Question 152487: Hi,
How would I solve the following: Applications of Linear Equations
Walt made an extra $9000 last year from a part-time job. He invested part of the money at 9% and the rest at 8%. He made a total of $770 in interest. How much was invested at 8%?
Thank you for taking the time to help me.
Sally
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! G'day Sally;
Let x = the amount invested at 8% and the remainder ($9000-x) is invested at 9%
The amount of interest earned on these amounts can be expressed as:
x(0.08) This is amount earned at 8%.
($9000-x)(0.09) This is the amount earned at 9%
The sum (+) of these two amounts is given as $770.00, so you can write the equation to solve for x, the amount invested at 8%
(0.08)x + (0.09)($9000-x) = $770 Simplify and solve for x.
0.08x + $810 - 0.09x = $770 Subtract $810 from both sides.
0.08x-0.09x = -$40 Combine the x's on the left side.
-0.01x = -$40 Finally, divide both sides by -0.01 to get x by itself.
x = $4000
So $4000 was invested at 8% and $9000-$4000 = $5000 was invested at 9%
Let's check the solution:
0.08($4000) + 0.09($5000) = $320 + $450 = $770 The total interest earned.
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