Find two numbers whose product is 12 such that one of the number is four less than eight times the other number.

Question

This is a number problem

Answer 1

xy = 12

y = 8x - 4

 

x(8x-4) = 12

8x² - 4x - 12 = 0

factor out a 4

4(2x² - x - 3) = 0

2x² - x - 3 = 0

 

we may be able to factor by grouping:

2(-3)= -6

Are there factors of -6 that add to -1?

(2)(-3) = -6 and 2-3=-1

Replace -x with 2x-3x and group

 

2x² + 2x - 3x - 3 = 0

(2x²+2x) + (-3x-3) = 0

2x(x+1) -3(x+1) = 0

(2x-3)(x+1) = 0

 

Either 2x-3= 0    or    x+1=0

          2x=3               x = -1

           x = 3/2

 

we have 2 values of x

Find associated y values

 

x=3/2

y = 8x-4 = 8(3/2)-4 = 8

 

x = -1

y = 8x-4 = 8(-1)-4 = -12

 

your answers are:

(3/2, 8) and (-1, -12)

Answer 2

Hi Harvey, let x be one of the numbers and y the other number. Set up two equations to solve.

 

Equation 1, product:

 

x * y = 12

 

Equation 2, difference:

 

x = 8y - 4

 

Solve equation 1 by substituting for x in equation 2:

 

x * y = 12

 

Becomes:

 

(8y - 4) * y = 12

 

Rearrange:

 

8y² - 4y - 12 = 0

 

Divide equation by 4:

 

2y² - y - 3 = 0

 

Factor

 

(2y - 3) * (y + 1) = 0

 

From here you can find the solutions, there are two, to the question.

 

questions?