Last year, Evelin's invested 10,000 dollars, part at 6% annual interest, and the rest at 8% annual interest. If she recieves $760 at the end of the year, how much did he invest at each rate?

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# 2 Answers

X - The part invests in 6%

10,000 - X -- The rest invests in 8 %

X ( 0.06) --gain in 6% investment

(10,000 - X ) 0.08 -- gain on 8%

Solve for X , if:

X ( 0.06 ) + ( 10,000 - X ) ( 0.08) = 760

X ( 0.06) - X ( 0.08 ) = -10,000( 0.08) + 760

X ( - 0.02 ) = - -800 + 760 = -40

X = -40/ -0.02 = 2,000 / invested in 6% Plan

10,000 - 2,000 = 8,000 / invested in 8% plan

This can be solved with a system of equations. Let x be the amount invested at 6% and let y be the amount invested at 8%.

You know that the total invested is 10,000. This yields one equation:

x + y = 10,000

You also know that the interest earned on x plus the interest earned on y equals $760. The interest earned on x is 0.06x and the interest earned on y is 0.08y. Now you have a second equation:

0.06x + 0.08y = 760

There are a variety of ways to solve this system. I will use elimination and back substitution. I will multiply the first equation by -0.06, then when I add that to the second equation I will eliminate the x variable.

-0.06x - 0.06y = -600

0.06x + 0.08y = 760

.02y = 160

y = 8000

Now I will substitute 8000 for y in the original first equation and solve for x.

x + 8000 = 10,000

x = 2000

Evelin invested $2000 at 6% and $8000 at 8%.