Lane invested $22,000, part at 8% and part at 7%. If the total interest at the end of the year is $1,710, how much did she invest at 8%?
2006-06-25 03:48:53 · 4 answers · asked by spontanious402000 1 in Science & Mathematics ➔ Mathematics
1710 = x*8*1/100 + (22000-x)*7/100
1710 * 100 = 154000 + x
x=171000-154000
x= $17000
2006-06-25 03:57:05 · answer #1 · answered by Anonymous · 1⤊ 0⤋
We have two unknowns, and we know to things, so we need to set up simultaneous equations.
a and b are the sums invested into two different accounts.
"a" is invested at 8% and "b" at 7%.
Eqn1: a + b = 22000
therefore a=22000-b
Eqn2: a*0.08 + b*0.07 = 1710
Substitute a into Eqn2,
therefore (22000-b)*0.08 + b*0.07 = 1710
therefore 1760 -0.08b + 0.07b = 1710
therefore 50 -0.08b +0.07 b = 0
therefore 50 = 0.08b-0.07b
therefore 50 = 0.01b
therefore b = 5000
Substitue into Eqn1
a + 5000 = 22000
there fore a = 22000-5000 = 17000
Therefore $17,000 was invested at 8%.
2006-06-25 10:57:43 · answer #2 · answered by superman2day 1 · 0⤊ 0⤋
x+y=22000
y=22000-x
The sum of the interests:
x * 8/100 * 1 + (22000-x)*7/100*1 =1710
Try to find the value of x from this equation.
2006-06-25 11:00:12 · answer #3 · answered by iyiogrenci 6 · 0⤊ 0⤋
Buddy no offense but you should study English and Spelling before trying to tackle aldebra as you like to call it.
2006-06-25 10:52:30 · answer #4 · answered by Førsâkëñ 5 · 0⤊ 0⤋