Question

beth has a total of $6000 to invest, part at 6% simple interest and part at 8% simple interest. Her return for one year is $410. How much did she invest at each rate?

Answer 1

Since interest=investment*interest rate*time in years for simple interest, we know that there are 2 separate investments. If the amount invested at 6% simple interest is x dollars and the amount invested at 8% is y dollars, then we know that x+y=6000 since she spent all 6000 dollars on only 2 separate investments. Subtracting y from both sides, we get that 6000-y=x. In addition, we know that her interest for the 6% interest rate is x*0.06*1 as well as y*0.08*1 for the 8% simple interest. Since the total interest is 410$, we know that x*0.06*1+y*0.08*1=410. Since y=6000-x, we plug that in to get x*0.06+(6000-x)*0.08=410. Using the distributive property to expand, we get x*0.06+480-0.08x=410=-0.02*x+480. Subtracting 480 from both sides, we get -0.02*x=-70. After that, we can multiply both sides by -50 (since -0.02 times -50 is 1) to get x=3500=amount invested at 6%. Since the amount invested at 8% is 6000-x, that equals 6000-3500=2500