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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.

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

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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?

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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 $$ \eqalign{ .05 w-.03 y&=-10\cr 2w-y&=1000. } $$

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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

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