Math problem help?

Question

The larger of two positive integers is five more than twice the smaller integer. The product of the integers is 52. What are the two integers?

Answer 1

x = 5+ 2y
xy = 52
y = 52/x
x = 5 + 104/x
x^2 = 5x + 104
x^2 -5x -104 = 0
( x+8) ( x-13)=0
x = 13 and y = 4

Answer 2

Equations: y = 5 + 2x and xy = 52

*Since "y" is isolated in the 1st equation > replace 5+2x with the "y" variable in the 2nd equation >

x(5 + 2x) = 52
x(5) + x(2x) = 52
5x + 2x^2 = 52

First: set the equation to "0" > subtract 52 from both sides >

5x + 2x^2 - 52 = 52 - 52
5x + 2x^2 - 52 = 0

*Rewrite the equation in alpha descending order ("x" variables first) > 2x^2 + 5x - 52 = 0

Sec: factor: Multiply the 1st and last coefficient to get -104. Find two numbers that give you -104 when multiplied and 5 (middle coefficient) when added/subtracted. The numbers are: 13, -8

*Rewrite the equation with the new middle numbers >
2x^2 + 13x - 8x - 52 = 0

*Group "like" terms and factor >

(2x^2 + 13x) - (8x - 52) = 0
x(2x + 13) - 4(2x + 13) = 0
(2x+13)(x - 4) = 0

Third: solve for the "x" variables (place both sets to equal "0") >

2x + 13 = 0
2x + 13 - 13 = 0 - 13
2x = -13
2x/2 = -13/2
x = -13/2

x - 4 = 0
x - 4 + 4 = 0 + 4
x = 4

*When you check the numbers in either equation, 4 is the only answer. Replace 4 with the "x" variable in the 2nd equation.

4y = 52
4y/4 = 52/4
y = 13

(4, 13)

Answer 3

2s + 5 = l
s * l = 52

l = 52/s
s = 52/l

104/l + 5 = l

l^2 - 5l - 104 = 0
(l + 8)(l - 13) = 0

l = 13

s = 52/l = 4

Answer 4

52=26*2=13*4. Choose between the two.

Answer 5

x times (5x+5)= 52

Answer 6

x=larger
y=smaller
x=2y+5
xy=52
(2y+5)y=52
2y^2+5y-52=0
(2y+13)(y-4)=0
since y>0
y-4=0
y=4
x=2y+5=2*4+5=13
x=13, y=4

Answer 7

the two integers would be 13, and 4. good luck!!! :)