Question

Money is invested at two rates of interest. One rate is 8%

and the other is 2%
. If there is $600
more invested at 8%
than at 2%
, find the amount invested at each rate if the total annual interest received is $470
. Let x=
amount invested at 8%
and y=
amount invested at 2%
. Then the system that models the problem is {x=y+6000.08x+0.02y=470
. Solve the system by using the method of addition.

Answer 1

• at 8% (x) = $4820
• at 2% (y) = $4220

Explanation:

You want the solution to the system of equation using the addition method:

• x = y + 600
• 0.08x + 0.02y = 470

Addition

We can multiply the first equation by 0.02 and add it to the second:

0.02(x) +(0.08x +0.02y) = 0.02(y +600) +(470)

0.10x +0.02y = 0.02y +482 . . . . . . . . . simplify

0.10x = 482 . . . . . . . . . . . . . . . . . . . . subtract 0.02y

x = 4820 . . . . . . . . . . . . . . . . . . . . multiply by 10

y = x -600 = 4820 -600 = 4220

The amount invested at 8% is $4820; the amount invested at 2% is $4220.

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Additional comment

You will notice that we solved the equations using "addition" by adjusting coefficients so one variable (y) had the same coefficient on each side of the equal sign. That term was then subtracted from both sides, eliminating the variable. The point of the "method of addition" is to eliminate one variable, which is what we did.

More conventionally, we might rearrange the first equation to x - y = 600 before multiplying by 0.02 and doing the addition. That also eliminates the y-variable.

The calculator result shown in the attachment is also a form of elimination by the addition method.