The sum of two numbers is 37. Their product is 312. What are the numbers?

2 Answers
Aug 6, 2018

x = 13, y = 24 and x = 24, y = 13

Explanation:

Let the numbers be represented by x and y

The sum of two numbers is 37

x + y = 37

Their product is 312

x xx y = 312

xy = 312

Solving simultaneously;

x + y = 37 - - - eqn1

xy = 312 - - - eqn2

From eqn2

xy = 312

Making x the subject formula;

(xy)/y = 312/y

(xcancely)/cancely = 312/y

x = 312/y - - - eqn3

Substitute eqn3 into eqn1

x + y = 37

(312/y) + y = 37

Multiply through by y

y(312/y) + y(y) = y(37)

cancely(312/cancely) + y^2 = 37y

312 + y^2 = 37y

y^2 - 37y + 312 = 0

Solving the Quadratic Equation..

y^2 - 37y + 312 = 0

Using Factorization Method

The factors are, -13 and -24

- 37y = -13y - 24y

312 = -13 xx - 24

Therefore;

y^2 - 13y - 24y + 312 = 0

By Grouping;

(y^2 - 13y) (- 24y + 312) = 0

Factorizing;

y(y - 13) -24 (y - 13) = 0

(y - 13) (y - 24) = 0

y - 13 = 0 or y - 24 = 0

y = 13 or y = 24

Substituting the values of y into eqn3

x = 312/y

When, y = 13

x = 312/13

x = 24

Similarly when, y = 24

x = 312/24

x = 13

Hence;

x = 13, y = 24 and x = 24, y = 13

Aug 6, 2018

The two numbers are : 13 and 24

Explanation:

Let x and y , (x < y ) be the two numbers,such that

sum =x+y=37=>y=37-xto (1)

and product x*y=312...to(2)

Subst. y=37-x into (2)

:.x(37-x)=312

:.37x-x^2=312

:.x^2-37x+312=0

Now ,

(-24)+(-13)=-37 and (-24)xx(-13)=312

:.x^2-24x-13x+312=0

:.x(x-24)-13(x-24)=0

:.(x-24)(x-13)=0

:.x-24=0 or x-13=0

:.x=24 or x=13

So, from (1)

y=13 or y=24

Hence the two numbers are : 13 and 24