	-3.5
4	14	-5

equation 1:
equation 2:

Explanation:

Problem 5:

Step 1: Find slope of Line \( q \)

Using points \((-9, 0)\) and \((-3, 18)\):
Slope \( m_q = \frac{18 - 0}{-3 - (-9)} = \frac{18}{6} = 3 \).
Using point-slope form with \((-9, 0)\):
\( y - 0 = 3(x + 9) \) → \( y = 3x + 27 \).

Step 2: Find slope of Line \( v \)

Using points \((-4, 10)\) and \((0, 0)\):
Slope \( m_v = \frac{0 - 10}{0 - (-4)} = \frac{-10}{4} = -\frac{5}{2} \).
Y-intercept is \( 0 \) (from \((0, 0)\)), so equation: \( y = -\frac{5}{2}x \).

Problem 6:

Step 1: Find slope of \( Y_1 \) (Line 1)

Using points \((-4, -10)\) and \((-2, -4)\):
Slope \( m_1 = \frac{-4 - (-10)}{-2 - (-4)} = \frac{6}{2} = 3 \).
Using point-slope with \((-4, -10)\):
\( y + 10 = 3(x + 4) \) → \( y = 3x + 2 \).

Step 2: Find slope of \( Y_2 \) (Line 2)

Using points \((-4, -1)\) and \((-2, -2)\):
Slope \( m_2 = \frac{-2 - (-1)}{-2 - (-4)} = \frac{-1}{2} = -\frac{1}{2} \).
Using point-slope with \((-4, -1)\):
\( y + 1 = -\frac{1}{2}(x + 4) \) → \( y = -\frac{1}{2}x - 3 \).

Answer:

s:

Problem 5:

Equation 1: \( \boldsymbol{y = 3x + 27} \)
Equation 2: \( \boldsymbol{y = -\frac{5}{2}x} \)

Problem 6:

Equation 1: \( \boldsymbol{y = 3x + 2} \)
Equation 2: \( \boldsymbol{y = -\frac{1}{2}x - 3} \)