Question

One integer is 4 less than twice another integer, and their product is 96.set up an alegebric equation and solve it to find the two integers.

Answer 1

Answer: 8 and 12

Explanation:

Let's first set one of the integers as x. The other integer is 4 less than 2 times that integer, so it would be 2x - 4. Their PRODUCT (multiplication) is 96. Thus, x * (2x-4) = 96.

If you solve it out:

2x^2 - 4x = 96

Divide each side by 2

x^2 - 2x = 48

Subtract 48

x^2 - 2x - 48 = 0

Factor the quadratic

(x+6)(x-8) = 0

So, the answer is x = 6 or x = -8

Let's plug it in to the equation...

x = -6

-6 * 2(-6) - 4 = -6* -20 which doesn't equal 96 so we can rule out that one

x = 8

8 * 2(8) - 4 = 8 * 12 = 96

Answer 2

Let x be the first integer and y be the second integer.

From the problem statement, we can write two equations:

x = 2y - 4 (one integer is 4 less than twice another integer)

xy = 96 (their product is 96)

Substitute the first equation into the second equation:

(2y - 4)y = 96

Simplify and rearrange:

2y^2 - 4y - 96 = 0

Divide both sides by 2:

y^2 - 2y - 48 = 0

Factor the quadratic equation:

(y - 8)(y + 6) = 0

From this, we get two possible values for y:

y - 8 = 0 (which gives y = 8)

or

y + 6 = 0 (which gives y = -6)

If y = 8, then x = 2y - 4 = 2(8) - 4 = 12.

If y = -6, then x = 2y - 4 = 2(-6) - 4 = -16.

Therefore, the two integers are either 12 and 8, or -16 and -6.