# I invest $x at 3%/annum & $y at 4%/ annum.Annual income fm this is $22. If I invested $y at 3% & $x at 4%, my income would be $20/annum. find x.

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The idea is to use both equations to solve for one variable and substitute it back for the other. So:

1) .03x + .04y = 22 and

2) .04x + .03y = 20 so combining (adding) equations gives you

3) .07x + .07y = 42 divide each term by .07 to reduce equation

4) x + y = 600 and set value for y

5) y = 600 - x

If you substitute 600 - x for y in equation 1, you should find that x = $200 and y = $400.

Let's do simultaneous equations:

0.3x+0.4y=22 -1

0.4x+0.3y=20 -2

Multiply both eqn by 10 to get whole number

3x+4y=220 -3

4x+3y=200 -4

Multiply eqn 3 by 4 and eqn 4 by 3

12x+16y=880 -5

12x+9y=600 -6

eqn5-eqn6

12x-12x+16y-9y= 880-600

7y=280

y=40

sub y=40

12x+9(40)=600

12x+360=600

12x=240

**x=20 dollars**