Given:
The system of equations are:
(a) [tex]y=3x+5[/tex]
[tex]y=-2x+30[/tex]
(b) [tex]x+y=-3[/tex]
[tex]2y=-2x+8[/tex]
To find:
The intersection points of the given system of equations.
Solution:
(a)
We have,
[tex]y=3x+5[/tex] ...(i)
[tex]y=-2x+30[/tex] ...(ii)
From (i) and (ii), we get
[tex]3x+5=-2x+30[/tex]
[tex]3x+2x=30-5[/tex]
[tex]5x=25[/tex]
[tex]x=5[/tex]
Putting x=5 in (i), we get
[tex]y=3(5)+5[/tex]
[tex]y=15+5[/tex]
[tex]y=20[/tex]
Therefore, the point of intersection is (5,20).
(b)
The given system of equations is:
[tex]x+y=-3[/tex]
[tex]2y=-2x+8[/tex]
Write these equation in slope intercept form [tex](y=mx+b)[/tex], where m is slope and b is y-intercept.
[tex]y=-x-3[/tex] ...(iii)
[tex]y=-x+4[/tex] ...(iv)
From the equation (iii) and (iv), it is clear that the slopes of both equations are same, i.e., -1 but the y-intercepts are different, -3 and 4 respectively.
It means the lines are parallel and parallel lines never intersect each other.
Therefore, the point of intersect does not exist because the lines are parallel.
