Answer:
- dependent system
- x = 2 -a
- y = 1 +a
- z = a
Step-by-step explanation:
Let's solve this by eliminating z, then we'll go from there.
Add 6 times the second equation to the first.
(3x -3y +6z) +6(x +2y -z) = (3) +6(4)
9x +9y = 27 . . . simplify
x + y = 3 . . . . . . divide by 9 [eq4]
Add 13 times the second equation to the third.
(5x -8y +13z) +13(x +2y -z) = (2) +13(4)
18x +18y = 54
x + y = 3 . . . . . . divide by 18 [eq5]
Equations [eq4] and [eq5] are identical. This tells us the system is dependent, and has an infinite number of solutions. We can find them in terms of z:
y = 3 -x . . . . solve eq5 for y
x +2(3 -x) -z = 4 . . . . substitute into the second equation
-x +6 -z = 4
x = 2 - z . . . . . . add x-4
y = 3 -(2 -z)
y = z +1
So far, we have written the solutions in terms of z. If we use the parameter "a", we can write the solutions as ...
x = 2 -a
y = 1 +a
z = a
_____
<em>Check</em>
First equation:
3(2-a) -3(a+1) +6a = 3
6 -3a -3a -3 +6a = 3 . . . true
Second equation:
(2-a) +2(a+1) -a = 4
2 -a +2a +2 -a = 4 . . . true
Third equation:
5(2-a) -8(a+1) +13a = 2
10 -5a -8a -8 +13a = 2 . . . true
Our solution checks algebraically.
are solutions of a given quadratic equation:
is the completely factored form of 
.
, then:
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