Douglas invests money in two simple interest accounts. He invests three times as much in an account paying 14% as he does in an account paying 5%. If he earns $152.75 in interest in one year from both accounts combined, how much did he invest altogether?

Douglas invested $ 1300 altogether

Solution:

Let x represent the amount invested in the account paying 14% interest.

Let y represent the amount invested in the account paying 5% interest

He invests three times as much in an account paying 14% as he does in an account paying 5%

Which means,

x = 3y

The simple interest is given by formula:

Where,

"p" is the principal

"n" is the number of years

"r" is the rate of interest

He earns $152.75 in interest in one year from both accounts combined

Therefore,

Combined S.I = 152.75

n = 1 year

Considering the account earning 14% interest:

Considering the account earning 5% interest:

Since, Combined S.I = 152.75

Therefore,

0.42y + 0.05y = 152.75

0.47y = 152.75

Divide both sides by 0.47

y = 325

Therefore,

x = 3y

x = 3(325)

x = 975

how much did he invest altogether?

Amount invested together = x + y = 975 + 325 = 1300

Thus he invested $ 1300 altogether

3. (5x + 1) - (-10x + 6)

simplify the answer choice. combine like terms. remember to distribute the negative to the terms inside the second parenthesis. also remember that two negatives = one positive, and one of each sign = negative.

- (-10x + 6) = + 10x - 6

simplify. combine like terms:

5x + 1 + 10x - 6

5x + 10x + 1 - 6

15x - 5

∴ it is your answer

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