Solve... y^2 = x + 1 ... 8y = x + 17?

Question

What is the solution?

Answer 1

y^2=x+1
8y=x+17

Let's solve for x in the second equation
8y=x+17
8y-17=x
x=8y-17

Let's substitue that in for x in the first equation
y^2=x+1
y^2=(8y-17)+1
y^2=8y-16
y^2-8y+16=0
(y-4)(y-4)=0
y=4; y=4

so y = 4
sub 4 in for y in the second equation

8y = x + 17
8(4) = x + 17
32 = x + 17
15 = x
x = 15

so x = 15, y = 4

Answer 2

8y = x + 17 ==> x = 8y - 17
Now, use x in the second equation:
y^2 = x +1 ==> y^2 = 8y - 17 +1 ==> y^2 - 8y + 16 = 0
( y - 4)^2=0 ==> y - 4=0 ==> y = 4
from the first equation we have:
x = 8y - 17 then
x=8(4) - 17 ==> x = 15

Answer 3

x = 15 and y = 4.

y^2 - 1 = x and substitute that in to the other equation.

8y = y^2 -1 + 17 =>y^2 - 8y + 16 = 0
(y - 4)(y - 4)=0
y=4

substitute into any of the equations and get x = 15.

Answer 4

Since 8y = x +17, then y = (x+17)/8. Since y^2 = x+1, then ((x+17)/8)^2 = x+1. By expanding the binomial, (x^2 + 34x + 289)/64 = x+1. Multiplying each side by 64, x^2 + 34x + 289 = 64x + 64. Subtract 64x + 64 from each side, x^2 - 30x + 225 = 0.
Factor, (x -15)(x-15) = 0. Therefore x = 15!

Since x =15, y = (15+17)/8 = 4.

x = 15, y = 4