The graph shows the solution to which system of equations?
a) y = x + 2 and y = -x - 4
b) y = 3x + 7 and y = x + 12
c) y = 2x - 3 and y = −2x + 1
d) y = -x + 4 and y = −2x + 5

So let's start by guesstimating the slopes:
the green line has a slope close to -x, but more negative than that, possibly -2; the pink line has a slope close to +x, but higher towards +2.

Next let's look at the solution: the two lines intersect at the point (1, -1).
**you could just simple plug that x (1) into all the equations, but let's rule out answers anyway. ;)

A) is incorrect because the slopes of -1 and +1 are off from out predicted -2 and +2
B) is incorrect because of a similar reason, the slopes of +3 and +1 don't make any sense
C) Ooh, we do have a +2 and -2 for the slopes, and... violà! plug in 1 for the x's and we get -1 for the y in both equations
D) slopes are closer than in A and B, but plugging in 1 doesn't get us -1

So the correct answer is:
C)y = 2x - 3 and y = −2x + 1

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

ac ≅ ab (given)< cax ≅ < bax   (given)ax ≅ ax (common side)

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

ac ≅ ab (given)< cay ≅ < bay (since < cax ≅ < bax is given and xy is the extension of line ax)ay ≅ ay (common side)

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)