Question

One positive number is 8 more than twice another. Their product is 10

Answer 1

the numbers are 10 and 1

let one of the numbers be X, if we take the second number to be y, then x - 2y = 8

X -2y = 8.... equation 1

also their products xy = 10

xy = 10.... equation 2

from first equation, making x the subject of the formula,

X = 8+2y

inserting this in equation 2

(8+2y)y = 10

open the brackets

8y+ 2y² = 10

we obtain a quadratic equation, next we arrange the equation following the reducing powers to make it easier to solve....

2y² +8y -10 = 0

divide by 2 to reduce the values for easier solving

y² +4y - 5 = 0

we can solve by factoring.....two numbers whose sum is +4 and product is -5. numbers are +5 and -1

Insert back in the equation in the place of 4y

y² - y +5y -5 = 0

y(y-1) +5(y- 1) = 0

(y+5)(y-1) =0

hence y = -5 or 1

since the question was talking about positive number we pick 1

y= 1

we solve for x using equation 2

X(1) = 10

X = 10

Answer 2

no.s are 1 and 10.

Explanation:

Let the positive no.s be x and y.

given that:

y = 2x + 8 -------- 1)

xy = 10 --------2)

Substituting 1 in 2:

⇒ x (2x + 8) = 10

⇒ 2x² + 8x = 10 (dividing both sides by 2)

⇒ x² + 4x = 5

⇒ x² + 4x -5 = 0

⇒ x² - x + 5x - 5 =0 (factorizing)

⇒ (x²-x) + (5x-5) = 0

⇒ x(x-1)+5 (x-1) = 0

⇒ (x+5) (x-1) = 0

⇒ x=5 = 0 or x -1 = 0

x = -5 or x = 1

since no.s are +ve, x = -5 is rejected

∴ x = 1.

⇒ y = 2*1 + 8 = 10