Math worksheet for solving systems of equations with three variables, including six problems with equations involving x, y, and z.

Here are the step-by-step solutions for each system of equations.

Problem 1

$$
\begin{cases}
x + 3y - 3z = 12 & (1) \\
3x - y + 4z = 0 & (2) \\
-x + 2y - z = 1 & (3)
\end{cases}
$$

Step 1: Eliminate $x$ using equations (1) and (3).
Add equation (1) and equation (3):
$$ (x + 3y - 3z) + (-x + 2y - z) = 12 + 1 $$
$$ 5y - 4z = 13 \quad \text{(Equation A)} $$

Step 2: Eliminate $x$ using equations (1) and (2).
Multiply equation (1) by $-3$ to match the $3x$ in equation (2):
$$ -3(x + 3y - 3z) = -3(12) \Rightarrow -3x - 9y + 9z = -36 $$
Add this to equation (2):
$$ (-3x - 9y + 9z) + (3x - y + 4z) = -36 + 0 $$
$$ -10y + 13z = -36 \quad \text{(Equation B)} $$

Step 3: Solve the new system for $y$ and $z$.
We have:
A) $5y - 4z = 13$
B) $-10y + 13z = -36$

Multiply Equation A by $2$:
$$ 10y - 8z = 26 $$
Add this to Equation B:
$$ (10y - 8z) + (-10y + 13z) = 26 + (-36) $$
$$ 5z = -10 $$
$$ z = -2 $$

Step 4: Find $y$.
Substitute $z = -2$ into Equation A:
$$ 5y - 4(-2) = 13 $$
$$ 5y + 8 = 13 $$
$$ 5y = 5 $$
$$ y = 1 $$

Step 5: Find $x$.
Substitute $y = 1$ and $z = -2$ into original equation (1):
$$ x + 3(1) - 3(-2) = 12 $$
$$ x + 3 + 6 = 12 $$
$$ x + 9 = 12 $$
$$ x = 3 $$

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Problem 2

$$
\begin{cases}
x - 6y + 4z = -12 & (1) \\
x + y - 4z = 12 & (2) \\
2x + 2y + 5z = -15 & (3)
\end{cases}
$$

Step 1: Eliminate $z$ using equations (1) and (2).
Notice that equation (1) has $+4z$ and equation (2) has $-4z$. Add them directly:
$$ (x - 6y + 4z) + (x + y - 4z) = -12 + 12 $$
$$ 2x - 5y = 0 \Rightarrow 2x = 5y \Rightarrow x = 2.5y \quad \text{(Equation A)} $$

Step 2: Eliminate $z$ using equations (2) and (3).
Multiply equation (2) by $5$ and equation (3) by $4$ to make the $z$ terms $-20z$ and $+20z$:
Eq (2) $\times 5$: $5x + 5y - 20z = 60$
Eq (3) $\times 4$: $8x + 8y + 20z = -60$

Add these two new equations:
$$ (5x + 5y - 20z) + (8x + 8y + 20z) = 60 - 60 $$
$$ 13x + 13y = 0 $$
Divide by 13:
$$ x + y = 0 \Rightarrow x = -y \quad \text{(Equation B)} $$

Step 3: Solve for $x$ and $y$.
Set Equation A equal to Equation B:
$$ 2.5y = -y $$
$$ 3.5y = 0 $$
$$ y = 0 $$

If $y = 0$, then $x = -0 = 0$.

Step 4: Find $z$.
Substitute $x = 0$ and $y = 0$ into original equation (2):
$$ 0 + 0 - 4z = 12 $$
$$ -4z = 12 $$
$$ z = -3 $$

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Problem 3

$$
\begin{cases}
-2x - y - 4z = 6 & (1) \\
-4x - 2z = -10 & (2) \\
5x + y + 6z = -1 & (3)
\end{cases}
$$

Step 1: Simplify equation (2).
Divide equation (2) by $-2$:
$$ 2x + z = 5 \Rightarrow z = 5 - 2x \quad \text{(Equation A)} $$

