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Solve the following system of equations algebraically for [tex]\( x \)[/tex], [tex]\( y \)[/tex], and [tex]\( z \)[/tex].

[tex]\[
\begin{aligned}
2x + 4y - 3z & = 12 \quad \rightarrow \text{Equation 1} \\
-3x - 2y + 2z & = -9 \quad \rightarrow \text{Equation 2} \\
-x + y - 3z & = 0 \quad \rightarrow \text{Equation 3}
\end{aligned}
\][/tex]

[tex]\[
2x + 4y - 3z = 12
\][/tex]

[tex]\[
-x + y - 3z = 0
\][/tex]

[tex]\[
\begin{aligned}
x + 5y & = 12 \\
3x - 2(x + 3z) + 2z & = -9 \\
y & = x + 3z
\end{aligned}
\][/tex]

Sagot :

GinnyAnsweridn
To solve the given system of equations, we have:

[tex]\[ \begin{aligned} 1: & \quad 2x + 4y - 3z = 12 \\ 2: & \quad -3x - 2y + 2z = -9 \\ 3: & \quad -x + y - 3z = 0 \end{aligned} \][/tex]

We need to find values for [tex]\( x \)[/tex], [tex]\( y \)[/tex], and [tex]\( z \)[/tex] that satisfy all three equations. Let's proceed step-by-step.

### Step 1: Solve for One Variable

Begin with equation [tex]\(3\)[/tex]:

[tex]\[ -x + y - 3z = 0 \][/tex]

Rearrange to solve for one variable, say [tex]\( y \)[/tex]:

[tex]\[ y = x + 3z \quad \text{(Equation 3')} \][/tex]

### Step 2: Substitute [tex]\( y \)[/tex] into Equations 1 and 2

Substitute [tex]\( y = x + 3z \)[/tex] into equations 1 and 2:

#### For Equation 1:

[tex]\[ \begin{aligned} 2x + 4y - 3z &= 12 \\ 2x + 4(x + 3z) - 3z &= 12 \\ 2x + 4x + 12z - 3z &= 12 \\ 6x + 9z &= 12 \\ \end{aligned} \][/tex]

Simplify:

[tex]\[ 6x + 9z = 12 \quad \text{(Equation 1')} \][/tex]

Divide the entire equation by 3:

[tex]\[ 2x + 3z = 4 \quad \text{(Equation 1'')} \][/tex]

#### For Equation 2:

[tex]\[ \begin{aligned} -3x - 2y + 2z &= -9 \\ -3x - 2(x + 3z) + 2z &= -9 \\ -3x - 2x - 6z + 2z &= -9 \\ -5x - 4z &= -9 \\ \end{aligned} \][/tex]

Simplify:

[tex]\[ -5x - 4z = -9 \quad \text{(Equation 2')} \][/tex]

### Step 3: Solve the Simplified System of Two Variables

Now we have a simpler system of two equations with two variables:

[tex]\[ \begin{aligned} 1'': & \quad 2x + 3z = 4 \\ 2': & \quad -5x - 4z = -9 \end{aligned} \][/tex]

Let's solve these equations simultaneously.

#### Solve Equation 1'' for [tex]\( x \)[/tex]:

[tex]\[ 2x + 3z = 4 \\ 2x = 4 - 3z \\ x = \frac{4 - 3z}{2} \quad \text{(Equation 4)} \][/tex]

#### Substitute Equation 4 into Equation 2':

[tex]\[ -5 \left( \frac{4 - 3z}{2} \right) - 4z = -9 \\ - \frac{20 - 15z}{2} - 4z = -9 \\ -10 + \frac{15z}{2} - 4z = -9 \\ -10 + \frac{15z}{2} - \frac{8z}{2} = -9 \\ -10 + \frac{7z}{2} = -9 \\ \frac{7z}{2} = 1 \\ 7z = 2 \\ z = \frac{2}{7} \][/tex]

### Step 4: Substitute [tex]\( z \)[/tex] into Equation 4 to Find [tex]\( x \)[/tex]:

[tex]\[ x = \frac{4 - 3z}{2} \\ x = \frac{4 - 3 \left( \frac{2}{7} \right)}{2} \\ x = \frac{4 - \frac{6}{7}}{2} \\ x = \frac{\frac{28}{7} - \frac{6}{7}}{2} \\ x = \frac{\frac{22}{7}}{2} \\ x = \frac{22}{14} \\ x = \frac{11}{7} \][/tex]

### Step 5: Substitute [tex]\( x \)[/tex] and [tex]\( z \)[/tex] into Equation 3' to Find [tex]\( y \)[/tex]:

[tex]\[ y = x + 3z \\ y = \frac{11}{7} + 3 \left( \frac{2}{7} \right) \\ y = \frac{11}{7} + \frac{6}{7} \\ y = \frac{17}{7} \][/tex]

### Final Solution:

The solutions to the system of equations are:

[tex]\[ x = \frac{11}{7}, \quad y = \frac{17}{7}, \quad z = \frac{2}{7} \][/tex]

These values satisfy all the given equations.
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