Substance 1 contains 13% of A, 62% of B and 25% of C. Substance 2 contains 5% of A, 77% of B, and 18% of C. Substance 3 contains 29% of A, 13% of B, and 58% of C. How many parts of each to get 10% A, 50% B and 40% C? How do I solve this problem?
+37157 x = amount of 1 y = amount of 2 z= amount of 3 final amount = x+y+z
.13x + .05y + .29z = .1 (x+y+z) This is for A
.62x + .77 y + .13 z = .5 (x+y+z) this is for B
.25x + .18y + .58 z = .4 (x+y+z) this is for C
I think this is correct......I used wolfram alpha and got no answer ! This problem has no solution that I can fathom !
Hope someone else can find an answer ....or the same conclusion !!! 
a=0;p=0; b=0;c=0;n=0.62*a + 0.77*b + 0.13*c +0.13*a + 0.05*b + 0.29*c+0.25*a + 0.18*b + 0.58* c ;if(n==0.10*(a+b+c) + 0.50*(a+b+c) +0.4*(a+b+c), goto loop, goto next);loop:p=p+1;printp," =",a,b,c,n;next:a++;if(a<150, goto4,0);a=0;b++;if(b<150, goto4, 0);a=0;b=0;c++;if(c<150, goto4,0)
OUTPUT: By adding up the 3 equations, there appear to be many solutions such as:
a = 20, b=140, c=40
+37157 I'm a little confused.... are a,b,c in your solution corresponding to Substance 1 2 and 3 ?
Take your first equation:
.13x + .05y + .29z = .1 (x+y+z) This is for A
There are numerous solutions to x, y, z:
such as x =70, y = 80 and z =10 . Sub these values for x, y, z and you get:
0.13 *70 + 0.05 * 80 + 0.29 * 10 =0.10*(70+80+10)
9.1 + 4 + 2.9 =0.10*(160)
16 =16
...of course there are numerous solutions for just ONE of the substances equations......but we are trying to mix substance A B and C to get the final substance of 10% A 50% B and 40 % C .....I don't believe there is a way to mix all three to get the final ....
In other words....you only solved ONE of the equations......we need solve the SYSTEM of equations to find x y and z that solves all THREE of the equations......
