Question

2
(a)
graph both equations on the same graph:
equation 1: $x + 2y = -4$
equation 2: $3x + 2y = 4$

Explanation:

Step1: Rewrite Equation 1 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 Equation 1: $x + 2y=-4$
Subtract $x$ from both sides: $2y=-x - 4$
Divide both sides by 2: $y=-\frac{1}{2}x-2$
To find two points on this line, we can use the y - intercept and another point.
When $x = 0$, $y=-2$. So one point is $(0,-2)$.
When $y = 0$, $0=-\frac{1}{2}x - 2$, add 2 to both sides: $2=-\frac{1}{2}x$, multiply both sides by - 2: $x=-4$. So another point is $(-4,0)$.

Step2: Rewrite Equation 2 in slope - intercept form

Starting with Equation 2: $3x + 2y = 4$
Subtract $3x$ from both sides: $2y=-3x + 4$
Divide both sides by 2: $y=-\frac{3}{2}x + 2$
To find two points on this line, use the y - intercept and another point.
When $x = 0$, $y = 2$. So one point is $(0,2)$.
When $y = 0$, $0=-\frac{3}{2}x+2$, subtract 2 from both sides: $-2=-\frac{3}{2}x$, multiply both sides by $-\frac{2}{3}$: $x=\frac{4}{3}\approx1.33$. So another point is $(\frac{4}{3},0)$ or we can use $x = 2$, then $y=-\frac{3}{2}(2)+2=-3 + 2=-1$, so the point $(2,-1)$ is also on the line.

Step3: Plot the points and draw the lines

For Equation 1 ($y =-\frac{1}{2}x-2$), plot the points $(0,-2)$ and $(-4,0)$ and draw a straight line through them.
For Equation 2 ($y=-\frac{3}{2}x + 2$), plot the points $(0,2)$ and $(2,-1)$ (or $(\frac{4}{3},0)$) and draw a straight line through them.

Answer:

To graph the lines:

• For $x + 2y=-4$ (or $y =-\frac{1}{2}x-2$), plot the points $(0,-2)$ and $(-4,0)$ and draw a line through them.
• For $3x + 2y = 4$ (or $y=-\frac{3}{2}x + 2$), plot the points $(0,2)$ and $(2,-1)$ (or $(\frac{4}{3},0)$) and draw a line through them.