x, y - two positive numbers
one positive number is 5 less than twice a second number
(1) x = 2y - 5
their product is 140
(2) xy = 140
substitute from (1) to (2)
(2y - 5)y = 140 use distributive property
(2y)(y) - (5)(y) = 140
2y² - 5y = 140 subtract 140 from both sides
2y² - 5y - 140 = 0
use quadratic formula
a = 2, b = -5, c = -140
b² - 4ac = (-5)² - 4(2)(-140) = 25 + 1120 = 1145
[tex]y_1=\dfrac{-b-\sqrt{b^2-4ac}}{2a}\\\\y_1=\dfrac{-(-5)-\sqrt{1145}}{2(2)}=\dfrac{5-\sqrt{1145}}{4} < 0[/tex]
[tex]y_2=\dfrac{-b+\sqrt{b^2-4ac}}{2a}\\\\y_2=\dfrac{-(-5)+\sqrt{1145}}{2(2)}=\dfrac{5+\sqrt{1145}}{4}[/tex]
substitute the value of y₁ to (1)
[tex]x=2\cdot\dfrac{5+\sqrt{1145}}{4}-5=\dfrac{5+\sqrt{1145}}{2}-\dfrac{10}{2}=\dfrac{5+\sqrt{1145}-10}{2}\\\\=\dfrac{-5+\sqrt{1145}}{2}[/tex]
Answer:
[tex]x=\dfrac{-5+\sqrt{1145}}{2},\ y=\dfrac{5+\sqrt{1145}}{4}[/tex]
