Intermediate Algebra Problem Solving Help?

Question

Hollis is paying off two student loans. One loan charges 7% interest per year. The other loan charges 9% interest per year. He owes $1500 more on the 7% loan than he does on the other. Last year he paid a total of $617 interest. How much does he owe on each loan?

Please help me make the two equations.

Answer 1

x = amount owed on 9% loan
x+1500 = amount owed on 7%loan
.09x + .07(x+1500) = 617
.09x +.07x + 105 = 617
.16x = 512
x = $3200
x+1500 = $4500
So he owes $4500+$3200 = $8700

Answer 2

Hollis is paying off two student loans. One loan charges 7% interest per year. The other loan charges 9% interest per year. He owes $1500 more on the 7% loan than he does on the other. Last year he paid a total of $617 interest. How much does he owe on each loan?

L₁ = loan at 7%
L₂ = loan at 9%
L₁ = L₂ + 1500
,07L₁ + .09L₂ = 617
NOTE: If all you want is the equations, PLEASE STOP HERE.

Substituting from the equation above into what we know about last year’s interest...
,07L₁ + .09L₂ = 617
.07(L₂+1500) + .09L2 = 617
7L₂ + 10500 + 9L₂=61700

Distributing, adding -10500 to both sides, and doing a little arithmetic.
16L₂ = 51200
Multiplying both sides by 1/16
L₂ = 3200

Substituting the value of L₂ and solving for L₁
L₁=L₂+1500=3200+1500=4700

Answer 3

Let Hollies owes on 9% loan=x
so (x+1500)*7/100+x*9/100=617
or 7x/100+1500*7/100+9x/100=617
or 16x/100+105=617
or4x/25=617-105=512
or x=512*25/4=128*25=3200
so he owes 3200 on 9% & 4700 on 7% ans

Answer 4

Set up interest equation:
617 = .07(x + 1500) + .09x, where x = amount on the 9% loan.
617 = .16x + 105, or .16x = 512. So x = 3200.
He owes 1500 more on the other, so = 4700.

Test it:
617 = .07*4700 + .09*3200 = 329 + 288 = 617.

Answer 5

Set up your two cases: C1: Speed = x Distance = 600 Time = y C1: Speed = x + 10 Distance = 600 Time = y - 2 Now form your two systems of equations (using speed = distance/time): x = 600/y x + 10 = (600) / (y - 2) Substitute (600/y) for x in the second equation: 600/y + 10 = (600)/(y - 2) Multiply everything by y(y - 2) to cancel off all denominators: 600(y - 2) + 10y(y - 2) = 600y Expand all brackets: 600y - 1200 + 10y^2 - 20y = 600y Combine all the like terms: 10y^2 + 580y - 1200 = 600y Move everything to one side: 10y^2 - 20y - 1200 = 0 Divide everything by 10 to simplify the equation: y^2 - 2y - 120 = 0 Factor it out: (y + 10) (y - 12) = 0 ------------------------- This gives you the possibility of the time it took the first train to travel being either 12 hours or -10 hours. Now obviously there is no such thing as a negative time (or going back in time), so the time it took the first train can only be 12 hours. Now go back to one of your cases: C1: Speed = x Distance = 600 Time = 12 Solve for the speed (using speed = distance/time): x = 600/12 x = 50 If the speed of the second train was x + 10, it's speed was 60. Overall: The speed of the two trains are 50mph and 60mph.

Answer 6

x = 7% loan
y = 9% loan

x = y + 1500
.07x + .09y = 617

.07(y + 1500) + 0.09y = 617
.07y + 105 + 0.09y = 617

.16y = 617 - 105 = 512
y = 3200


x = y + 1500
x = 3200 + 1500 = 4700

7% loan = 4700
9% loan = 3200

Answer 7

let x represent the amount in the 7% loan
let y represent the amount in the 9% loan

then,

0.07x + 0.09y = 617

and

y= x-1500


those are your two equations and you have to sub the second equation into the first one to solve for "x"

i believe the answers are $4700 in the 7% loan and $3200 in the 9% loan

Answer 8

loan #2=x
loan#1=x+1500

loan1*.07 +loan2*,09=617

.07*(x+1500)+.09*(x)=617
.07x+105+.09x=617
.16x+105=617
.16x=512
x=3200

loan 2=3200
loan 1=4700

check it
3200 *9%=288
4700*7%=329
total interest=617