3 Variable System of Equations Problems and Answers
Systems of equations with three variables are fundamental in algebra and have wide applications in fields like engineering, economics, and physics. These systems consist of three equations involving three unknowns (typically denoted as x, y, and z), and solving them requires a combination of algebraic techniques to find values that satisfy all equations simultaneously. Below, we explore three representative problems with detailed solutions, demonstrating methods such as substitution and elimination.
Introduction to 3-Variable Systems
A 3-variable system of equations typically takes the form:
[ \begin{cases} a_1x + b_1y + c_1z = d_1 \ a_2x + b_2y + c_2z = d_2 \ a_3x + b_3y + c_3z = d_3 \ \end{cases} ]
The goal is to find values of x, y, and z that satisfy all three equations. These systems can have one unique solution, no solution (inconsistent system), or infinitely many solutions (dependent system).
Problem 1: Solving a 3-Variable System Using Substitution
Problem:
Solve the following system of equations:
- ( 2x + y - z = 3 )
- ( x - y + 2z = 1 )
- ( 3x + 2y + z = 11 )
Solution:
Step 1: Solve one equation for one variable.
From Equation 1:
[
2x + y - z = 3 \implies y = 3 - 2x + z
]
Step 2: Substitute ( y ) into the other two equations.
Substitute into Equation 2:
[
x - (3 - 2x + z) + 2z = 1 \implies x - 3 + 2x - z + 2z = 1 \implies 3x + z = 4
]
New Equation 2: ( 3x + z = 4 )
Substitute into Equation 3:
[
3x + 2(3 - 2x + z) + z = 11 \implies 3x + 6 - 4x + 2z + z = 11 \implies -x + 3z = 5
]
New Equation 3: ( -x + 3z = 5 )
Step 3: Solve the reduced system of two equations.
From Equation 2: ( 3x + z = 4 \implies z = 4 - 3x )
Substitute ( z = 4 - 3x ) into Equation 3:
[
-x + 3(4 - 3x) = 5 \implies -x + 12 - 9x = 5 \implies -10x = -7 \implies x = \frac{7}{10}
]
Step 4: Find ( z ).
[
z = 4 - 3x = 4 - 3\left(\frac{7}{10}\right) = 4 - \frac{21}{10} = \frac{19}{10}
]
Step 5: Find ( y ).
[
y = 3 - 2x + z = 3 - 2\left(\frac{7}{10}\right) + \frac{19}{10} = 3 - \frac{14}{10} + \frac{19}{10} = 3 + \frac{5}{10} = \frac{35}{10} = \frac{7}{2}
]
Final Answer:
[ x = \frac{7}{10}, \quad y = \frac{7}{2}, \quad z = \frac{19}{10} ]
Problem 2: Solving via Elimination Method
Problem:
Solve the system:
- ( x + 2y - z = 1 )
- ( 2x - y + 3z = 10 )
- ( 3x + y + 2z = 15 )
Solution:
Step 1: Eliminate one variable by combining equations.
Add Equations 1 and 2 to eliminate ( y ):
[
(x + 2y - z) + (2x - y + 3z) = 1 + 10 \implies 3x + y + 2z = 11 \quad \text{(Equation 4)}
]
Step 2: Subtract Equation 4 from Equation 3 to eliminate ( y ):
[
(3x + y + 2z) - (3x + y + 2z) = 15 - 11 \implies 0 = 4
]
Wait—this yields an inconsistency. Let’s recheck calculations.
Revised Step 1: Let’s instead eliminate ( z ). Multiply Equation 1 by 3 and Equation 2 by 1:
Equation 1 × 3: ( 3x + 6y - 3z = 3 )
Equation 2: ( 2x -
Continuing Problem 2 – Solving via Elimination
After revisiting the algebraic manipulations, the correct path is to recognise that the third equation is actually the sum of the first two Most people skip this — try not to..
Corrected Step 1:
Multiply the first equation by 3 and add it to the second:
[ 3,(x+2y-z)=3\cdot1;\Longrightarrow;3x+6y-3z=3, ] [ 2x-y+3z=10. ]
Adding these two results eliminates (y) and yields
[ 5x+3z=13\qquad\text{(call this Equation 4)}. ]
**Corrected Step 2
