# Algebra system of equation?

How would I solve this system of equation?

\begin{align*}&.05(w+2000)=.03(y+3000)\\ &4w=\frac{y}2+500 \end{align*}

I end up setting them up like this but I am not sure if it is correct.

\begin{align*}&5w-3y=-10\\ &-2w-y=500 \end{align*}

-

The first one expands to $0.05w+100=0.03y+90$, which you can rewrite as $0.05w-0.03y=-10$; if you want to get rid of the decimals, you’ll need to multiply both sides by $100$, and you’ll get $5w-3y=-1000$. You made a similar error in manipulating the second equation: if you multiply both sides by $2$ to get rid of the fraction, you should have $2w=y+1000$. You forgot to multiply the $500$ by $2$. You also made a sign error in getting both variables on the same side of the equation. Can you see now what it ought to be?

-

It's off a bit.

For the first equation, $$.05(w+2000) = .03(y+3000)$$ do the multiplications first, using the distributive law: $$.05 w +100 =.03 y +90.$$ Rewrite this a bit to get $$\tag{1} .05 w-.03 y=-10.$$

For the second equation $$w={y\over 2}+500,$$ multiply both sides by 2 (so multiply each term by 2) to get $$2w=y+1000.$$ Rewrite this a bit to get $$\tag{2} 2w-y=1000.$$

So, your system becomes \eqalign{ .05 w-.03 y&=-10\cr 2w-y&=1000. }

-
i got w=4000 I think that is correct thanks for the help – El Cholo May 21 '12 at 23:34
@ElCholo That is correct. $w=4000$ (and $y=7000$). – David Mitra May 21 '12 at 23:38