Solve the system by the substitution method. Show your work.
2y - x = 5
x2 + y2 - 25 = 0
I solved the problem but I got it wrong for some reason..can you solve it step by step?
next time, show your work, so we can find your error.
From equation 1, x = 2y-5
plug that into #2, and you have
x^2+y^2 = 25
(2y-5)^2 + y^2 = 25
5y^2 - 20y + 25 = 25
5y(y-4) = 0
y = 0 or 4
now you can find x values as needed, from either equation.
http://www.wolframalpha.com/input/?i=plot+2y+-+x+%3D+5,+x2+%2B+y2+-+25+%3D+0
That is exactly what I had for my answer but I was confused that I had an answer of 2 solutions.
Certainly! Let's solve the system of equations using the substitution method step by step.
Step 1:
We start by solving one equation for one variable and substituting it into the other equation. Let's solve the first equation for x in terms of y:
2y - x = 5
-x = -2y + 5
x = 2y - 5
Step 2:
Now we substitute the value of x in the second equation:
x2 + y2 - 25 = 0
(2y - 5)2 + y2 - 25 = 0
4y2 - 20y + 25 + y2 - 25 = 0
5y2 - 20y = 0
5y(y - 4) = 0
Step 3:
We have two possible solutions:
y = 0 or y - 4 = 0
Case 1: y = 0
Substitute y = 0 back into the first equation to find x:
2y - x = 5
2(0) - x = 5
-x = 5
x = -5
So one solution is (x, y) = (-5, 0).
Case 2: y - 4 = 0
Solve for y:
y - 4 = 0
y = 4
Substitute y = 4 back into the first equation to find x:
2y - x = 5
2(4) - x = 5
8 - x = 5
-x = 5 - 8
-x = -3
x = 3
So another solution is (x, y) = (3, 4).
Therefore, the system of equations has two solutions: (-5, 0) and (3, 4).