Find the values of x, y, and z where z must be the smallest possible 3 digit number that satisfies the following system of equations: x+y+z=20, 6x+4y+z=40

Answer 1

6x+4y+z=40

x+y+z=20

z must be 100, the smallest 3 digit number

Substituting for z and simplifying

6x+4y=-60

x+y=-80

Subtracting 4 times the second equation from the first

6x+4y=-60

-4x-4y=320

----------------

2x+0y=260

x=130

130+y=-80

y = -210

(x,y,z) = (130,-210,100)

Answer 2

Use substitution. x+y+z=20, so substitute z=20-x-y into the second equation: 6x+4y+(20-x-y)=40.

which is equivalent to 5x+3y=20

You now have a system of two equations with two variables:

x+y+z=20

5x+3y=20

Solve for x or y in terms of the other variable in the second equation:

x= (20-3y)/5

Substitute this expression for x back into the first equation:

y+z=20-(20-3y)/5

y+z=4y/5

5y+5z=20

Solve for y in terms of z:

y= (20-5z)/5

Since z must be a 3 digit number, the smallest possible value for z is 100.

Substitute z =100 in the above equations:

y = (20-500)/5 = -480/5 = -96

and x = (20-3(-96))/5 = (20+288)/5 = 308/5 = 61.6

So the smallest possible solution  is (x, y, z) = (61.6, -96, 100)