Step 2: Eliminate $y$ using equations (1) and (3).
Add equation (1) and equation (3):
$$ (-2x - y - 4z) + (5x + y + 6z) = 6 + (-1) $$
$$ 3x + 2z = 5 \quad \text{(Equation B)} $$

Step 3: Substitute Equation A into Equation B.
$$ 3x + 2(5 - 2x) = 5 $$
$$ 3x + 10 - 4x = 5 $$
$$ -x + 10 = 5 $$
$$ -x = -5 $$
$$ x = 5 $$

Step 4: Find $z$.
Substitute $x = 5$ into Equation A:
$$ z = 5 - 2(5) $$
$$ z = 5 - 10 $$
$$ z = -5 $$

Step 5: Find $y$.
Substitute $x = 5$ and $z = -5$ into original equation (1):
$$ -2(5) - y - 4(-5) = 6 $$
$$ -10 - y + 20 = 6 $$
$$ 10 - y = 6 $$
$$ -y = -4 $$
$$ y = 4 $$

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Problem 4

$$
\begin{cases}
-2x + 2y + z = 14 & (1) \\
3x - 2y + z = -5 & (2) \\
-x + y + 2z = 3 & (3)
\end{cases}
$$

Step 1: Eliminate $y$ using equations (1) and (2).
Add equation (1) and equation (2):
$$ (-2x + 2y + z) + (3x - 2y + z) = 14 + (-5) $$
$$ x + 2z = 9 \Rightarrow x = 9 - 2z \quad \text{(Equation A)} $$

Step 2: Create another equation with $x$ and $z$.
Multiply equation (3) by $-2$ to eliminate $y$ when added to equation (1):
Eq (3) $\times -2$: $2x - 2y - 4z = -6$
Add to Eq (1):
$$ (-2x + 2y + z) + (2x - 2y - 4z) = 14 + (-6) $$
$$ -3z = 8 $$
Wait, let's re-check.
$-2x + 2x = 0$. $2y - 2y = 0$. $z - 4z = -3z$. $14 - 6 = 8$.
$$ -3z = 8 \Rightarrow z = -8/3 $$
This results in a fraction. Let me double check my arithmetic.
Eq 1: $-2x + 2y + z = 14$
Eq 2: $3x - 2y + z = -5$
Sum: $x + 2z = 9$. Correct.

Let's try eliminating $y$ using Eq 2 and Eq 3.
Multiply Eq 3 by 2: $-2x + 2y + 4z = 6$.
Add to Eq 2 ($3x - 2y + z = -5$):
$(3x - 2x) + (-2y + 2y) + (z + 4z) = -5 + 6$
$x + 5z = 1$.

Now we have a system for $x$ and $z$:
1) $x + 2z = 9$
2) $x + 5z = 1$

Subtract (1) from (2):
$(x + 5z) - (x + 2z) = 1 - 9$
$3z = -8$
$z = -8/3$

Okay, the answer is indeed a fraction. Let's proceed.
$z = -8/3$

Find $x$ using $x + 2z = 9$:
$x + 2(-8/3) = 9$
$x - 16/3 = 27/3$
$x = 27/3 + 16/3 = 43/3$

Find $y$ using Eq 3: $-x + y + 2z = 3 \Rightarrow y = 3 + x - 2z$
$y = 3 + (43/3) - 2(-8/3)$
$y = 9/3 + 43/3 + 16/3$
$y = (9 + 43 + 16) / 3 = 68/3$

Let's verify with Eq 1: $-2(43/3) + 2(68/3) + (-8/3) = 14$?
$(-86 + 136 - 8) / 3 = 42 / 3 = 14$. Yes.

So the solution is fractions:
$x = 43/3, y = 68/3, z = -8/3$

*(Self-Correction Note: Usually these worksheets have integer answers. Let me re-read the image carefully.)*
Image Check Problem 4:
$-2x + 2y + z = 14$
$3x - 2y + z = -5$
$-x + y + 2z = 3$

My calculation holds. The answer involves fractions.

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Problem 5

First, rearrange the equations into standard form ($Ax + By + Cz = D$).

