Answer
2 units of Mix A, 3 units of Mix B, and 1 unit of Mix C
Work Step by Step
Mix A: 4 protein, 6 fat, 3 carb (represented by $x$)
Mix B: 6 protein, 1 fat, 2 carb (represented by $y$)
Mix C: 4 protein, 1 fat, 12 carb (represented by $z$)
Must have 30 protein, 16 fat, 24 carb
$4x+6y+4z=30$
$6x+y+z=16$
$3x+2y+12z=24$
If we multiply the third equation by $-2$ and add the second equation, we will have the following
$4x+6y+4z=30$
$6x+y+z=16$
$(3x+2y+12z)*-2+(6x+y+z)=24*-2+16$
$-6x-4y-24z+6x+y+z=-48+16$
$-3y-23z=-32$
$4x+6y+4z=30$
$6x+y+z=16$
$-3y-23z=-32$
If we multiply the top equation by -1.5 and add the second equation, we will have the following
$(4x+6y+4z)*-1.5+6x+y+z=30*-1.5+16$
$6x+y+z=16$
$-3y-23z=-32$
$(4x+6y+4z)*-1.5+6x+y+z=30*-1.5+16$
$-6x-9y-6z+6x+y+z=-45+16$
$-8y-5z=-29$
$-8y-5z=-29$
$6x+y+z=16$
$-3y-23z=-32$
$-8y-5z=-29$
$-8y-5z+5z+29=-29+5z+29$
$-8y+29=5z$
$(-8y+29)/5 =5z/5$
$-8/5*y+29/5 = z$
$-3y-23z=-32$
$-3y-23(-8/5*y+29/5)=-32$
$-3y-23(-8/5*y+29/5)=-32$
$-3y+184y/5-667/5=-32$
$-3y+36.8y-133.4=-32$
$33.8y-133.4+32=-32+32$
$33.8y-101.4=0$
$33.8y-101.4+101.4=0+101.4$
$33.8y = 101.4$
$33.8y/33.8 = 101.4/33.8$
$y=3$
$-3y-23z=-32$
$-3*3-23z=-32$
$-9-23z=-32$
$-9-23z+32+23z=-32+32+23z$
$23 = 23z$
$23/23 = 23z/23$
$1=z$
$4x+6y+4z=30$
$4x+6*3+4*1=30$
$4x+18+4=30$
$4x+22=30$
$4x+22-22=30-22$
$4x=8$
$4x/4 =8/4$
$x=2$
