Solving nonlinear systems by substitution

hatjuice

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Apr 19, 2014
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I seem to be on the wrong track whenever I try to solve a nonlinear system. This is what I have been trying to do:
y + x = 17
x^2 + y^2 = 169
Solving for y in the first equation gives me:
y = 17 -x
Then solving for y in the second equation gives me:
y^2 = (13 - x)(13+x)
so then:
y = 13 - x or y = 13+x
since y = y
17-x = 13 +x
Which gives me:
x = 2
HOWEVER, the book says the solutions are (5, 12) and (12, 5). I can't seem to figure out how!
Help? Many thanks in advance!
 
I seem to be on the wrong track whenever I try to solve a nonlinear system.

Don't worry about this; learning math is a process of making mistakes. You will improve, with more practice.


y + x = 17
x^2 + y^2 = 169
Solving for y in the first equation gives me:
y = 17 -x
Then solving for y in the second equation gives me:
y^2 = (13 - x)(13+x)
so then:
y = 13 - x or y = 13+x

The last conclusion above is a mistake. We cannot conclude that y equals either 13-x or 13+x. Consider the following.

IF y were to be 13-x, then y^2 would be (13-x)^2

IF y were to be 13+x, then y^2 would be (13+x)^2

You already know that y^2 equals 169-x^2, but neither of the squares above equals this (use FOIL, to convince yourself).


Try this approach.

You determined that y = 17-x

Substitute that expression for y in the equation x^2 + y^2 = 169 and simplify.

You will end up with a quadratic equation in x.

Cheers :)

 
\(\displaystyle x^2= ab\) does NOT give "x= a or x= b". If \(\displaystyle x^2= 12= (3)(4)\) then \(\displaystyle x= \pm\sqrt{12}= \pm2\sqrt{3}\) not 3 and 4.
 
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