Question

graph the line.

y + 3x = 9

Explanation:

Step1: Rewrite the equation in slope - intercept form

The slope - intercept form of a line is $y = mx + b$, where $m$ is the slope and $b$ is the y - intercept.
Starting with the equation $y+3x = 9$, we solve for $y$ by subtracting $3x$ from both sides:
$y=-3x + 9$

Step2: Identify the y - intercept and plot the point

The y - intercept $b = 9$. But looking at our graph, the y - axis only goes from - 8 to 8. Maybe it's better to find the x - intercept and y - intercept within the range of the graph.
For the y - intercept, set $x = 0$:
Substitute $x = 0$ into $y=-3x + 9$, we get $y=9$. But since 9 is outside the graph's y - range (up to 8), we find the x - intercept.
For the x - intercept, set $y = 0$:
$0=-3x + 9$
Add $3x$ to both sides: $3x=9$
Divide both sides by 3: $x = 3$. So the x - intercept is the point $(3,0)$.
Now, let's find another point. Let's choose $x = 0$, even though $y = 9$ is outside the graph, but we can also choose $x = 2$.
When $x = 2$, $y=-3(2)+9=-6 + 9=3$. So the point is $(2,3)$.
Or $x = 1$, $y=-3(1)+9 = 6$. The point is $(1,6)$.
Now, plot the points $(3,0)$ and $(1,6)$ (or other points we found) and draw a straight line through them.

Answer:

To graph the line $y + 3x=9$ (or $y=-3x + 9$):

1. Find two points on the line:

• When $x = 0$, $y=9$ (outside the given graph's y - range, but we can use other points).
• When $y = 0$, $x = 3$, so the point is $(3,0)$.
• When $x = 1$, $y=-3(1)+9 = 6$, so the point is $(1,6)$.

1. Plot the points (e.g., $(3,0)$ and $(1,6)$) on the coordinate plane and draw a straight line passing through them.