Question 0cec2

Dec 12, 2017

$15 \text{ and } 3$

Explanation:

"let the 2 numbers be "x" and "ycolor(white)(x);x>y#

$\text{then } x + y = 18 \to \left(1\right)$

$\text{and } x - y = 12 \to \left(2\right)$

$\text{adding the equations together will eliminate y}$

$\text{adding term by term on both sides}$

$\left(x + x\right) + \left(y - y\right) = \left(18 + 12\right)$

$\Rightarrow 2 x = 30$

$\text{divide both sides by 2}$

$\frac{\cancel{2} x}{\cancel{2}} = \frac{30}{2}$

$\Rightarrow x = 15$

$\text{substitute this value into equation } \left(1\right)$

$\Rightarrow 15 + y = 18 \Rightarrow y = 18 - 15 = 3$

$\text{the numbers are "15" and } 3$

$15 + 3 = 18 \text{ and } 15 - 3 = 12$

Dec 12, 2017

$x = 15$
$y = 3$

Explanation:

Alright, lets think of this problem as two different equations: $x + y = 18$ and $x - y = 12$. Your trying to solve for $x$ and $y$.

We need to get rid of one of those variables. Since there is a positive and a negative $y$, all we have to do is add both equations!

$x + y = 18$
$x - y = 12$

$2 x = 30$

$x = 15$

Now we just plug this $x$ value into one of the original equations to get $y$.

$15 + y = 18$
$y = 3$

Now to check, we are going to plug the $x$ and $y$ into $x - y = 12$ to make sure out answer is correct.

$15 - 3 = 12$

$12 = 12$

Hope this helped!
~Chandler Dowd