Question

what is the graph of the system $y = -2x + 3$ and $2x + 4y = 8$?

Explanation:

Step1: Rewrite the second equation in slope - intercept form

The second equation is \(2x + 4y=8\). We want to solve for \(y\) to get it in the form \(y = mx + b\) (where \(m\) is the slope and \(b\) is the y - intercept).
First, subtract \(2x\) from both sides: \(4y=-2x + 8\).
Then, divide each term by 4: \(y=\frac{-2x + 8}{4}=-\frac{1}{2}x + 2\).
The first equation is \(y=-2x + 3\).

Step2: Analyze the slopes and y - intercepts

• For the first line \(y=-2x + 3\): The slope \(m_1=-2\) and the y - intercept \(b_1 = 3\) (so it crosses the y - axis at \((0,3)\)).
• For the second line \(y =-\frac{1}{2}x+2\): The slope \(m_2=-\frac{1}{2}\) and the y - intercept \(b_2 = 2\) (so it crosses the y - axis at \((0,2)\)).

Now, let's analyze the options:

• The first line has a steeper slope (since \(\vert - 2\vert>\vert-\frac{1}{2}\vert\)) and a higher y - intercept.
• The second line has a less steep slope and a lower y - intercept.

Looking at the graphs, the third option (the one with the two lines where one has a y - intercept at 3 and a steeper slope, and the other has a y - intercept at 2 and a less steep slope) matches the equations.

Answer:

The third graph (the one with the line \(y = - 2x+3\) (steeper, y - intercept 3) and \(y=-\frac{1}{2}x + 2\) (less steep, y - intercept 2))