Question

1 (a) solve the system of equations by graphing. equation 1: $y + x = 2$ equation 2: $-8 - 2y + x = 0$

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: $y + x=2$.
Subtract $x$ from both sides: $y=-x + 2$.
For this line, the slope $m=-1$ and the y - intercept $b = 2$. To find two points on this line, when $x = 0$, $y=2$ (the y - intercept). When $y = 0$, $0=-x + 2$, so $x = 2$. So two points on the line are $(0,2)$ and $(2,0)$.

Step2: Rewrite Equation 2 in slope - intercept form

Starting with Equation 2: $-8-2y + x = 0$.
First, isolate $y$. Add $2y$ to both sides: $x-8=2y$.
Then divide both sides by 2: $y=\frac{1}{2}x-4$.
For this line, the slope $m = \frac{1}{2}$ and the y - intercept $b=-4$. To find two points on this line, when $x = 0$, $y=-4$. When $y = 0$, $0=\frac{1}{2}x-4$, add 4 to both sides: $\frac{1}{2}x=4$, multiply both sides by 2: $x = 8$. So two points on the line are $(0,-4)$ and $(8,0)$.

Step3: Graph the two lines

Plot the points for each line and draw the lines. The first line (from Equation 1) passes through $(0,2)$ and $(2,0)$. The second line (from Equation 2) passes through $(0,-4)$ and $(8,0)$.

Step4: Find the intersection point

The intersection point of the two lines is the solution to the system of equations. By graphing (or by solving the equations algebraically), we can see that the two lines intersect at the point $(4,-2)$. We can verify this:
For Equation 1: $y+x=-2 + 4=2$, which satisfies the equation.
For Equation 2: $-8-2y+x=-8-2\times(-2)+4=-8 + 4 + 4=0$, which satisfies the equation.

Answer:

The solution to the system of equations is $x = 4,y=-2$ (or the ordered pair $(4,-2)$).