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anonymous
 5 years ago
Linda invests $13,000 for one year. Part is invested at 4%, another part at 5%, and the rest at 7%. The total income from all 3 investments is $705. The combined income from the 4% and 5% investments is $75 more than the income from the 7% investment. Find the amount invested at each rate.
anonymous
 5 years ago
Linda invests $13,000 for one year. Part is invested at 4%, another part at 5%, and the rest at 7%. The total income from all 3 investments is $705. The combined income from the 4% and 5% investments is $75 more than the income from the 7% investment. Find the amount invested at each rate.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x*0.04+y*0.05+(13000xy)*0.07=705 x*0.04+y*0.05(13000xy)*0.07=75 solve for x and y

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i still dont understand how to solve it...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let money invested for 4% be x let money invested for 5% be y so money invested for 7% is money left i.e 13000xy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now total income is the sum of income from all so it's 1 eq now use second condition

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ugh... im never gonna get this...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0try it i will help u to get it never lose hope always try and u will get it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok let me try, give me a few

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im not gettin it because whatever I have it is cancellin x and y

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so I cant solve for neither

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0have u solved 2 eq which i gave u?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok i made a mistake, I got x=13000, and y=9250

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0an z=16750 which are all probably wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(1)(2) will give u value of z i.e 5571.4285..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can somebody please help me solve this problem...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Interest Rate * Amount Loaned = return or return on interest or revenue Borrowing from Jas's work we have the following: Let x = amount loaned at 4% or 4/100 Note: 4/100 is 4% converted to a fraction. Let y = amount loaned at 5% or 5/100 The remaining cash is loaned at 7% or 7/100 The total income from all 3 investments is $705.\[\left(\frac{4}{100}x\right)+\left(\frac{4}{100}y\right)+\left(\frac{7}{100}(13000(x+y))\right)=705 \] Solve the above equation for y now; we will need it later. Multiply through by 100, gather all of the y terms and put them on the left hand side and all of the other terms on the right hand side and then simplify. \[y= \frac{1}{2} (205003 x) \] The combined income from the 4% and 5% investments is $75 more than the income from the 7% investment. Since the left hand side is 75 more than the right, we add in an additional 75 to the right hand side to make both sides equal to each other. \[\left(\frac{4}{100}x\right)+\left(\frac{5}{100}y\right)\text{= }75+\frac{7}{100} (13000(x+y))\] Solve for y. \[y= \frac{1}{12} (9850011 x) \] Set the right hand sides of the y= equations above to each other as follows: \[\frac{1}{2} (205003 x)\text{=}\frac{1}{12} (9850011 x)\] and solve for x. \[x=3500, y=\frac{1}{12} (9850011*3500)=4500, 1300035004500 =5000\text{ } \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There is an error in the first equation expression. \[\left(\frac{4}{100}y\right)\text{ should be changed t}\text{o }\left(\frac{5}{100}y\right)\] The computations were used with the latter.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is the same as i answered earlier

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Jas, The reason I put forth a solution was because after all of yesterday's back and forth on this problem, hermommy1101 in the last problem posting yesterday, issued a plea, "can somebody please help me solve this problem..." Just trying to help out.
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