One number is 44 more than twice another. Their product is 44 more than twice their sum. Find the numbers x and y. (Assume x

Question

One number is 44 more than twice another. Their product is 44 more than twice their sum. Find the numbers x and y. (Assume x < y.)

anonymous · Answer

attachment

amistre64 · Answer

m = 44+2n

mn = 44+2(m+n)

anonymous · Answer

y=44+2x
y=2(x+y)+44

Solve system using substitution or elimination

amistre64 · Answer

mn = 44 +2m+2n  ; yet 2n = m-44

mn = 44 +2m +2m -88

mn = -44 +4m

44 = mn +4m

44 = m(n+4)

m = 44/(n+4)  ; n!= -4

amistre64 · Answer

m = 44+2n = 44/(n+4)

(44+2n)(n+4) = 44

44n +176 +2n^2 +8n = 44

2n^2 +52n +176-44 = 0

2n^2 +52n +132  = 0

amistre64 · Answer

2(n^2 +26n + 66) = 0

n^2 +26 = -66

(n+13) = sqrt(-66 + 169)

n = -13 +- sqrt(103)

amistre64 · Answer

htats prolly wrong lol

anonymous · Answer

it is. i know there's no square root for any of the problems. this one is killing me! haha

anonymous · Answer

one of the numbers is 0.

amistre64 · Answer

m = 44+2n
mn = 44+2(m+n)

these are good equations right?

amistre64 · Answer

if one of the numbers is 0; then their product is 0

anonymous · Answer

y=44+2x
y=2(x+y)+44 

The solution to this system is 
x=-22
y=0

amistre64 · Answer

how is y a product in that?

amistre64 · Answer

-22 is a good solution, I just cant see how you interpret the information..

anonymous · Answer

we'll skip that one. lol