The sum of two number is 13, and their product is 36. What are the numbers?

Answer 1

let X = 1st number
let y = 2nd number

Equation 1: x + y = 13 (sum of the two numbers is 13)
Equation 2: xy = 36 (product of the two numbers is 36)

Solve for x in the 2nd equation and replace the answer into the 1st equation, wherever x is.

xy = 36
x = 36/y

x + y = 13
36/y + y = 13 (multiply the "y" by y/y)
36/y + (y^2)/y = 13
(y^2 + 36)/y = 13 (cross-multiply)
13y = y^2 + 36
y^2 - 13y + 36 = 0 (factorize)
(y - 4)(y - 9) = 0 (set each factor to 0)
y - 4 = 0 and y - 9 = 0
y = 4 and y = 9

Put these values into one of the previous equations (either equation 1 or 2) Lets say equation 1:

x + y = 13, when y = 4:
x + 4 = 13
x = 9

x + y = 13, when y = 9:
x + 9 = 13
x = 4

So the answers will be as follows:
When x = 9, y = 4
When y = 9, x = 4

Check: 9 + 4 = 13 and 9 x 4 = 36 so the answer is correct.

These are ur two sets of solutions. Hope this helps! =)

Answer 2

Number 1 = A
Number 2 = B

A + B = 13
A x B = 36

A + B = 13, so A = 13 - B

A x B = 36, since A = 13 - B, so
(13 - B ) x B = 36

13 B - B2 (B square) = 36, or

B2 - 13 B + 36 = 0
(B - 9) x (B - 4) = 0

B - 9 = 0, so B = 9
or
B - 4 = 0, so B = 4

if B = 9, A + B = 13, then A = 4
if B = 4, A + B = 13, then A = 9

Answer 3

9 and 4

Answer 4

9 and 4

Answer 5

9 and 4

Answer 6

9 and 4

Answer 7

4 x 9 = 36
4 + 9 = 13

Answer 8

Two Number

Answer 9

Let x be one number.
Let y be the other number.

x+y =13
=> y = 13-x ---- Eq1

x*y = 36 ---- Eq2

Sub Eq1 into Eq2:
x * (13-x) = 36
13x -x^2 = 36
x^2 -13x +36 = 0
(x-4)(x-9) = 0
x = 4 or x = 9

sub x=4 into Eq1:
y = 13-4 = 9

or
sub x=9 into Eq2:
y = 13-9 = 4

Answer 10

Let numbers be x and y
x + y = 13
x.y = 36
x = 36/y
36/y + y = 13
36 + y² = 13y
y² - 13y + 36 = 0
(y - 9).(y - 4) = 0
y = 9,y = 4
When y = 9,x = 4
When y = 4,x = 9
The numbers are 4 and 9