One integer is 8 times another. If the product of two integers is 128, then find the integers

Answer 1

Let's say the first integer is x and the other is y.

Then make a system of equations representing the situation

y=8x

xy=128

Then substitute on variable in the form of the other.(since y is equivalent to 8x then we can sub that in in the second equation

x*(8x)=128

8x^2=128

8x^2-128=0

8(x^2-16)=0

8(x+4)(x-4)=0

X now could either be -4 or +4

Now sub in 4 and -4 into one of the equations and solve for y

y=8*4=32

y=8*-4=-32

So x is 4 and y is 32, or -4 and -32 repsectively hope that helps

Answer 2

Call the integers a and b.

a = 8b One integer is 8 times another.

ab = 128 The product is 128

Since a = 8b, substitute 8b for a in the second equation.

8b * b = 128

8b^2 = 128

b^2 = 128/8 = 16

b = sqrt(16) = 4 or -4

a = 8*4 = 32

a = 8 * -4 = -32