8.4 - Investment Problem

# 8.4 - Investment Problem - Interest at 6 =.06y Our second...

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Investment Problems

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A college student earned \$5000 during summer vacation working as a waiter in a popular restaurant. The student invested part of the money at 8% and the rest at 6%. If the student received a total of \$380 in interest at the end of the year, how much was invested at 8%? We’re asked to find the amount invested at each rate, so … Let x = the amount invested at 8% Let y = the amount invested at 6%
Now we need to find out two relationships between these amounts: 1) A college student earned \$5000 during summer vacation . The amount invested at 8% + the amount invested at 6% = 5000 Our first equation is x + y = 5000

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We need one more relationship between the amounts: 2) The student received a total of \$380 in interest at the end of the year. Interest = Principal (amount invested) x rate x time Interest at 8% = amount invested x .08 x 1 year

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Unformatted text preview: Interest at 6% = .06y Our second equation is .08x + .06y = 380 Now we have two equations in two unknowns x + y = 5000 .08x + .06y = 380 I would solve this by elimination x + y = 5000 .08x + .06y = 380 multiply by 100 to clear the decimals x + y = 5000 8x + 6y = 38000 Multiply the top equation by -6 to clear the y column: x + y = 5000 8x + 6y = 38000 -6x - 6y = -30000 8x + 6y = 38000 Now just add the equations together. The y’s will add up to zero and disappear! 2x = 8000 x = 4000 Now let’s find the amount invested at 6% and check our work x + y = 5000 x = 4000 4000 + y = 5000 y = 1000 Now we have \$4000 at 8% and \$1000 at 6%. Does this fit our story? Does \$4000 at 8% + \$1000 at 6% = 380 4000(.08) + 1000(.06) 320 + 60 = 380 We have a winner! Be sure you answer the question! How much was invested at 8%? \$4000 was invested at 8%...
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