student takes out two loans 1 at 10% and the other at 12 percent interest the 10 percent loan is $2000 more then the 12% loan and the total interest for one year is $430 how much is each loan?
I1=.1(M+2000)
I2=.12(M)
I1+I2=430=.1M+200+.12M
solve for M, then M+2000
To solve this problem, we can set up a system of equations based on the information given. Let's let x represent the amount of the loan with a 12% interest rate and y represent the amount of the loan with a 10% interest rate.
We're told that the 10% loan is $2000 more than the 12% loan, so we can write the equation:
y = x + 2000
We're also told that the total interest for one year is $430. The interest on the 12% loan would be 12% of x, and the interest on the 10% loan would be 10% of y. The sum of these two interests should equal $430, so we can write the equation:
0.12x + 0.10y = 430
Now we have a system of two equations in two variables. We can solve it using substitution or elimination method.
Let's solve it using the substitution method:
Step 1: Substitute the value of y from the first equation into the second equation.
0.12x + 0.10(x + 2000) = 430
Step 2: Distribute the 0.10 to both terms in the parentheses.
0.12x + 0.10x + 200 = 430
Step 3: Combine like terms.
0.22x + 200 = 430
Step 4: Subtract 200 from both sides.
0.22x = 230
Step 5: Divide both sides by 0.22 to solve for x.
x = 230 / 0.22
Calculating this division, we find that x is approximately 1045.45.
Step 6: Substitute the value of x back into the first equation to solve for y.
y = 1045.45 + 2000
Calculating this, we find that y is approximately 3045.45.
So, the loan with a 12% interest rate is approximately $1045.45, and the loan with a 10% interest rate is approximately $3045.45.