word problems?

Question

Jane invested some amount at the rate of 12% simple interest and some other amount at the rate of 10% simple interest. She received yearly interest of $130. Randy also invested in the same scheme, but he interchanged the amounts invested, and received $4 more as interest. how much amount did each of them invest at different rates.

Answer 1

.12 * x + .10 * y = 130
.10 * x + .12 * y = 130 + 4 = 134

That gives you two equations to solve in two variables.

One way would be to multiply (both sides of) the first equation by 5 and multiply both sides of the second equation by 6, then subtract (each side of) the new second equation from the new first one. That lets you solve for y, after which you can easily solve for x.

Answer 2

Following the lead of the previous post-er, Curt Monash, let's define:
|--: x = amount invested by Jane in option 1 (12% return)
|--: y = amount invested by Jane in option 2 (10% return)

J's total R.O.I.:
(1) $130 = .12x + .10y

R's total R.O.I.:
(2) $134 = .10x + .12y

Multiplying both sides of (2) by 6 and both sides of (1) by 5:
6*(2): 6*$134 = .60x + .72y
5*(1): 5*$130 = .60x + .50y

Subtracting:
$804-$650 = .22y
And so:
y = $154/.22 = $700

Substituting into (1):
$130 = .12x + .10*700
$130-$70 = .12x
$60 = .12 x
And so:
x = $500.

Doublechecking (1):
|--? $130 = .12($500) + .10($700)
Yes: $130 = $60 + $70

Doublechecking (2):
|--? $134 = .10($500) + .12($700)
Yes: $134 = $50 + $84.