A college student earned $8200 during summer vacation working as a waiter in a restaurant. The student invested part of the money at 7% and the rest at 6%. If the student received a total of $528 in interest at the end of the year, how much was invested at 7%?
So far, I have x+y=8200
y=8200-x
0.07x + 0.06y=528
0.07x + 0.06(8200-y)=528
0.07x + 492 - 0.06y=528
Where do I go from there?
Thanks so much for your help!
So far, I have x+y=8200
y=8200-x
0.07x + 0.06y=528
0.07x + 0.06(8200-y)=528
0.07x + 492 - 0.06y=528
Where do I go from there?
Thanks so much for your help!
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This is where I would use substitution.
x + y = 8200
0.07x + 0.06y = 528
Rewrite the first equation in terms of one of the variable.s
y = 8200 - x
Now SUBSTITUTE for that in the other equation.
0.07x + 0.06(8200 - x) = 528
0.07x + 492 - 0.06x = 528
0.01x = 36
x = 3600
So the student invested 3600 at 7% and 8200 - 3600 = 4600 at 6%
x + y = 8200
0.07x + 0.06y = 528
Rewrite the first equation in terms of one of the variable.s
y = 8200 - x
Now SUBSTITUTE for that in the other equation.
0.07x + 0.06(8200 - x) = 528
0.07x + 492 - 0.06x = 528
0.01x = 36
x = 3600
So the student invested 3600 at 7% and 8200 - 3600 = 4600 at 6%