Question

Solve the system and round to the nearest hundredth. Enter the

smallest x coordinate first.
{x + y = 50 {xy = 300

Answer 1

x ≈ 6.97, y ≈ 43.03; x ≈ 43.03, y ≈ 6.97

Explanation:

System: x + y = 50, xy = 300

Alter 1st equation: y = 50 - x

Substitute: x(50 - x) = 300

Distribute: -x² + 50x = 300

Subtract: -x² + 50x - 300 = 0

Quadratic formula: x = (-50 ± √(2500 - 4 * -1 * -300))/(-2)

Simplify: x = (-50 ± √1300)/-2

Distribute: x = 25 ± (10√13)/-2

Simplify: x = 25 - 5√13 --> x ≈ 6.97, x = 25 + 5√13 --> x ≈ 43.03

Substitute: 6.97 + y = 50 --> y ≈ 43.03, 43.04 + y = 50 --> y ≈ 6.97

Answer 2

Explanation:

xy = 300

x + y = 50 Solve for y

y = 50 - x substitute into xy = 300

x(50 - x) = 300 Remove the brackets.

50x - x^2 = 300 Bring the left to the right.

0 = x^2 - 50x + 300

This is a quadratic. It will have 2 solutions.

a=1

b = - 50

c = 300

Put these into the quadratic equation.

It turns out that x has two values -- both plus

x1 = 42.03

x2 = 6.97

x1 + y1 = 50

42.03 + y = 50

y = 50 - 42.03

y = 7,97

(42.03 , 7.97)

====================

x2 + y2 = 50

6.97 + y2 = 50

y2 = 50 - 6.97

y2 = 43.03

(6.97 , 43.03)