Answer:
A solution to a system of equations means the point must work in both equations in the system. So, we test the point in both equations. It must be a solution for both to be a solution to the system.
Step-by-step explanation:
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Xy = -150
x + y = 5
x + y = 5
x - x + y = -x + 5
y = -x + 5
xy = -150
x(-x + 5) = -150
x(-x) + x(5) = -150
-x² + 5x = -150
-x² + 5x + 150 = 0
-1(x²) - 1(-5x) - 1(-150) = 0
-1(x² - 5x - 150) = 0
-1 -1
x² - 5x - 150 = 0
x = -(-5) ± √((-5)² - 4(1)(-150))
2(1)
x = 5 ± √(25 + 600)
2
x = 5 ± √(625)
2
x = 5 ± 25
2
x = 2.5 ± 12.5
x = 2.5 + 12.5 or x = 2.5 - 12.5
x = 15 or x = -10
x + y = 5
15 + y = 5
- 15 - 15
y = -10
(x, y) = (15, -10)
or
x + y = 5
-10 + y = 5
+ 10 + 10
y = 15
(x, y) = (-10, 15)
The two numbers that add up to 5 and multiply to -150 are 15 and -10.
The equivalent fraction of 9/15 would be 3/5.
