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You invested ​$18,000 in two accounts paying 7% and 8% annual​ interest, respectively. If the total interest earned for the year was $1290 how much was invested at each​ rate?

Sagot :

sqdancefan

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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