Step 1) Because the second equation is already solved for yy we can substitute (2x - 5)(2x−5) for yy in the first equation and solve for xx:
5x - 4y = -105x−4y=−10 becomes:
5x - 4(2x - 5) = -105x−4(2x−5)=−10
5x + (-4 xx 2x) + (-4 xx - 5) = -105x+(−4×2x)+(−4×−5)=−10
5x + (-8x) + 20 = -105x+(−8x)+20=−10
5x - 8x + 20 = -105x−8x+20=−10
(5 - 8)x + 20 = -10(5−8)x+20=−10
-3x + 20 = -10−3x+20=−10
-3x + 20 - color(red)(20) = -10 - color(red)(20)−3x+20−20=−10−20
-3x + 0 = -30−3x+0=−30
-3x = -30−3x=−30
(-3x)/color(red)(-3) = (-30)/color(red)(-3)−3x−3=−30−3
(color(red)(cancel(color(black)(-3)))x)/cancel(color(red)(-3)) = 10
x = 10
Step 2) Substitute 10 for x in the second equation and calculate y
y = 2x - 5 becomes:
y = (2 xx 10) - 5
y = 20 - 5
y = 15
The Solution Is: x = 10 and y = 15 or (10, 15)
