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Sagot :
Answer:
- 8%: $3000
- 7%: $15000
Step-by-step explanation:
Let x represent the amount invested at 8%. Then the amount invested at 7% is (18000-x) and the interest earned is ...
0.08x +0.07(18000 -x) = 1290
0.01x +1260 = 1290 . . . simplify
0.01x = 30 . . . . . . . . . . subtract 1260
x = 3000 . . . . . . . . . . multiply by 100
$3000 was invested at 8%; $15000 was invested at 7%.
Answer:
$15,000 was invested at 7%
$3,000 was invested at 8%
Step-by-step explanation:
Let [tex]x[/tex] be the amount of money you spend on the 7% account and [tex]y[/tex] be the amount of money you spend on the 8% account. With the information given, you can set up two equations:
[tex]x+y=18000[/tex]
[tex]0.07x+0.08y=1290[/tex]
Rearrange the first equation and multiply the second equation to get:
[tex]x=18000-y[/tex]
[tex]7x+8y=129000[/tex]
Then, substitute [tex]x[/tex] for [tex]18000-y[/tex] in the second equation to get:
[tex]7(18000-y)+8y=129000[/tex]
Use the distributive property ([tex]a(b-c)=ab-ac[/tex]) to get:
[tex]126000-7y+8y=129000[/tex]
Subtract [tex]126000[/tex] from both sides and combine like terms to get:
[tex]y=3000[/tex]
Thus, substitute [tex]y[/tex] for [tex]3000[/tex] in the first original equation to get:
[tex]x+3000=18000[/tex]
Finally, subtract [tex]3000[/tex] from both sides to reach:
[tex]x=15000[/tex]
Our answer is $15,000 was invested at 7% and $3,000 was invested at 8%.
Hope this helps :)
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