{y=x^2+5x-3
{y-x=2
I think it's (5,1) and (-3,3)
correct me if I'm wrong
4 answers
What is the solution of the linear-quadratic system of equations?
please write the full response so I can understand it further on in the future
eq #2 says that x+2
Use that to get
x+2 = x^2+5x-3
x^2 + 4x - 5 = 0
(x+5)(x-1) = 0
Looks like your solution is wrong.
Too bad you didn't show your work... or even check your solution to see whether it worked.
Use that to get
x+2 = x^2+5x-3
x^2 + 4x - 5 = 0
(x+5)(x-1) = 0
Looks like your solution is wrong.
Too bad you didn't show your work... or even check your solution to see whether it worked.
y - x = 2
y = 2 + x
y = x² + 5 x - 3
y = y
x² + 5 x - 3 = 2 + x
x² + 5 x - 3 - x - 2 = 0
x² + 4 x - 5 = 0
The solutions of this equation are:
x = 1 and x = - 5
So:
y = 2 + x = 2 + 1 = 3
and
y = 2 + x = 2 - 5 = - 3
Final solution:
( 1 , 3 ) , ( - 5 , - 3 )
y = 2 + x
y = x² + 5 x - 3
y = y
x² + 5 x - 3 = 2 + x
x² + 5 x - 3 - x - 2 = 0
x² + 4 x - 5 = 0
The solutions of this equation are:
x = 1 and x = - 5
So:
y = 2 + x = 2 + 1 = 3
and
y = 2 + x = 2 - 5 = - 3
Final solution:
( 1 , 3 ) , ( - 5 , - 3 )
2 answers