Solve the system by the substitution method. x2 + y2 = 113 x + y = 15

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).

step-by-step explanation:

hans did not square the 40-unit side; he merely multiplied it by 2.   incorrect.