1.)A box contains 6.55 nickels, dimes, and quarters. There are 38 coins in all,

Question

1.)A box contains 6.55 nickels, dimes, and quarters. There are 38 coins in all, and the sum of the numbers of nickels and dimes is 2 less than the number of quarters. how many coins of each kind are there? nickels? dimes? quarters?

2.)Part of $4000 is invested at 12%, another part at 13%, and the remainder at 14%. The total yearly income from the three investments is $534. The sum of the amounts invested at 12% and 13% equals the amount invested at 14%. Determine how much is invested at each rate. At 12% interest..At13% interest..At 14% interest...

Please show all work and equations..Thanks
Im no good at these especially solving a system of three linear equations in three variables...

Explanation / Answer

1)The three equations defined by the problem are as follows: a) 655=5n+10d+25q b) n+d+q=38 c) q-2=n+d Add equations b and c to remove variable q: n+d-q=-2 n+d+q=38 +_________ 2n+2d=36 (d) Then, add equations (a) and (b) to remove q first divide equation (a) by 5 , and multiply equation (b) by 5 to match coefficients of q 131= n+ 2d+5d -190=-5n+-5d+-5d + ____________ -59=-4n+-3d (e) Now, you have 2 equations with 2 variables, so we will add them. First multiply equation d by 2, so the n variable will cancel. 4n+ 4d= 72 -4n+-3d=-59 + ___________ d=13 plug variable d into equation d: 2n+2(13)=36 n=5 Then, using equation b: 5+13+q=38 q=20 PROBLEM #2: The 3 equations from the problem are: .12a+.13b+.14c=534 (a) a+b+c=4000 (b) a+b=c (c) Add equations (b) and (c) to eliminate variable c: a+b+c=4000 a+b-c=0 + _______ 2a+2b=4000 reduce to: a+b=2000 (d) Add equation (a) and (b) to eliminate variable c: First multiply equation (b) by -.14: .12a+ .13b+ .14c= 534 -.14a+-.14b+-.14c=-560 + _____________________ -.02a+-.01b=-26 (e) Now multiply equation d by .02 so the variable a will cancel, and add. .02a+ .02b= 40 -.02a+-.01b=-26 + _______________ .01b=14 b=1400 using equation (d) a+1400=2000 a=600 and equation (b) a+b+c=4000 600+1400+c=4000 c=2000

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