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Vikentia [17]
3 years ago
10

Mr. Azu invested an amount at rate of 12% per annum and invested another amount, 580 ghana cedis more than the first at 14% . if

Mr. Azu had total accumulated amount of 2,358.60, how much was his total investment? ​
Mathematics
1 answer:
MariettaO [177]3 years ago
5 0

Answer:

  2082.12 was the total invested

Step-by-step explanation:

Let x represent the amount invested at 14%. Then the amount invested at 12% was (x-580). The total accumulated amount was ...

  112%(x -580) +114%(x) = 2358.60

  2.26x -649.60 = 2358.60

  2.26x = 3008.20 . . . add 649.60

  x = 1331.06 . . . . . . divide by 2.26

  x -580 = 751.06

The total invested was 1331.06 +751.06 = 2082.12 cedis.

__

<em>Check</em>

The investment at 12% was 751.06, so the accumulated amount of that investment was 751.06×1.12 = 841.19.

The investment at 14% was 1331.06, so the accumulated amount of that investment as 1331.06×1.14 = 1517.41.

The accumulated total amount was 841.19 +1517.41 = 2358.60.

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Step-by-step explanation:

This one is simple substitution... at least, substitution is the easiest method. The first equation is 3<em>x</em> – 30 = <em>y</em>  and the second is 7<em>y</em> – 6 = 3<em>x</em>

As I look, I see 3 ways to use substitution to solve this:

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We're going to only use 1 method for the sake of time. Try the other two on your own. Assuming you don't make any mistakes, they will work.

<u>Method 1</u>:

3<em>x</em> – 30 = <em>y</em>

7<em>y</em> – 6 = 3<em>x</em>  — initial system of equations

7<em>y</em> – 6 – 30 = <em>y</em>  — substitute 7<em>y</em> – 6 for 3<em>x</em>

<u>7</u><u><em>y</em></u> – 6 – 30 = <u><em>y</em></u>  — marking like terms, bold for constants, <u>underlined</u> for variables

7<em>y</em> – 36 = <em>y</em>  — combining the constants and simplifying

Here, you could diverge into multiple paths: add 36 to both sides, subtract <em>y</em> from both sides, divide by 6 OR subtract 7<em>y</em> from both sides and divide by –6 . For the sake of time, I'm subtracting 7<em>y</em>, though I don't like dealing with negatives.

7<em>y</em> – 7<em>y</em> – 36 = <em>y</em> – 7<em>y</em>  — subtract 7<em>y</em> from both sides

–36 = –6<em>y</em>  — simplify

–36 ÷ –6 = –6<em>y</em> ÷ –6  — divide by –6 on both sides

<em>y</em> = 6  — simplify

Again, we can diverge here: substitute <em>y</em> into 3<em>x</em> – 30 = <em>y</em> or substitute <em>y</em> into 7<em>y</em> – 6 = 3<em>x</em>

I'm going to choose 3<em>x</em> – 30 = <em>y</em> but it will work either way, should you take the time (if you have it) to chase down every path this problem can take.

3<em>x</em> – 30 = <em>y</em>  — initial equation

3<em>x</em> – 30 = 6  — substitute 6 for <em>y</em>

3<em>x</em> – 30 + 30 = 6 + 30  — add 30 to both sides to isolate 3<em>x</em>

3<em>x</em> = 36  — simplify the expression

3<em>x</em> ÷ 3 = 36 ÷ 3  — divide both sides by 3 to isolate <em>x</em>

<em>x</em> = 12  — simplify

So, we have <em>x</em> = 12 and <em>y</em> = 6 . We know they work for 3<em>x</em> – 30 = <em>y</em>  but not if they work for 7<em>y</em> – 6 = 3<em>x</em> . Let's substitute those in to see if (12, 6) really is the solution point.

7<em>y</em> – 6 = 3<em>x</em>  — original equation

7(6) – 6 ≟ 3(12)  — substitute 6 for <em>y</em> and 12 for <em>x</em>

42 – 6 ≟ 36  — simplify by multiplying

36 = 36 ✔  — simplify by combining like terms on left side

Success! It works! We have found our solution!

I hope this helps increase your understanding of the concept. Have a great day!

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