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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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.

Mathematics
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x*0.04+y*0.05+(13000-x-y)*0.07=705 x*0.04+y*0.05-(13000-x-y)*0.07=75 solve for x and y
theres no z to it?
im so lost...

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no there is no need
as z=13000-x-y
i still dont understand how to solve it...
let money invested for 4% be x let money invested for 5% be y so money invested for 7% is money left i.e 13000-x-y
ok.....
now total income is the sum of income from all so it's 1 eq now use second condition
ugh... im never gonna get this...
try it i will help u to get it never lose hope always try and u will get it
ok let me try, give me a few
im not gettin it because whatever I have it is cancellin x and y
so I cant solve for neither
have u solved 2 eq which i gave u?
ok i made a mistake, I got x=-13000, and y=9250
an z=16750 which are all probably wrong
k iet me check again
ok
(1)-(2) will give u value of z i.e 5571.4285..
now solve others
ok now u lost me...
can somebody please help me solve this problem...
Interest 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. \[y= \frac{1}{2} (20500-3 x) \] 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. \[y= \frac{1}{12} (98500-11 x) \] Set the right hand sides of the y= equations above to each other as follows: \[\frac{1}{2} (20500-3 x)\text{=}\frac{1}{12} (98500-11 x)\] and solve for x. \[x=3500, y=\frac{1}{12} (98500-11*3500)=4500, 13000-3500-4500 =5000\text{ } \]
There 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.
this is the same as i answered earlier
Jas, 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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