Solve for x, y and 4x in 4x+x+y=280^0

[-Physician]

Homework Statement

We have $4x+x+y=280^0$(degrees), find $x$, $y$ and $4x$.

Homework Equations

$4x+x+y=280^0$

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!

 

[tiny-tim]

Hi -Physician!

erm …

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

 

[-Physician]

So the $y=0$?

 

[Dadface]

So the $y=0$?

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.

 

[-Physician]

This is what the book says. Not my own tasks

 

[Dadface]

Look at tiny-tims post above.He went with the method you started and corrected your mistakes.He expressed x and 4x in terms of a number and y.Alternatively and just as good,y can be expressed in terms of x and a number.Without extra information that's the best that can be done.

 

[-Physician]

I see, thank you.