# Question cb146

Jan 11, 2018

11 and 1

#### Explanation:

Well, if $a + b = 12$ and
$a - b = 10$, then you have 2 equations in 2 unknowns, which, if they are linearly independent and there is a real solution, you can algebraically solve for.

We'll use the substitution method.

Take the second equation, $a - b = 10$. You can add 'b' to both sides, giving you an equivalent definition for a, that you can substitute in the other equation.

So, $a = b + 10$
...substitute this into the first equation:

$\left(b + 10\right) + b = 12$

$2 b + 10 = 12$
...subtract 10 from both sides:

$2 b = 2$

$b = 1$

now, substitute this back into either of the equations:

$a + 1 = 12$
$a = 11$

GOOD LUCK

Jan 12, 2018

$\text{The numbers are 11 and 1}$

#### Explanation:

There are a couple of ways to solve this problem.

You can solve it by logic

List all the pairs of numbers that add to 12 and find the pair whose difference is 10.

$11 + 1$ $\leftarrow$ Here it is on the first try.
$10 + 2$
$\textcolor{w h i t e}{.}$ 9 + 3
$\textcolor{w h i t e}{.}$etc

$11 + 1 = 12$
$11 - 1 = 10$

$\text{The numbers are 11 and 1}$
..................................

You can solve it by math

Let $a$ and $b$ represent the two numbers

$\textcolor{w h i t e}{,}$$a + b = 12$
$\textcolor{w h i t e}{.}$$a - b = 10$
.......................
$2 a \textcolor{w h i t e}{\ldots . .} = 22$

$a = 11$, so $b$ must be $1$

color(white)(.)a  +  b = 12
 11  +  1 = 12

color(white)(.)a -  b = 10
11 -  1 = 10#

$\text{The numbers are 11 and 1}$ $\leftarrow$ same answer