# Peter invested some money at 6% annual interest, and Martha invested some at 12%. If their combined investment was $6,000 and their combined interest was$450, how much money did Martha invest?

Dec 12, 2016

Peter invested $.4500 Martha invested $.1500

#### Explanation:

Peter invested $.x Martha invested $.y

Interest from $.x= x xx 6/100=(6x)/100 Interest from $.y= y xx 12/100=(12y)/100

Then -

$\frac{6 x}{100} + \frac{12 y}{100} = 450$

To do away with fraction, let us multiply both sides by 100

$6 x + 12 y = 45000$----------(1)

$x + y = 6000$-----------------(2)

Let us solve the 2nd equation for $x$

$x = 6000 - y$

Plug in the value of $x = 6000 - y$ in equation (1)

$6 \left(6000 - y\right) + 12 y = 45000$
$36000 - 6 y + 12 y = 45000$
$6 y = 45000 - 36000 = 9000$
$y = \frac{9000}{6} = 1500$

Substitute $y = 1500$ in equation (2) and simplify

$x + 1500 = 6000$
$x = 6000 - 1500 = 4500$

Peter invested $.4500 Martha invested $.1500