Question

use the graph method to solve the system of linear equations:
y - x = -2 and 2x + y = 7
a (0,7)
b (2,0)
c (3.5,0)
d (3,1)

Explanation:

Step1: Rewrite equations in slope - intercept form

For the equation \(y - x=-2\), we can rewrite it as \(y=x - 2\). The slope \(m = 1\) and the \(y\) - intercept \(b=-2\).
For the equation \(2x + y=7\), we can rewrite it as \(y=-2x + 7\). The slope \(m=-2\) and the \(y\) - intercept \(b = 7\).

Step2: Find the intersection point (solution)

We can also solve the system by substitution or elimination. Let's use elimination.
We have the system:
\(

$$\begin{cases}y=x - 2\\y=-2x + 7\end{cases}$$

\)
Set the two expressions for \(y\) equal to each other:
\(x-2=-2x + 7\)
Add \(2x\) to both sides: \(x + 2x-2=-2x+2x + 7\), which simplifies to \(3x-2 = 7\)
Add 2 to both sides: \(3x-2 + 2=7 + 2\), so \(3x=9\)
Divide both sides by 3: \(x = 3\)
Substitute \(x = 3\) into \(y=x - 2\), we get \(y=3-2=1\)

We can also check each option:

• Option A: Substitute \(x = 0,y = 7\) into \(y - x=-2\): \(7-0=7

eq - 2\), so A is wrong.

• Option B: Substitute \(x = 2,y = 0\) into \(y - x=-2\): \(0 - 2=-2\), substitute into \(2x + y=7\): \(2\times2+0 = 4

eq7\), so B is wrong.

• Option C: Substitute \(x = 3.5,y = 0\) into \(y - x=-2\): \(0-3.5=-3.5

eq - 2\), so C is wrong.

• Option D: Substitute \(x = 3,y = 1\) into \(y - x\): \(1-3=-2\) (satisfies the first equation), substitute into \(2x + y\): \(2\times3+1=7\) (satisfies the second equation).

Answer:

D. \((3,1)\)