You'd find this problem easier to understand and do if you'd please list the defining equations vertically and line up variables:
1. x - 1y + 2 z = -7
2. y + 1z = 1
3. x - 2 y - 3 z = 0 Now eliminate the line numbers:
x - 1y + 2 z = -7
1y + 1 z = 1
x - 2 y - 3 z = 0
Let's use the elimination method to eliminate variable z: Seeing that z = 1 - y, we transform the first equation into 1x - 1y + 2(1-y) = -7
and the third into x - 2y - 3(1-y) = 0.
Simplifying 1x - 1y + 2(1-y) = -7
and x - 2y - 3(1-y) = 0,
we get
1x - 2y - 3 + 3y) = 0 and 1x - 1y + 2 - 2y = -7
which in turn simplify to
1x + y = 3 and 1x - 3y = -9
Having eliminated the variable z, we now focus on eliminating x. Mult. the 1st equation by -1, obtaining -1x - 1y = -3. Add this result to 1x - 3y = -9:
0 - 4y = -12, which tells us that y = 3. Subbing 3 for y in 1x + 1y = 3 tells us that x = 0.
All we have left to determine is the vaue of z.
Borrowing Equation 3, from above, we get x - 2 y - 3 z = 0, and into this equation we substitute x = 0 and y = 3: 0 -2(3) - 3z = 0.
Thus, -3z = 6, and z = -2.
The solution set is (0, 3, -2). You should check this by substitution.
. If there are 50 fish, how many grams of fish food will you need?
