
Annette C.
asked 03/06/13word problem
Henry has $15,000 to invest. He invests x dollars at 4% and the rest at 3.5%. Suppose at the end of the year Henry has $15,575. How much money did he start with in each account?
3 Answers By Expert Tutors
PV1 = x
PV2 = 15000-x
r1=.04
r2=.035
FV = PV1(1+r1) + PV2(1+r2)
15575 = 1.04x + 1.035(15000-x)
15575 = 1.04x + 15525 - 1.035x
50 = .005x
x = $10,000 = PV1
$15,000-x = $5000 = PV2

Anthony P. answered 03/06/13
Experienced tutor in earth sciences and basic math to trigonometry
This can be solved using a systems of equations approach, the system being two eqns with two unknowns. Te first eqn will represent the total amt of money in each acct, and the 2nd eqn will represent the interest earned in each acct.
Let x = amt invested in the 1st acct
Let y = amt invested in the second acct
The sum between the accts is 15,000, so
x + y = 15,000 (eqn 1)
The interest earned in acct 1 is 0.04x and in acct 2 is 0.035y, and the total interest earned is the sum of those two. In addition, we know the total interest earned to be equal to (15,575 - 15,000) = 575
I1 + I2 = 575
0.04x + 0.035y = 575 (eqn 2)
Solve by substitution. Solve for either variable in eqn 1 and substitute into eqn 2.
x = (15,000 - y)
0.04x + 0.035y = 575 (eqn 2)
0.04(15,000 - y) + 0.035y = 575
solve for y
600 - 0.04y + 0.035y = 575
600 - 0.005y = 575
-0.005y = -25
y = 5000
So the amt in acct y is $5,000
Now substitute that value into eqn 1, or simply
x = (15,000 - y) = (15,000 - 5,000) = 10,000
So the amt in acct 1 is $10,000
You can also find the amt of money earned in each acct by substituting those into eqn 2,
0.04(10,000) + 0.035(5,000) = 575
400 + 175 = 575
575 = 575
Q.E.D.

Matthew S. answered 03/06/13
Statistics, Algebra, Math, Computer Programming Tutor
Henry invests $15,000 into two accounts. Let's write an equation relating x and y, the amounts invested in each of the two accounts. We know these sum to $15,000, so:
x + y = 15000
The first account earns interest at the rate of 4%. So after one year his initial investment, x, has grown to x + 0.04x = 1.04x, where 0.04x is the 4% earned.
Similarly, the second account grows from y to y + 0.035y = 1.035y.
The sum of the two accounts after one year is $15,575. We can write a second equation to describe Henry's investments after one year:
1.04x + 1.035y = 15575
You now have two equations and two unknowns. We can solve this problem!

Matthew S.
Let's use the first equation, and solve for x. This is easily done since all we need to do is subtract y from both sides.
x + y = 15000
x + y - y = 15000 - y
x = 15000 - y
Now, substitute this expression for x in the second equation:
1.04(15000 - y) + 1.035y = 15575
Multiply the terms in parentheses by 1.04:
1.04*15000 - 1.04y + 1.035y = 15575
15600 - 1.04y + 1.035y = 15575
15600 - 0.005y = 15575
Subtract 15600 from both sides:
15600 - 0.005y - 15600 = 15575 - 15600
-0.005y = -25
Divide both sides by -0.005:
-0.005y/-0.005 = -25/-0.005
y = 5000
Go back to our simple equation relating x to y and substitute in y's value:
x = 15000 - y
x = 15000 - 5000
x = 10000
So Henry invested $10,000 in the 4% interest account and $5,000 in the 3.5% interest account.
03/06/13

Matthew S.
Let's use the first equation, and solve for x. This is easily done since all we need to do is subtract y from both sides.
x + y = 15000
x + y - y = 15000 - y
x = 15000 - y
Now, substitute this expression for x in the second equation:
1.04(15000 - y) + 1.035y = 15575
Multiply the terms in parentheses by 1.04:
1.04*15000 - 1.04y + 1.035y = 15575
15600 - 1.04y + 1.035y = 15575
15600 - 0.005y = 15575
Subtract 15600 from both sides:
15600 - 0.005y - 15600 = 15575 - 15600
-0.005y = -25
Divide both sides by -0.005:
-0.005y/-0.005 = -25/-0.005
y = 5000
03/06/13

Matthew S.
Henry invested $10,000 in the 4% interest account and $5,000 in the 3.5% interest account.
03/06/13

Matthew S.
03/06/13

Anthony P.
03/06/13

Anthony P.
03/06/13
Still looking for help? Get the right answer, fast.
Get a free answer to a quick problem.
Most questions answered within 4 hours.
OR
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.
Matthew S.
Henry invests $15,000 into two accounts. Let's write an equation relating x and y, the amounts invested in each of the two accounts. We know these sum to $15,000, so:
x + y = 15000
The first account earns interest at the rate of 4%. So after one year his initial investment, x, has grown to x + 0.04x = 1.04x, where 0.04x is the 4% earned.
Similarly, the second account grows from y to y + 0.035y = 1.035y.
The sum of the two accounts after one year is $15,575. We can write a second equation to describe Henry's investments after one year:
1.04x + 1.035y = 15575
You now have two equations and two unknowns. We can solve this problem!
Let's use the first equation, and solve for x. This is easily done since all we need to do is subtract y from both sides.
x + y = 15000
x + y - y = 15000 - y
x = 15000 - y
Now, substitute this expression for x in the second equation:
1.04(15000 - y) + 1.035y = 15575
Multiply the terms in parentheses by 1.04:
1.04*15000 - 1.04y + 1.035y = 15575
15600 - 1.04y + 1.035y = 15575
15600 - 0.005y = 15575
Subtract 15600 from both sides:
15600 - 0.005y - 15600 = 15575 - 15600
-0.005y = -25
Divide both sides by -0.005:
-0.005y/-0.005 = -25/-0.005
y = 5000
Go back to our simple equation relating x to y and substitute in y's value:
x = 15000 - y
x = 15000 - 5000
x = 10000
So Henry invested $10,000 in the 4% interest account and $5,000 in the 3.5% interest account.
03/06/13