Linear Equations?

Question

An alloy used in eletrical transfermers contains nickel(Ni), Iron(Fe), and Molybdenum(Mo). The percent of Ni is 1% less than five times the percent of Fe. The percent of Fe is 1% more than three time the percent of Mo. find the percent of each in the alloy

Answer 1

Let m be the percentage of Mo, n be the percentage of Ni, and f be the percentage of Fe. Obviously,
m + n + f = 100
Also, from the second sentence,
n - 5f = -1
And from the third sentence,
3m - f = -1

Now, you can use these equations to solve for m, n, and f. I would do this with matrices if I were you; substitution is a pain except for the simplest cases. Typing a matrix solution is also a pain. The easiest process for solving this with a matrix is called Gaussian elimination; here's a decent site with information about it (it's probably in your book too): http://www.sosmath.com/matrix/system1/system1.html

Anyway, you can use this site to solve it for you if you need to:
http://www.quickmath.com/
This one may be easier:
http://hopper.unco.edu/lss/index.html
Good night!

Answer 2

Mo=x
Fe=3x+1
Ni=5(3x+1)-1

Mo+Fe+Ni=100
x+3x+1+(5(3x+1)-1=100
4x+1+(15x+4)=100
19x+5=100
19x=95
x=5

Mo=5%
Fe=16%
Ni=79%

5+16+79=100

Answer 3

Ni + 1 = 5Fe
Fe = 3Mo +1
Mo + Fe + Ni = 100
IS THE CORRECT ANSWER. the something something 1960 dude is wrong

Answer 4

n = 5f - 1
f = 3m + 1
n + f + m = 100
5f - 1 + f + (f - 1)/3 = 100
15f - 3 + 3f + f - 1 = 300
19f = 304
f = 304/19 = 16
n = 5*16 - 1 = 79
m = (16 -1 )/3 = 5

Answer 5

solve
Ni = 5Fe + 1
Fe +1 = 3Mo
Mo + Fe + Ni = 100

Answer 6

Ni =A
Fe=B
Mo=C

A+B+C=100 in percent

A+B+C=100 .......................i)
A=5B-1 ....................ii)
B=3C+1 ...................iii)


from ii & iii

A=5B-1
A=5(3C+1)-1
A=15C+5-1
A=15C+4 ...............iv)


from i,iii & iv
A+B+C=100
(15C+4)+(3C+1)+C=100
15C+4+3C+1+C=100
19C+5=100
19C=100-5
19C=95
C=95/19
C=5

B=3C+1
B=3(5)+1
B=16

A=15C+4
A=15(5)+4
A=75+4
A=79

A+B+C=100
79+16+5=100

so the answers is
A=79 %
B=16%
C=5%