You invested $6000 between two accounts paying 3% and 9% annual interest, respectively. If the total interest earned for the year was $480, how much was invested at each rate?

$1,000 and $5,000

Step-by-step explanation:

let the 2 amounts be x and y (which add  to $6,000)

i.e,

x + y = 6000 (eq1)

assume x is invested at 3% and y is invested at 9%

Assuming simple interest, the following formula is applicable.

I = Prt

where

P = principal ( i.e either $x or $y)

r = rate (i.e 3% or 9%)

t = time elapsed (given as 1 year)

since t = 1 year, the equation reduces to

I = Pr

For $x invested at 3% (i.e 0.03) ---> I = 0.03x

For $y invested at 9% (i.e 0.09) ---> I = 0.09y

given that the total interest for the year = $480,

0.03x + 0.09y = 480   (multiplying both sides by 100)

3x + 9y = 48,000 (eq2)

solving the system of equations made up of eq 1 and eq 2 by elimination,

eq 1 x 3

3x + 3y = 18,000 (eq 3)

eq2 - eq3

9y - 3y = 48,000 - 18,000

6y = 30,000

y = 30,00 / 6 = $5,000 (answer)

substitute y = 5,000 into eq 1.

x + 5000 = 6000

x = $1,000 (answer)

step-by-step explanation:

30% of $420 is 0.30($420) = $126 and $420 + $126 = $546.

or, 130% of $420 is 1.3($420) = $546.

notice that this is not the $600 original list price with which we started.

why doesn't this work?

the method does not work because 30% of $420 is not the same amount as the 30% of $600. it may seem obvious when stated this way, but it is a very common error made on problems of this type. we cannot just use the same percent and get back to the original amount because we are taking that percent of a different value.

to work problems of this type correctly, it is generally best to write an equation for the situation and then solve.

example: since (100% – 30%)(original price) = (sale price)

70% of the   (original price) is $420.

0.7x   = 420

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x = 600

the original cost was $600.

we may also solve this problem using the two-step method. working the problem in two steps requires us to use more algebra skills. we have to know that x = 1x (identity property of multiplicationtext annotation indicator) and that 1x – 0.3x = 0.7x (subtracting like terms). then we solve the problem as follows:

(original price) – 30%(original price) = sale price

x – 0.3x = 420

0.7x = $420

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x = 600

the original cost was $600.  

note that with both of the above methods we do obtain the correct original amount.

example: after a 5% pay raise, hermione is earning $22,680 per year. what was she earning before the pay raise?

105% of (original salary) = current salary

1.05x = 22,680

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x = 21,600

hermione earned $21,600 per year before the pay raise.

this is an example so maybe you will get how to at least do it!

it's d.

step-by-step explanation:

because: -1(4x + 1y) +2 (1x-3y) distribute each number into the parenthesis.

then you would get: -4x - 1y + 2x -6y combine like terms

and get: 2x -7y

hope my answer has you if not then i'm sorry.