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Solve the system x^2 + y^2 = 25, x - 3y = 15

 Jul 23, 2020
 #1
avatar+359 
+5

We will solve this system by substitution. 

 

First, isolate the variable x for the second equation to get \(x=15+3y\)

 

Next, for \(x^2+y^2=25\), substitute x with \(15+3y\) that we obtained from the first question.

 

We get \((3y+15)^2 +y^2=25\)

 

We expand the equation to get \(10y^2+90y+225=25\)

 

Subtract 25 from both sides \(10y^2+90y+200=0\)

 

Solve with the quadratic formula 

 

Simplify (it seems like a lot, but it actually wasn't so much computation)

 

After I found y, you should be able to plug it into the equation x-3y = 15, I don't have time to do that since I have to go, but it should be quite easy. :)

 

y = -5 = \(\frac{-90\pm \sqrt{90^2-4\cdot \:10\cdot \:200}}{2\cdot \:10}\)

y = -4 = \(\frac{-90\pm \sqrt{90^2-4\cdot \:10\cdot \:200}}{2\cdot \:10}\)

 Jul 23, 2020
edited by iamhappy  Jul 23, 2020
 #2
avatar+359 
+5

sorry, I really tried, but someone needs to check this. :(

iamhappy  Jul 23, 2020
 #3
avatar+110758 
+2

Nice answer Iamhappy,  (although you did not take 25 from both sides)

You do not need to use the quadratic formula at the end though. It is not necessary.

A quick sketch would help too.

 

The asker can finish it by him/herself now anyway.   cool   It is better if they do some of their own thinking.

Melody  Jul 23, 2020
edited by Melody  Jul 23, 2020
 #4
avatar+359 
+4

Thank you Melody for clarifying my slightly hurried and slightly incorrect answer. :) I was about to go to online class. :) I really needed some more reliance since I can't always rely on myself. And yes you are correct that Guest should be able to finish it him/herself. \(:)\)

iamhappy  Jul 23, 2020
edited by iamhappy  Jul 23, 2020

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