# Question #9133b

##### 3 Answers
Nov 30, 2017

The two numbers are $10 \mathmr{and} 14$.

#### Explanation:

Given that: the sum of two numbers is 24

If one number is assumed to be $x$, then the other number will be $24 - x$

Also given that the product of the two numbers is 140 :

$\implies x \times \left(24 - x\right) = 140$

$\implies 24 x - {x}^{2} = 140$

$\implies - {x}^{2} + 24 x - 140 = 0$

$\implies {x}^{2} - 24 x + 140 = 0$-------is a quadratic equation.

To solve it, we need two such numbers, which sum up to give the coefficient of the middle term, i.e.$- 24$ and their product must be equal to the product of coefficients of first term and last term, i.e. $1 \times 140 = 140$.

Two such numbers are$- 14 \mathmr{and} - 10$

$\implies {x}^{2} - 14 x - 10 x + 140 = 0$

$\implies x \left(x - 14\right) - 10 \left(x - 14\right) = 0$

$\implies \left(x - 14\right) \left(x - 10\right) = 0$

$\implies x = 14 \mathmr{and} x = 10$

If we take $x = 14$ the other number is $24 - x = 24 - 10$
and
If we take $x = 10$ then the other number is $24 - 10 = 14$

That means, the two numbers are $10 \mathmr{and} 14$.

Nov 30, 2017

Formulate the problem

#### Explanation:

These kinds of problems are easy to solve when you understand how to formulate them.

Let the numbers be $x$ and $y$. Then you have

$x + y = 24$
$x y = 140$

You can solve these in a couple of ways.

Method 1.

${\left(x - y\right)}^{2} = {\left(x + y\right)}^{2} - 4 x y$

Then you have

${\left(x - y\right)}^{2} = {24}^{2} - 4 \setminus \times 140$
${\left(x - y\right)}^{2} = 576 - 560$
${\left(x - y\right)}^{2} = 16$
$\left(x - y\right) = \setminus \pm 4$

Now you have 2 scenarios. $x - y = 4$, $x - y = - 4$. Based on which of the equation is suitable for you, you can solve either of these and get a solution.

Solve $x + y = 24$ and $x - y = 4$ then you get $x = 14 , y = 10$
Solve $x + y = 24$ and $x - y = - 4$ then you get $x = 10 , y = 14$

Method 2

Substitute and calculate.

$x + y = 24$ then $y = 24 - x$. Substitute in the other equation

$x y = 140$
$x \left(24 - x\right) = 140$
$- {x}^{2} + 24 x = 140$
${x}^{2} - 24 x + 140 = 0$
${x}^{2} - 14 x - 10 x + 140 = 0$
$x \left(x - 14\right) - 10 \left(x - 14\right) = 0$
$\left(x - 14\right) \left(x - 10\right) = 0$
$x = 14 , 10$

Then you will get $y = 10 , 14$ after substituting back in $y = 24 - x$.

Nov 30, 2017

10 and 14

#### Explanation:

There must be two equation to solve for two variables.

Let x equal one number
Let y equal the other number

One equation is

$x + y = 24$

The other equation is

$x \times y = 140$

Solve the first equation for y

$x + y = 24$ subtract x from both sides

$x - x + y = 24 - x$ this gives

$y = 24 - x$

Substitute this value into the second equation.

$\left(24 - x\right) \times x = 140$ distribute the x across the parenthesis

$24 x - {x}^{2} = 140$ Subtract 24 x and add x^2

$24 x - 24 x - {x}^{2} + {x}^{2} = 140 - 24 x + {x}^{2}$ This results in

${x}^{2} - 24 x + 140 = 0$ Factor this trinomial into binomials

$\left(x - 10\right) \times \left(x - 14\right) = 0$

Solve for x in both binomials

$x - 10 = 0$ add 10 to both sides

$x - 10 + 10 = 0 + 10$ which gives

$x = 10$

$x - 14 = 0$ add both 14 to both sides

$x - 14 + 14 = 0 + 14$ which gives

$x = 14$

The two numbers are 10 and 14