I do not know how to get the following answer

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

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- Aug 28th 2008, 06:55 PMmcdanielnc89Applications and problem solving@
I do not know how to get the following answer

reloading image.

The answer that is tehre i think is the right one, but i have no clue... - Aug 28th 2008, 06:58 PMChris L T521
You have the system $\displaystyle \left\{\begin{array}{lcr}x-y&=&6\\x+y&=&10\end{array}\right.$

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 $\displaystyle 2x=16\implies x=\dots$

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

Can you take it from here?

--Chris - Aug 28th 2008, 07:02 PMmcdanielnc89
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...

http://img222.imageshack.us/img222/2918/sssssses3.png - Aug 28th 2008, 07:09 PMChris L T521
Determine the equation that represents the area of the picture:

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

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

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

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

Factor out the left side, and then the equation becomes

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

Solve this quadratic equation.

Can you take it from here?

--Chris - Aug 28th 2008, 07:12 PMmcdanielnc89
Thank you so much Chris!!!! the answer is 5.3