+0

# algebra

+1
66
4

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

Jul 23, 2020

#1
+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 $$10y^2+90y+200=0$$

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
+359
+5

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

iamhappy  Jul 23, 2020
#3
+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.      It is better if they do some of their own thinking.

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