Interest Worded Problems: An investor received $1400 interest per annum from a sum of money....

x = amount invested at 10%
y = amount invested at 7%

.10x + .07y = 1400
.07x + .10y = 1490

OK?

Denis said:

x = amount invested at 10%
y = amount invested at 7%

.10x + .07y = 1400
.07x + .10y = 1490

OK?

Click to expand...

Awesome thanks a lot Denis. These are my steps for the answer -

x = amount invested at 10%
y = amount invested at 7%

1) 0.10x + 0.07y = 1400 (1)
2) 0.07x + 0.10y = 1490 (2)
3) (1) x 7, (2) x 10
4) 0.7x + 0.49y = 9800 (3)
5) 0.7x + 1y = 14900 (4)
6) (4) - (3)
7) 0.51y = 5100
8) y = 10000
9) Sub y into (1)
10) 0.10x + 0.07(10000) = 1400
11) 0.10x + 700 = 1400
12) 0.10x = 700
13) x = 7000
14) Therefore total amount = x + y = 10000 + 7000 = $17000

kazafz said:

1) 0.10x + 0.07y = 1400 (1)
2) 0.07x + 0.10y = 1490 (2)
3) (1) x 7, (2) x 10
4) 0.7x + 0.10y = 9800 (3)
5) 0.7x + 1y = 14900 (4)
6) (4) - (3)

Click to expand...

Good work!

Same as your way, except I find it easier to get rid of decimals:
10x + 7y = 140000 [1]
7x + 10y = 149000 [2]

then multiply in a way so I can add the equations ([1] by -7, [2] by 10):

-70x - 49y = -980000 [3]
70x + 100y = 1490000 [4]

[3] + [4]
51y = 510000
y = 10000

Get my drift?

Denis said:

kazafz said:

1) 0.10x + 0.07y = 1400 (1)
2) 0.07x + 0.10y = 1490 (2)
3) (1) x 7, (2) x 10
4) 0.7x + 0.10y = 9800 (3)
5) 0.7x + 1y = 14900 (4)
6) (4) - (3)

Click to expand...

Good work!

Same as your way, except I find it easier to get rid of decimals:
10x + 7y = 140000 [1]
7x + 10y = 149000 [2]

then multiply in a way so I can add the equations ([1] by -7, [2] by 10):

-70x - 49y = -980000 [3]
70x + 100y = 1490000 [4]

[3] + [4]
51y = 510000
y = 10000

Get my drift?

Click to expand...

Haha thanks man! Yeah i get you. You just multiple out the decimals to make it whole numbers to make it easier to calculate. The only problem I'm facing with worded question is to actually find the equation out for the question. After that i can do it. Thanks anyways man! Great help!

But i dont get why you times 0.07 with y when you have times it with x and why times 0.1 with x when you have timed it with y.
please explain.

ella.895246 said:

i dont get why you times 0.07 with y when you have times it with x and why times 0.1 with x when you have timed it with y

Click to expand...

Hi ella. The invested amounts were interchanged because there are two situations. The original post contains typographical errors. Here is the corrected statement describing the second situation.

kazafz said:

She [found] that [if] she interchanged the amounts she had invested she could increase her return by $90.

Click to expand...

The word 'interchanged' means the invested amounts were switched. In other words, instead of x dollars earning 10% and y dollars earning 7%, the second situation has x dollars earning 7% and y dollars earning 10%. Denis set up the equations in post#3, but I'll add the adjective "originally" to his variable definitions below, for clarity:

x = amount originally invested at 10%
y = amount originally invested at 7%

.10x + .07y = 1400
.07x + .10y = 1490

The second equation above models the second situation. Instead of earning $1400 total interest, the investor discovered that she could earn $90 more by switching the invested amounts.
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