1. System of Equations

Solve the following system of equation for y.

2x - y = 13

2x = y

I came up with no solution for y.

The book's answer is y = -39.

Who's right?

2. It would appear that, in this case, you are right.

Because if $\displaystyle 2x=y$ and then we substitute for $\displaystyle y$ in the othe equation we get $\displaystyle 2x-2x=13$ I don't think so! because $\displaystyle 2x-2x=0\neq13$

Are you sure you copied the problem down right?

3. You could also substitute $\displaystyle -39$ for $\displaystyle y$wherever it occurs in both equations. You will find that
$\displaystyle 2x-(-39)=1$
$\displaystyle 2x+39=13$
$\displaystyle 2x=-26$
$\displaystyle x=-13$

and now for the other equation

$\displaystyle 2x=-39$
$\displaystyle x=\frac{-39}{2}$

and since the coordinates $\displaystyle (\frac{-39}{2},-39)\neq(-13,-39)$there is no solution.

So, you got a bad book baby!

4. yes...

Originally Posted by VonNemo19
It would appear that, in this case, you are right.

Because if $\displaystyle 2x=y$ and then we substitute for $\displaystyle y$ in the othe equation we get $\displaystyle 2x-2x=13$ I don't think so! because $\displaystyle 2x-2x=0\neq13$

Are you sure you copied the problem down right?
I thank you for the quick reply. Yes, the problem is copied down right. It is probably a typo.

5. ok...I see...

Originally Posted by VonNemo19
You could also substitute $\displaystyle -39$ for $\displaystyle y$wherever it occurs in both equations. You will find that
$\displaystyle 2x-(-39)=1$
$\displaystyle 2x+39=13$
$\displaystyle 2x=-26$
$\displaystyle x=-13$

and now for the other equation

$\displaystyle 2x=-39$
$\displaystyle x=\frac{-39}{2}$

and since the coordinates $\displaystyle (\frac{-39}{2},-39)\neq(-13,-39)$there is no solution.

So, you got a bad book baby!
The book is Math A by Lawrence S. Leff. I have found many typos in his book. He is also the author of Math B. There are WAY TOO MANY errors in his books. If I can spot them, anyone can.