1. Feb 14, 2012

-Physician

1. The problem statement, all variables and given/known data
We have $4x+x+y=280^0$(degrees), find $x$, $y$ and $4x$.

2. Relevant equations
$4x+x+y=280^0$

3. The attempt at a solution
$4x+x+y=280^0$
$5x+y = 280^0$
$5x=280^0-y$
$x=\frac{280^0-y}{5}$
$x=56^0-y$
---------------------
$4x=4(56^0-y)=224^0-y$
------------------------------------
$(224^0-y+56^0-y)+y=280^0$
$y=280^0-(224^0-y+56^0-y)$
$y=280^0-280^0-y$
$y=0$
Is that correct? thanks!

2. Feb 14, 2012

tiny-tim

Hi -Physician!

erm
$x=\frac{280^0-y}{5}$
$x=56^0-y/5$
---------------------
$4x=4(56^0-y/5)=224^0-4y/5$
------------------------------------
$(224^0-y/5+56^0-4y/5)+y=280^0$
$280^0=280^0$​

3. Feb 14, 2012

-Physician

So the $y=0$?

4. Feb 14, 2012

No ,y could be zero but that is just one out of many possible answers.The question as you presented it does not have enough information to get numerical values for x and y.

5. Feb 14, 2012

-Physician

This is what the book says. Not my own tasks

6. Feb 14, 2012