Michael Perez deposited a total of $2000 with two savings institutions. One pays interest at a rate of 6%/year, whereas the other pays interest at a rate of 7%/year

Question

Formulate but do not solve the problem.

Michael Perez deposited a total of $2000 with two savings institutions. One pays interest at a rate of 6%/year, whereas the other pays interest at a rate of 7%/year. If Michael earned a total of $126 in interest during a single year, how much did he deposit in each institution? (Let x and y denote the amount of money, in dollars, invested at 6% and 7%, respectively.)

		= 126
		= 2000

Answer 1

If x and y are the amounts invested at 6% and 7%, we will do is set up a system of linear equations which we will solve. We know the total amount deposited is $2000 so we know that x + y =2000. Now we will look at this from the perspective of the interest. The total amount of interest is $126. The interest from the savings institution charging 6% will be .06x. The interest from the savings institution charging 7% will be .07y. So we put all that together to get .06x + .07y = 126.

So this is our system:

.06x + .07y = 126

x + y =2000

If you want to avoid using decimals, we can multiply both sides of the first equation by 100 and get:

6x+7y=12600

x + y =2000

There are two main approaches we can take from here: The addition method or the substitution method. In either case our goal is going to be to obtain a single equation with one unknown (which we can solve for)

Let's try the addition method. Multiply the second equation by -6 and we get

6x+7y=12600

-6x -6 y =-12000 then add the two equations together:

6x + 7y - 6x - 6y = 12600 - 12000

which gives us y=600.

Then you can plug it in some other equation containing both variables to solve for x. The most convenient is probably

x + y = 2000

So: x + 600 = 2000

And we can subtract 600 from both sides to get

x=1400.

So we invest $1400 at 6% and $600 at 7%

In instead you chose to use the substitution method you would still start with the same system:.

6x+7y=12600

x + y =2000

But instead of adding two equations together, we will solve for one variable in terms of the other, and then insert or substitute the right expression in the other equation.

For instance, take x +y = 2000. We can solve for x in terms of y. x = 2000-y. So now we can rewrite the first equation all in terms of y.

6x + 7y = 12600 becomes

6(2000-y) +7y = 12600

We distribute and get

12000 - 6y + 7y = 12600 and combine like terms to get

12000 + y = 12600 and subtract by 12000 on both sides to get as before

y=600.

Now, since we already have a nice formula for x in terms of y, namely x= 2000 - y we can plug in the fact that y=600 and obtain that

x = 2000 - 600 = 1400

So the substitution method and addition method give the same answer.