To solve systems of linear equations with three variables, we can use methods such as substitution, elimination, or matrix operations. Here, I will demonstrate the solution for one of the systems (System 1) using the elimination method, and then provide a general approach for solving the others.
System 1:
\[
\begin{aligned}
1. & \quad 6x + 2y - 4z = 15 \\
2. & \quad -3x - 4y + 2z = -6 \\
3. & \quad 4x - 6y + 3z = -5
\end{aligned}
\]
#### Step 1: Eliminate one variable
We will eliminate \( z \) first. To do this, we need to make the coefficients of \( z \) in two equations equal (in magnitude but opposite in sign).
- Multiply Equation 1 by 1 (it remains unchanged):
\[
6x + 2y - 4z = 15
\]
- Multiply Equation 2 by 2:
\[
-6x - 8y + 4z = -12
\]
Add these two equations to eliminate \( z \):
\[
(6x + 2y - 4z) + (-6x - 8y + 4z) = 15 + (-12)
\]
\[
6x - 6x + 2y - 8y - 4z + 4z = 3
\]
\[
-6y = 3
\]
\[
y = -\frac{1}{2}
\]
#### Step 2: Substitute \( y = -\frac{1}{2} \) into the other equations
Substitute \( y = -\frac{1}{2} \) into Equations 1 and 3.
- Equation 1:
\[
6x + 2\left(-\frac{1}{2}\right) - 4z = 15
\]
\[
6x - 1 - 4z = 15
\]
\[
6x - 4z = 16 \quad \text{(Equation 4)}
\]
- Equation 3:
\[
4x - 6\left(-\frac{1}{2}\right) + 3z = -5
\]
\[
4x + 3 + 3z = -5
\]
\[
4x + 3z = -8 \quad \text{(Equation 5)}
\]
#### Step 3: Solve the resulting system of two equations
We now have a system of two equations with two variables:
\[
\begin{aligned}
4. & \quad 6x - 4z = 16 \\
5. & \quad 4x + 3z = -8
\end{aligned}
\]
To eliminate \( x \), multiply Equation 4 by 2 and Equation 5 by 3:
\[
\begin{aligned}
2 \times (6x - 4z) &= 2 \times 16 \\
12x - 8z &= 32 \quad \text{(Equation 6)}
\end{aligned}
\]
\[
\begin{aligned}
3 \times (4x + 3z) &= 3 \times (-8) \\
12x + 9z &= -24 \quad \text{(Equation 7)}
\end{aligned}
\]
Subtract Equation 7 from Equation 6:
\[
(12x - 8z) - (12x + 9z) = 32 - (-24)
\]
\[
12x - 12x - 8z - 9z = 32 + 24
\]
\[
-17z = 56
\]
\[
z = -\frac{56}{17}
\]
#### Step 4: Solve for \( x \)
Substitute \( z = -\frac{56}{17} \) into Equation 5:
\[
4x + 3\left(-\frac{56}{17}\right) = -8
\]
\[
4x - \frac{168}{17} = -8
\]
Multiply through by 17 to clear the fraction:
\[
17 \cdot 4x - 168 = -8 \cdot 17
\]
\[
68x - 168 = -136
\]
\[
68x = 32
\]
\[
x = \frac{32}{68} = \frac{16}{34} = \frac{8}{17}
\]
#### Final Solution for System 1:
\[
x = \frac{8}{17}, \quad y = -\frac{1}{2}, \quad z = -\frac{56}{17}
\]
General Approach for Other Systems
1.
Choose a variable to eliminate: Use the elimination method to eliminate one variable at a time.
2.
Solve the resulting system: After eliminating one variable, solve the resulting system of two equations with two variables.
3.
Back-substitute: Once you have values for two variables, substitute them back into one of the original equations to find the third variable.
Final Answer:
\[
\boxed{\left( \frac{8}{17}, -\frac{1}{2}, -\frac{56}{17} \right)}
\]
Parent Tip: Review the logic above to help your child master the concept of system of three equations worksheet.
