stuck on this word problem with percentages

[drummer8622]

Word problem:
Pat invested a total of 3000 dollars in two parts, one part earns 10% interest, one part earns 8% interest. At the end of the year he had earned a total of 256 dollars in interest, how much money was invested at each interest rate.

So far i only got to x+y=3000, and .08x + .1y = 256. from there i have tried substitution but maybe i am just doing something wrong because of the interests
below is a sum of one attempt
.08x +.1y = 256 -> divide both sides by .1
.08x +y = 2560
rearrange
y= 2560 -.08x
substitute
.08x + .1(2560 + 0.08x) =256
0.08x + 256 +.008x = 256
256 + .088x = 256
.088x = 0

thanks

 

[wjm11]

Word problem:
Pat invested a total of 3000 dollars in two parts, one part earns 10% interest, one part earns 8% interest. At the end of the year he had earned a total of 256 dollars in interest, how much money was invested at each interest rate.

So far i only got to x+y=3000, and .08x + .1y = 256. from there i have tried substitution but maybe i am just doing something wrong because of the interests
below is a sum of one attempt
.08x +.1y = 256 -> divide both sides by .1
.08x +y = 2560
rearrange
y= 2560 -.08x
substitute
.08x + .1(2560 + 0.08x) =256
0.08x + 256 +.008x = 256
256 + .088x = 256
.088x = 0

thanks

Good job setting up your two equations. You just have an algebra error later on. When you divide both sides by .1 (same thing as multiplying by 10), you forgot to divide .08x by .1.

Your approach is fine. Give it another try.

 

[Mrspi]

Your two equations were fine. It would have been easier to solve that first equation for y:

x + y = 3000
Subtract x from both sides, and you have
x + y - x = 3000 - x
y = 3000 - x

You can now substitute (3000 - x) for y in the second equation.

Generally speaking, it is a good idea to pick the equation that is EASIEST to solve for one of the variable, solve for that variable, and then substitute the resulting expression into the other equation.