# Thread: Applications and problem solving@

1. ## Applications and problem solving@

I do not know how to get the following answer

The answer that is tehre i think is the right one, but i have no clue...

2. Originally Posted by mcdanielnc89
I do not know how to get the following answer

The answer that is tehre i think is the right one, but i have no clue...
You have the system $\left\{\begin{array}{lcr}x-y&=&6\\x+y&=&10\end{array}\right.$

Use the method of elimination to solve the system.

Adding the two equations together, we see that the y term canceled out.

So we get $2x=16\implies x=\dots$

Then substitute this value of x into either equation in the system to find y.

Can you take it from here?

--Chris

3. I'm terribly sorry. I'm so ticked i put the wrong one. the electricity is ging on and off.. hehres the one i don't know...

4. Originally Posted by mcdanielnc89
I'm terribly sorry. I'm so ticked i put the wrong one. the electricity is ging on and off.. hehres the one i don't know...

Determine the equation that represents the area of the picture:

The dimensions are given to you: $(39-2x)~cm$ by $(26-2x)~cm$

So we see that area is $(39-2x)(26-2x)~cm^2$

However, the area has a value of $440~cm^2$

Thus, $(39-2x)(26-2x)=440$

Factor out the left side, and then the equation becomes

$4x^2-130x+1014=440\implies4x^2-130x+574=0$