Question

solve the system by graphing: \\(\

$$\begin{cases} x + y = 4 \\\\ -x + y = 1 \\end{cases}$$

\\)

Explanation:

Step1: Rewrite equations in slope - intercept form

For the first equation \(x + y=4\), we can rewrite it as \(y=-x + 4\). The slope \(m_1=- 1\) and the \(y\) - intercept \(b_1 = 4\).
For the second equation \(-x + y=1\), we can rewrite it as \(y=x + 1\). The slope \(m_2 = 1\) and the \(y\) - intercept \(b_2=1\).

Step2: Find two points for each line

For \(y=-x + 4\):

• When \(x = 0\), \(y=4\) (so the point is \((0,4)\)).
• When \(x = 4\), \(y=-4 + 4=0\) (so the point is \((4,0)\)).

For \(y=x + 1\):

• When \(x = 0\), \(y=1\) (so the point is \((0,1)\)).
• When \(x = 3\), \(y=3 + 1=4\) (so the point is \((3,4)\)).

Step3: Graph the lines and find the intersection

When we graph the line \(y=-x + 4\) (using points \((0,4)\) and \((4,0)\)) and the line \(y=x + 1\) (using points \((0,1)\) and \((3,4)\)), we find that the two lines intersect at the point \((\frac{3}{2},\frac{5}{2})\) or \((1.5,2.5)\). We can also solve the system algebraically to verify:
Set \(-x + 4=x + 1\) (since both equal \(y\)).
\(-x-x=1 - 4\)
\(-2x=-3\)
\(x=\frac{3}{2}\)
Substitute \(x = \frac{3}{2}\) into \(y=x + 1\), we get \(y=\frac{3}{2}+1=\frac{3 + 2}{2}=\frac{5}{2}\)

Answer:

The solution of the system is \(x=\frac{3}{2},y=\frac{5}{2}\) (or \((\frac{3}{2},\frac{5}{2})\))