wyatt invests $6000 in two accounts, the first pays 10 percent simple interest, the second pays 11 percent simple interest. At the end of the year he has earned $640 in interest. How much is invested at each rate?
this is pre-algebra? It sure looks like algebra to me.
If he invests $x at 10%, then the rest (6000-x) is at 11%. So, add up the interest:
0.10x + 0.11(6000-x) = 640
Now just solve for x.
To solve this problem, we can set up a system of equations.
Let x represent the amount invested at 10% interest, and y represent the amount invested at 11% interest.
We know that the total amount invested is $6000, so we can write the equation:
x + y = 6000 -----(equation 1)
We also know that the interest earned from the first account, which pays 10% interest, is 10% of x. Similarly, the interest earned from the second account, which pays 11% interest, is 11% of y. Since the total interest earned is $640, we can write the equation:
0.1x + 0.11y = 640 -----(equation 2)
Now we can solve this system of equations using substitution or elimination method.
Let's use the substitution method to solve this system:
From equation 1, we can express x in terms of y as:
x = 6000 - y
Substituting this value of x into equation 2, we have:
0.1(6000 - y) + 0.11y = 640
600 - 0.1y + 0.11y = 640
0.01y = 40
y = 4000
Now, substitute the value of y back into equation 1 to find x:
x + 4000 = 6000
x = 2000
Therefore, Wyatt invested $2000 at 10% interest and $4000 at 11% interest.