Question

Solve the system by the substitution method.
2y-x=5
x^2 + y^2 - 25 = 0

Answer 1

i know I have to do something with the first equation and then place it in the second but I am stuck there

Answer 2

solve for x in the first equation... then use that x in the second equation.

Answer 3

so that would make it
x=-5+2y

Answer 4

yes. now plug (-5+2y) in for x in the second equation.

Answer 5

okay so now i have
-5 + 2y + y^2 - 25 = 0

Answer 6

no. (-5+2y)^2+y^2-25-0 you have to square the value of x

Answer 7

okay now I am stuck

Answer 8

(-5+2y)^2+y^2-25=0 expand the squared value

Answer 9

the squares always get me

Answer 10

okay I tried it and I got 5y^2 - 50

Answer 11

\[(-5+2y)^2+y^2-25=0\]
\[25-20y+4y^2+y^2-25=0\]
\[5y^2-20y=0\]
factor out the y
\[y(5y-20)=0\]
so y=0 or 5y-20=0
y=0 or y=4

Answer 12

now use those two values of y and plug them back in where you had x=-5+2y
you get x=-5 or x=(-5+8)=3

Answer 13

but that don't match my choices
I have
(-5,0)

Answer 14

yea your solutions are (-5,0) or (3,4)

Answer 15

oh so
x = -5 and 3
y= 0 and 4

Answer 16

okay i see it now I forgot the plugging in of y to get x

Answer 17

its all good. well as long either (-5,0) or (3,4) match your choices ur fine

Answer 18

I was looking at what you did and I thought we found x first I was just confused

Answer 19

thank you

Answer 20

ur welcome