Set up an algebraic equation and then solve the following problems. an integer is one more than four times another. if the product of the two integers is 39, then find the integers.

Let x represent the smaller integer. Then the other one is 4x+1 (1 more than 4 times the first one). Their product will be ...

... x(4x+1) = 39 . . . . . . an algebraic equation to solve

In standard form, this is ...

... 4x² +x -39 = 0

... (x -3)(4x +13) = 0 . . . . . factored form

... x = 3 or x = -13/4

The only integer solution here is x = 3. Then the other is 4·3+1 = 13.

The two integers are 3 and 13.

Comment on factoring

There are a couple of different ways you can factor the equation. They basically involve looking for factors of 4×39 that differ by 1. Of course 39 = 3×13, so you are looking for factors of 4×3×13 = 12×13 that are different by 1. Guess what? You've found them! (12 and 13).

One method of factoring requires you use these to rewrite the equation as ...

... 4x² -12x +13x -39 = 0

then factor by grouping.

... 4x(x -3) +13(x -3) = 0 = (4x+13)(x -3)

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Another method of factoring lets you use the leading coefficient in both factors, then divide by that coefficient in some convenient way:

... (4x-12)(4x+13)/4 = 0 = (x -3)(4x +13) . . . . . note the use of 4x in both factors

a) y = -3x +2

b) y = 1/2x -6

these 2 can be combined into one equation.

-3x +2 = .5x -6

8 = 3.5x

x = 8 / 3.5 = 2.2857142857

y = -3 * (8 / 3.5) +2

y = (-24 / 3.5) + 2

y = -4.8571428571

step-by-step explanation: