OpenStudy (anonymous):

A person invested $8100 for 1 year

3 years ago
OpenStudy (anonymous):

3 years ago
OpenStudy (anonymous):

I got to the stop of 2x+2y=7200

3 years ago
OpenStudy (anonymous):

y=3600-x

3 years ago
OpenStudy (anonymous):

so is my total invested $4500?

3 years ago
OpenStudy (anonymous):

I feel like i did it wrong, 4%=1500, 10%=2100, 12%=540 is this correct?

3 years ago
OpenStudy (anonymous):

@ranga could you help me out and tell me if I did that correctly?

3 years ago
OpenStudy (anonymous):

Let X be amount at 4%, Y be the amount at 10% and Z the amount at 12%. What equation are you using for the total amount of money? ... = $8100 What equation for the income? ...=$810 What equation for the relationship among the investments? Z = X + Y + 900 Three equations for three unknowns. Doable.

3 years ago
OpenStudy (anonymous):

z=8100 and the 2x and 2y?

3 years ago
OpenStudy (anonymous):

so 8100=2x+2y+900

3 years ago
OpenStudy (anonymous):

8100 = x + y + z

3 years ago
OpenStudy (ranga):

X + Y + Z = 8100 (1) .04X + .1Y + .12Z = 810 (2) Z = X + Y + 900 (3) From (3): X + Y = Z - 900. Replace X + Y in (1) by Z - 900 and you can solve for Z. Then put that Z value in (1) and (2) and solve for X and Y.

3 years ago
OpenStudy (anonymous):

Yea I did that and got 4%=1500, 10%=2100, 12%=540, is this correct though?

3 years ago
OpenStudy (ranga):

Those three numbers don't add up to 8,100. It is not correct.

3 years ago
OpenStudy (anonymous):

oh ok, hold on

3 years ago
OpenStudy (anonymous):

1) z=7200

3 years ago
OpenStudy (ranga):

Try again.

3 years ago
OpenStudy (anonymous):

oh 9000?

3 years ago
OpenStudy (ranga):

2Z = 9000 So Z = ?

3 years ago
OpenStudy (anonymous):

4500

3 years ago
OpenStudy (ranga):

Yes. Put Z = 4500 in (1) and (2) and solve for X and Y.

3 years ago
OpenStudy (anonymous):

.04x+.1y=270

3 years ago
OpenStudy (anonymous):

and 3600=x+y

3 years ago
OpenStudy (ranga):

Yes.

3 years ago
OpenStudy (anonymous):

so could I just divide .1 by 270 and .04 by 270?

3 years ago
OpenStudy (ranga):

.04x + .1y = 270 --- (1) x + y = 3600 --- (2) Multiply (2) by .04 and subtract from (1). Solve for Y.

3 years ago
OpenStudy (anonymous):

.04(.-2)+.1y=270

3 years ago
OpenStudy (anonymous):

.1y=3375 y=33750?

3 years ago
OpenStudy (ranga):

.04x + .1y = 270 --- (1) x + y = 3600 --- (2) multiply (2) by .04: .04x +.04y = 144 subtract from (1) .04x + .1y = 270 .04x +.04y = 144 subtract: .06y = 126 y = 126/.06 = 2100 X = 3600 - y = 3600 - 2100 = 1500 X = 1500, Y = 2100 and Z = 3600. Get some practice solving simultaneous equations.

3 years ago
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