System of Equations

Maths
1

:abacus: System of Equations

\begin{align*} \text{(1)}\quad & 2x + y - z = 8 \\ \text{(2)}\quad & x - 3y + 2z = -3 \\ \text{(3)}\quad & -x + y + 5z = 7 \end{align*}

:white_check_mark: Step 1: Solve Equation (2) for x

x = 3y - 2z - 3

:white_check_mark: Step 2: Substitute into Equations (1) and (3)

Substitute into (1):

2(3y - 2z - 3) + y - z = 8
6y - 4z - 6 + y - z = 8
7y - 5z = 14 \quad \text{(Equation 4)}

Substitute into (3):

-(3y - 2z - 3) + y + 5z = 7
-3y + 2z + 3 + y + 5z = 7
-2y + 7z = 4 \quad \text{(Equation 5)}

:white_check_mark: Step 3: Solve the 2-variable System

\text{Equation 4: } 7y - 5z = 14 \\ \text{Equation 5: } -2y + 7z = 4

Multiply to eliminate y:

2(7y - 5z) = 14 \cdot 2 \Rightarrow 14y - 10z = 28 \\ 7(-2y + 7z) = 4 \cdot 7 \Rightarrow -14y + 49z = 28

Add:

(14y - 10z) + (-14y + 49z) = 28 + 28 \\ 39z = 56 \Rightarrow z = \frac{56}{39}

:white_check_mark: Step 4: Solve for y

From:

7y - 5z = 14 \Rightarrow y = \frac{14 + 5z}{7}

Substitute z = \frac{56}{39} :

y = \frac{14 + 5 \cdot \frac{56}{39}}{7} = \frac{14 + \frac{280}{39}}{7} = \frac{\frac{546 + 280}{39}}{7} = \frac{826}{273}

:white_check_mark: Step 5: Solve for x

From:

x = 3y - 2z - 3

Substitute y = \frac{826}{273},\ z = \frac{56}{39} :

3y:

3 \cdot \frac{826}{273} = \frac{2478}{273}

2z:

2 \cdot \frac{56}{39} = \frac{112}{39} = \frac{784}{273}

Then:

x = \frac{2478 - 784 - 819}{273} = \frac{875}{273}

:green_square: Final Answer

x = \frac{875}{273}, \quad y = \frac{826}{273}, \quad z = \frac{56}{39}