Question

writing and solving systems using substitution: mastery test

table 1:
\

$$\begin{array}{|c|c|} \\hline x & y \\\\ \\hline 0 & 2 \\\\ \\hline -4 & 0 \\\\ \\hline 6 & 5 \\\\ \\hline -6 & -1 \\\\ \\hline \\end{array}$$

table 2:
\

$$\begin{array}{|c|c|} \\hline x & y \\\\ \\hline 0 & 1 \\\\ \\hline -2 & 0 \\\\ \\hline -4 & -1 \\\\ \\hline 2 & 2 \\\\ \\hline \\end{array}$$

options:

\

$$\begin{aligned} y &= -\\frac{1}{2}x - 2 \\\\ y &= -\\frac{1}{2}x + 1 \\end{aligned}$$

\

$$\begin{aligned} y &= -x - 2 \\\\ y &= -x + 1 \\end{aligned}$$

\

$$\begin{aligned} y &= \\frac{1}{2}x + 2 \\\\ y &= \\frac{1}{2}x + 1 \\end{aligned}$$

\

$$\begin{aligned} y &= x + 2 \\\\ y &= x + 1 \\end{aligned}$$

Explanation:

⚡ Using what you learned: introduction to systems of equations · 🆕 New: writing linear equations from tables
Finding equations from points in a table

Step 1: Analyze the first table

Find the \(y\)-intercept and slope for the first table:

• When \(x = 0\), \(y = 2\), so the \(y\)-intercept is \(b_1 = 2\).
• Using points \((0, 2)\) and \((-4, 0)\):

$$ m_1 = \frac{2 - 0}{0 - (-4)} = \frac{2}{4} = \frac{1}{2} $$

• Equation 1:

$$ y = \frac{1}{2}x + 2 $$

Step 2: Analyze the second table

Find the \(y\)-intercept and slope for the second table:

• When \(x = 0\), \(y = 1\), so the \(y\)-intercept is \(b_2 = 1\).
• Using points \((0, 1)\) and \((2, 2)\):

$$ m_2 = \frac{2 - 1}{2 - 0} = \frac{1}{2} $$

• Equation 2:

$$ y = \frac{1}{2}x + 1 $$

Answer:

The correct system of equations is:

$$
LATEXBLOCK0
$$

(This corresponds to the third option in the image).