Given equations: x^2 + y^2 = 113 ---------------equation(1)
x + y = 15 --------------------- eqaution(2).
Solution: We need to solve it by substitution.
In order to solve by substituion, we need to solave second equation for a variable and substitute in first equation.
x+y = 15.
Subracting x from both sides, we get
x-x + y = 15-x.
y= 15-x.
Substituting y=(15-x) in first equation x^2 + y^2 = 113.
x^2 + (15-x)^2 = 113.
Expanding (15-x)^2 = (15)^2 + (x)^2 -2*15 *x = 225 +x^2 -30x.
x^2 + (15-x)^2 = 113 would become
x^2 + 225 + x^2 -30x =113.
Combining like terms x^2+x^2, we get 2x^2.
2x^2 -30x +225 =113.
Subtracting 113 from both sides, we get
2x^2 -30x +225-113 =113-113.
2x^2 -30x + 112 = 0
2 is the greatest common factor (gcf) there. Dividing whole equation by 2.
2x^2/2 -30x/2 + 112/2 = 0
x^2 -15x + 56 =0.
Factoring out above quadratic equation by product sum rule.
We have a=1, b=-15 and c=56.
Product of a and c= 56 and b=-15.
So, we need to find two numbers that add upto -15 and product = 56.
We get -7 and -8 in factors of 56.
Sum of -7 and -8 = -15 and product of -7 * -8 = +56.
So, we could factor out above quadratic as
(x-7)(x-8) =0.
By product sum rule, we need to put those factors equal to 0 and solve for x.
x-7=0
Adding 7 on both sides we get
x-7+7=0+7
x=7.
x-8=0.
Adding 8 on both sides, we get
x-8+8 = 0+8
x=8.
Therefore, x=7 and 8.
Plugging those values of x's in firsr equation y=15-x, we get
y=15-7 = 8 and y=15-8 = 7.
Therefore, we got two solutions x=7, y=8 and x=8,y=7.
(7,8) and (8,7).