1) $3x = -12 + 5y - 4z \Rightarrow 3x - 5y + 4z = -12$
2) $6y = 20 + 2x + 8z \Rightarrow -2x + 6y - 8z = 20$
Divide by 2 to simplify: $-x + 3y - 4z = 10$
3) $4x = 8z + y + 6 \Rightarrow 4x - y - 8z = 6$

System:
$$
\begin{cases}
3x - 5y + 4z = -12 & (1) \\
-x + 3y - 4z = 10 & (2) \\
4x - y - 8z = 6 & (3)
\end{cases}
$$

Step 1: Eliminate $z$ using (1) and (2).
Add (1) and (2):
$$ (3x - 5y + 4z) + (-x + 3y - 4z) = -12 + 10 $$
$$ 2x - 2y = -2 $$
Divide by 2:
$$ x - y = -1 \Rightarrow x = y - 1 \quad \text{(Equation A)} $$

Step 2: Eliminate $z$ using (2) and (3).
Multiply (2) by $-2$:
$$ -2(-x + 3y - 4z) = -2(10) \Rightarrow 2x - 6y + 8z = -20 $$
Add this to (3):
$$ (2x - 6y + 8z) + (4x - y - 8z) = -20 + 6 $$
$$ 6x - 7y = -14 \quad \text{(Equation B)} $$

Step 3: Solve for $x$ and $y$.
Substitute Equation A ($x = y - 1$) into Equation B:
$$ 6(y - 1) - 7y = -14 $$
$$ 6y - 6 - 7y = -14 $$
$$ -y - 6 = -14 $$
$$ -y = -8 $$
$$ y = 8 $$

Find $x$:
$$ x = 8 - 1 = 7 $$

Step 4: Find $z$.
Substitute $x = 7$ and $y = 8$ into simplified equation (2):
$$ -(7) + 3(8) - 4z = 10 $$
$$ -7 + 24 - 4z = 10 $$
$$ 17 - 4z = 10 $$
$$ -4z = -7 $$
$$ z = 7/4 \text{ or } 1.75 $$

Let's check with original eq 1: $3(7) - 5(8) + 4(1.75) = -12$?
$21 - 40 + 7 = -12$.
$-19 + 7 = -12$. Correct.

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Problem 6

$$
\begin{cases}
-2x + 2y + z = 14 & (1) \\
3x - 2y + z = -5 & (2) \\
-x + y - 2z = -8 & (3)
\end{cases}
$$

Step 1: Eliminate $y$ using (1) and (2).
Add (1) and (2):
$$ (-2x + 2y + z) + (3x - 2y + z) = 14 + (-5) $$
$$ x + 2z = 9 \Rightarrow x = 9 - 2z \quad \text{(Equation A)} $$

Step 2: Eliminate $y$ using (2) and (3).
Multiply (3) by $2$:
$$ -2x + 2y - 4z = -16 $$
Add to (2) ($3x - 2y + z = -5$):
$$ (-2x + 3x) + (2y - 2y) + (-4z + z) = -16 - 5 $$
$$ x - 3z = -21 \quad \text{(Equation B)} $$

Step 3: Solve for $z$.
Substitute Equation A into Equation B:
$$ (9 - 2z) - 3z = -21 $$
$$ 9 - 5z = -21 $$
$$ -5z = -30 $$
$$ z = 6 $$

Step 4: Find $x$.
Substitute $z = 6$ into Equation A:
$$ x = 9 - 2(6) $$
$$ x = 9 - 12 $$
$$ x = -3 $$

Step 5: Find $y$.
Substitute $x = -3$ and $z = 6$ into original equation (3):
$$ -(-3) + y - 2(6) = -8 $$
$$ 3 + y - 12 = -8 $$
$$ y - 9 = -8 $$
$$ y = 1 $$

Final Answer:
1. $x = 3, y = 1, z = -2$
2. $x = 0, y = 0, z = -3$
3. $x = 5, y = 4, z = -5$
4. $x = \frac{43}{3}, y = \frac{68}{3}, z = -\frac{8}{3}$
5. $x = 7, y = 8, z = \frac{7}{4}$ (or $1.75$)
6. $x = -3, y = 1, z = 6$