One integer is 9 less than twice another. If their product is 56, find the two integers. I have no idea how to even start, any help would be appreciated. Thanks
2007-10-06 18:30:13 · 5 answers · asked by ErikaLuvsTim 2 in Science & Mathematics ➔ Mathematics
2x+9=y
x*y=56
simplest thing to do is find the factors of 56 and see if they satisfy the first eqn.
2007-10-06 18:38:12 · answer #1 · answered by Anonymous · 0⤊ 0⤋
let x=the first number and y=the second number
2x-9=y (one integer is 9 less than the twice the other)
xy=56 (their product is 56)
substitute 2x-9 to the y in the second equation.
x(2x-9)=56
2x^2 - 9x = 56
2x^2 - 9x - 56 = 0
(2x+7)(x-8)=0
so either 2x+7=0 or x-8=0
so x=-7/2 or x=8
since you are asking for integers, -7/2 cannot be an answer! :)
so the two integers are 8 and 7 (2*8-9)..!
hope this helps..!! =]
2007-10-07 01:49:15 · answer #2 · answered by wheeeeeeee..! 2 · 0⤊ 0⤋
one integer is 9 less than twice another
you would set it up as 2x-9 if there is two integers that means you are using the 2x also and if it has a product that means the 56 comes from multiplication so
2x(2x-9)=56
you foil so you have
4x^2-18x=56
I would than divide by 2
2x^2-9x=28
then expand by removing the x's
x(x-9)=28
so x=28 and x-9=28, you divide and x=28/9
2007-10-07 01:46:42 · answer #3 · answered by Tom L 3 · 0⤊ 0⤋
let integers be x and y
y = 2x - 9
x y = 56
x (2x - 9) = 56
2x² - 9x - 56 = 0
(2x + 7) (x - 8) = 0
x = 8 is acceptable answer
y = 7
Integers are 7 and 8
2007-10-07 03:30:52 · answer #4 · answered by Como 7 · 1⤊ 0⤋
set up a system of equations.
u know one on top of another and figure it out, its not that hard.
good luck, have fun
2007-10-07 01:38:38 · answer #5 · answered by SUPERMAN 4 · 0⤊ 0⤋