Paul has $40000 to invest. his intent is to earn 9% interest on his investment. he can invest part of his money at 6% and part at 10% interest. how much does paul need to invest in each option to make a total of 9% return on his $40000 ?

Given

Paul has $40000 to invest &he can invest part of his money at 6% and part at 10% interest.

Find out  Paul  invest in each option to make a total of 9% return on  $40000 .

To proof

Paul invest = $ 40000

let us assume that x be the part of money he invested at 6%.

let us assume that (40000 - x) be the part of money he invested at 10 %.

6% written in the simpler form              

= 0.06

similarly,

10% written in the simpler form = 0 .1

9% written in the simpler form = 0.09

than the equation becomes

0.09 (40000) = (40000 - x ) 0.1 +  0.06x

3600 = 4000 - 0.1x + 0.06x

0.1x - 0.06x = 4000 - 3600

0.04x = 400

x = 10000

$ 10000 be the part of money he invested at 6%.

$30000 be the part of money he invested at 10 %.

Hence proved

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18

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