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