SOLUTION: 1. Find two integers whose product is 14 such that one of the integers is three less than five times the other integer. 2. The perimeter of a rectangle is 50 inches, and the ar

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: 1. Find two integers whose product is 14 such that one of the integers is three less than five times the other integer. 2. The perimeter of a rectangle is 50 inches, and the ar      Log On


   

Question 1065941: 1. Find two integers whose product is 14 such that one of the integers is three less than five times the other integer.
2. The perimeter of a rectangle is 50 inches, and the area is 136 square inches. Find the length and width of the rectangle.

3. 9x3 + x2 + 7 from 5x3 − x − 8

Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
EDIT: MISTAKE HAS BEEN ADJUSTED.
RESULT NOW CORRECT

This is for #1 only.

x and y integers
system%28xy=14%2Cy=-3%2B5x%29

x%285x-3%29=14
5x%5E2-3x-14=0
and since you expect x and y to be integers, you should try factorization.

%285x%2B7%29%28x-2%29=0
You want the INTEGER, so this will be highlight%28x=2%29 meaning that highlight%28y=7%29.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
1. Find two integers whose product is 14 such that one of the integers is three less than five times the other integer.
2. The perimeter of a rectangle is 50 inches, and the area is 136 square inches. Find the length and width of the rectangle.

3. 9x3 + x2 + 7 from 5x3 − x − 8 
No. 1 can be answered without algebra, since 14 has only 2 PAIRS of factors. They are: 14 & 1, and 7 and 2. 

Now use the 2nd clue to determine which pair pertains to this problem.

By the way, last time I checked, 2 times - 7 DID NOT EQUAL 14, and I don't believe this has changed.