Interest Equation?

Question

How can I set up an equation for this and then solve it. I need to show my work.

Tim invested $6000 for one year. He invested part of it at 9% and the rest at 11%. At the end of the year he earned $624 in interest. How much did he invest at each rate?

I don't understand Raj how or why you would multiply by 100 ?

Answer 1

let the amount invested at 9% be x
amount invested at 11% will be 6000-x
the equation
x*0.09+(6000-x)0.11=624
multiplying by 100
9x+66000-11x=62400
-2x=-3600
x=1800
so heinvested $1800 @9% and $4200 @11%

Answer 2

being x the sum of money he invested at 9% and y the sum of money he invested at 11% then
x + y = 6000 (the sum of both inversions is 6000)
0.09x + 0.11y = 624 (the sum of the interests he earned is 624)

now expressing those equations as matrices is easy to solve them as Ax = b, solving the equation set as x = A^-1*b gives x= 1800 and y = 4200 (He put 1800 at 9% and 4200 at 11%)

[1_____1] [x] = [6000]
[0.09 0.11] [y] _ [_624]

[x] = [5.5 -50] [6000]
[y] _ [-4.5 50] [_624]

*Note: the _ is a space keeper

Answer 3

Raj multiplied by 100 to simplify the equation. Recall that you can multiply an equation by a constant and it remains an 'equivalent' equation. Equivalent means that it has the same solution.

a+b = 6000

9a + 11b = 62400

are the two equations to solve.