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Q:
Mike invested $7000 for one year. He invested part of it at 8% and the rest at 12%. At the end of the year he earned $764 in interest. How much did he invest at each rate?
A:
Let
x = amount of money Mike invests at 8%
y = amount of money Mike invests at 12%
Since "Mike invested $7000 for one year.", this means that
Also, because "He invested part of it at 8% and the rest at 12%. At the end of the year he earned $764 in interest.", this translates to

. Multiplying every term by 100 gets us
So we have the system of equations:
Let's solve this system by substitution
Now in order to solve this system by using substitution, we need to solve (or isolate) one variable. I'm going to solve for y.
So let's isolate y in the first equation

Start with the first equation

Subtract

from both sides

Rearrange the equation
---------------------
Since

, we can now replace each

in the second equation with

to solve for

Plug in

into the second equation. In other words, replace each

with

. Notice we've eliminated the

variables. So we now have a simple equation with one unknown.

Distribute

to

Multiply

Combine like terms on the left side

Subtract 84000 from both sides

Combine like terms on the right side

Divide both sides by -4 to isolate x

Divide
So this means that Mike invested $1,900 at 8%
-----------------------------
Since we know that

we can plug it into the equation

(remember we previously solved for

in the first equation).

Start with the equation where

was previously isolated.

Plug in

Combine like terms
So Mike invested $5,100 at 12%
===============================================
Answer:
So Mike invested $1,900 at 8% and $5,100 at 12%