<PRE><B>Question 147027</B>
x^2 + 3y = 25
x  +  y  =  7
:
Multiply the 2nd equation by 3 and subtract from the 1st equation:
x^2 + 3y = 25
3x  + 3y = 21
---------------Subtracting eliminates y, leaving us with
x^2 + 3x = 4
:
A quadratic equation that we can factor:
x^2 + 3x - 4 = 0
(x + 4)(x - 1) = 0
Two solutions:
x = -4
x = +1
:
Find the value for y using both solutions using the 2nd equation (x + y = 7)
x=-4
-4 + y = 7
y = 7 + 4
y = 11
and
x = +1
1 + y = 7
y = 7 - 1
y = 6
:
We have two sets of solutions: x=-4,y=11 and x=1,y=6
:
But only one set will work in the 1st equation, substitute both in x^2 + 3y = 25
You do this and you will see that x=1;y=6 is the solution that satisfies both equations.
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