Question

Solve using either substitution or elimination process for the nonlinear equations: x^2+y^2=36 and x+y=6

Answer 1

do i use substitution or elimination?

Answer 2

\[x^2+y^2=36\]
\[x+y=6\]
\[y=6-x\]

\[x^2+(6-x)^2=36\]

Answer 3

that would be "substitution"

Answer 4

If one of the equations given to you is very simple to put in a (for example) "x=" or "y=" form, use substitution.

Answer 5

so how do i take this info and find the solutions

Answer 6

by solving the quadratic equation
\[x^2+(6-x)^2=36\]

Answer 7

\[x ^{2}+(6-x)^{2}=36 \] meaning i would subtract 36?

Answer 8

\[x^2+(6-x)^2=36\]
\[x^2+(6-x)(6-x)=36\]
\[x^2+36-12x+x^2=36\]
\[2x^2-12x=0\]
\[2x(x-6)=0\]
\[x=0,x=6\]

Answer 9

omg thank you :)

Answer 10

so i would have 3 real solutions for this equation?

Answer 11

no just 2

Answer 12

0 and 6

Answer 13

okay because y='s that equation which isnt an answer

Answer 14

if x = 0, since
\[x+y=6\] that means
\[0+y=6\implies y =6\]

Answer 15

and if x = 6 we get
\[6+y=6\implies y=0\]

Answer 16

right the answers are written: Two real solutions \[(\pm6, 0) \And (0,\pm6) \]

Answer 17

remove the plus and minus sign.