Question

1. consider the following equations:

4x - 2y = 2
-2x + 4y = 2
5a plot the two lines.

Explanation:

Step1: Rewrite first equation in slope - intercept form

The first equation is \(4x - 2y=2\). We solve for \(y\):
Subtract \(4x\) from both sides: \(- 2y=-4x + 2\).
Divide both sides by \(-2\): \(y = 2x-1\).
To find two points on this line, when \(x = 0\), \(y=-1\); when \(y = 0\), \(2x-1=0\Rightarrow x=\frac{1}{2}\).

Step2: Rewrite second equation in slope - intercept form

The second equation is \(-2x + 4y=2\). We solve for \(y\):
Add \(2x\) to both sides: \(4y = 2x+2\).
Divide both sides by \(4\): \(y=\frac{1}{2}x+\frac{1}{2}\).
To find two points on this line, when \(x = 0\), \(y=\frac{1}{2}\); when \(y = 0\), \(\frac{1}{2}x+\frac{1}{2}=0\Rightarrow x=- 1\).

Step3: Plot the lines

For the line \(y = 2x - 1\), plot the points \((0,-1)\) and \((\frac{1}{2},0)\) and draw a straight line through them.
For the line \(y=\frac{1}{2}x+\frac{1}{2}\), plot the points \((0,\frac{1}{2})\) and \((-1,0)\) and draw a straight line through them. The graph provided in the problem shows the two lines plotted with their intersection point (which can be found by solving the system, but for plotting, the key is to use the slope - intercept form to find points).

Answer:

The two lines are plotted as follows: One line (from \(y = 2x-1\)) passes through \((0, - 1)\) and \((\frac{1}{2},0)\), and the other line (from \(y=\frac{1}{2}x+\frac{1}{2}\)) passes through \((0,\frac{1}{2})\) and \((-1,0)\), which matches the given graph.