So i am doing "solve systems of equations algebraically." i'm stuck on three of them. they are:
5x+8y=7300 and 1250=x+y
y=4x+45 and x=4y
y+11=2x and 3y-6x=-33
include steps !

Due to the more than usual numbber of questions in a question, we will have shortened explanations

questions an answes are as follows:
1. 5x+8y=7300 and 1250=x+y
2. y=4x+45 and x=4y
3. y+11=2x and 3y-6y=-33

1.
5x+8y=7300 and 1250=x+y
multily 2nd equation by -5 and add to othe equation
3y=1050
divide both sides by 3
y=350
sub back
x+y=1250
x+350=1250
minus 350 both sides
x=900

(900,350)

2.
sub 4y for x
y=4(4y)+45
y=16y+45
minus y both sides
0=15y+45
minus 45 both sides
-45=15y
divide both sides by 15
-3=y
sub back
x=4y
x=4(-3)
x=-12

(-12,-3)

3.
y+11=2x
3y-6x=-33
divide last equation by 3
y-2x=-11
add 2x to both sides
y=2x-11
add 11
y+11=2x
same equation
infinite solutions
all points on y=2x-11 are solutions

1.(900,350)
2. (-12,-3)
3. infinite solutions, any point on y=2x-11

In all honesty, there really is no “work” per say to show. mainly it’s just using the process of elimination. the question is asking which domain(s) would be the most appropriate for the function c(n). n is the number of observed vehicles so right away, 1, 2, and 3 can be eliminated.why? 1, 2 start in the negatives and therefore it is impossible to observe negative cars. so, #4 would be the best answer to the given problem.