Answer:
• (x,y)=(2,2)
,
• Quadrant I
Explanation:
Given the system of equations:
[tex]\begin{gathered} x+y-4=0 \\ x-y=0 \end{gathered}[/tex]
We are required to solve the system graphically.
To do this, find two points on each of the lines.
(a)x+y-4=0
When x=0
[tex]\begin{gathered} x+y-4=0 \\ 0+y-4=0 \\ y=4 \\ \implies(0,4) \end{gathered}[/tex]
When x=1
[tex]\begin{gathered} x+y-4=0 \\ 1+y-4=0 \\ y-3=0\implies y=3 \\ \implies(1,3) \end{gathered}[/tex]
Join the points (0,4) and (1,3) to plot the first equation.
(b)x-y=0
When x=0, y=0 ==>(0,0)
When x=2, y=2 ==>(2,2)
Join the points (0,0) and (2,2) to plot the second equation.
The graph is shown below:
The two lines intersect at (2,2).
Therefore, the solution to the system of equations is:
[tex](x,y)=(2,2)[/tex]
The solution (2,2) is in Quadrant I.
