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

Aug 6, 2018

$x = 13 , y = 24 \mathmr{and} x = 24 , y = 13$

Explanation:

Let the numbers be represented by $x \mathmr{and} y$

The sum of two numbers is $37$

$x + y = 37$

Their product is $312$

$x \times y = 312$

$x y = 312$

Solving simultaneously;

$x + y = 37 - - - e q n 1$

$x y = 312 - - - e q n 2$

From $e q n 2$

$x y = 312$

Making $x$ the subject formula;

$\frac{x y}{y} = \frac{312}{y}$

$\frac{x \cancel{y}}{\cancel{y}} = \frac{312}{y}$

$x = \frac{312}{y} - - - e q n 3$

Substitute $e q n 3$ into $e q n 1$

$x + y = 37$

$\left(\frac{312}{y}\right) + y = 37$

Multiply through by $y$

$y \left(\frac{312}{y}\right) + y \left(y\right) = y \left(37\right)$

$\cancel{y} \left(\frac{312}{\cancel{y}}\right) + {y}^{2} = 37 y$

$312 + {y}^{2} = 37 y$

${y}^{2} - 37 y + 312 = 0$

${y}^{2} - 37 y + 312 = 0$

Using Factorization Method

The factors are, $- 13 \mathmr{and} - 24$

$- 37 y = - 13 y - 24 y$

$312 = - 13 \times - 24$

Therefore;

${y}^{2} - 13 y - 24 y + 312 = 0$

By Grouping;

$\left({y}^{2} - 13 y\right) \left(- 24 y + 312\right) = 0$

Factorizing;

$y \left(y - 13\right) - 24 \left(y - 13\right) = 0$

$\left(y - 13\right) \left(y - 24\right) = 0$

$y - 13 = 0 \mathmr{and} y - 24 = 0$

$y = 13 \mathmr{and} y = 24$

Substituting the values of $y$ into $e q n 3$

$x = \frac{312}{y}$

When, $y = 13$

$x = \frac{312}{13}$

$x = 24$

Similarly when, $y = 24$

$x = \frac{312}{24}$

$x = 13$

Hence;

$x = 13 , y = 24 \mathmr{and} x = 24 , y = 13$

Aug 6, 2018

The two numbers are : 13 and 24

Explanation:

Let $x \mathmr{and} y , \left(x < y\right)$ be the two numbers,such that

sum =$x + y = 37 \implies y = 37 - x \to \left(1\right)$

and product $x \cdot y = 312. . . \to \left(2\right)$

Subst. $y = 37 - x$ into $\left(2\right)$

$\therefore x \left(37 - x\right) = 312$

$\therefore 37 x - {x}^{2} = 312$

$\therefore {x}^{2} - 37 x + 312 = 0$

Now ,

$\left(- 24\right) + \left(- 13\right) = - 37 \mathmr{and} \left(- 24\right) \times \left(- 13\right) = 312$

$\therefore {x}^{2} - 24 x - 13 x + 312 = 0$

$\therefore x \left(x - 24\right) - 13 \left(x - 24\right) = 0$

$\therefore \left(x - 24\right) \left(x - 13\right) = 0$

$\therefore x - 24 = 0 \mathmr{and} x - 13 = 0$

$\therefore x = 24$ $\mathmr{and} x = 13$

So, from $\left(1\right)$

$y = 13 \mathmr{and} y = 24$

Hence the two numbers are : 13 and 24