Question

graph each equation. determine t
x + 3y = 12
5x + 3y = 0

use the graphing tool to graph the sy
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graph

Explanation:

Step1: Rewrite first equation to slope-intercept form

Solve $x + 3y = 12$ for $y$:

$$\begin{align*} 3y &= -x + 12\\ y &= -\frac{1}{3}x + 4 \end{align*}$$

Step2: Find intercepts for first line

• x-intercept (set $y=0$): $x = 12$, so point $(12, 0)$
• y-intercept (set $x=0$): $y = 4$, so point $(0, 4)$

Step3: Rewrite second equation to slope-intercept form

Solve $5x + 3y = 0$ for $y$:

$$\begin{align*} 3y &= -5x\\ y &= -\frac{5}{3}x \end{align*}$$

Step4: Find points for second line

• x-intercept (set $y=0$): $x = 0$, so point $(0, 0)$
• Test $x=3$: $y = -\frac{5}{3}(3) = -5$, so point $(3, -5)$

Step5: Find intersection of the two lines

Subtract the first original equation from the second:

$$\begin{align*} (5x + 3y) - (x + 3y) &= 0 - 12\\ 4x &= -12\\ x &= -3 \end{align*}$$

Substitute $x=-3$ into $x + 3y = 12$:

$$\begin{align*} -3 + 3y &= 12\\ 3y &= 15\\ y &= 5 \end{align*}$$

Answer:

1. For $x + 3y = 12$: Graph the line through points $(12, 0)$ and $(0, 4)$ (slope $-\frac{1}{3}$).
2. For $5x + 3y = 0$: Graph the line through points $(0, 0)$ and $(3, -5)$ (slope $-\frac{5}{3}$).
3. The intersection point of the two lines is $(-3, 5)$.