How to Solve System of Equations with Three Variables
Solving a system of equations with three variables involves finding the values of three unknowns that satisfy all equations simultaneously. These systems are fundamental in algebra, physics, engineering, and economics, where multiple variables interact to determine outcomes. The general form of a system with three variables ( x ), ( y ), and ( z ) is:
[ \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 determine the values of ( x ), ( y ), and ( z ) that make all three equations true. And this process can be accomplished using substitution, elimination, or matrix methods. Below, we explore each method in detail, along with examples and practical applications Practical, not theoretical..
Introduction
Systems of equations with three variables represent scenarios where three conditions must be met simultaneously. Now, for instance, in economics, three products might have different costs, revenues, and profit margins, and you might need to find quantities that balance all three. In engineering, stress, strain, and temperature might interact in a system. Solving such systems requires systematic methods to reduce complexity and isolate variables.
Method 1: Substitution
The substitution method involves solving one equation for one variable and substituting the result into the other equations. This reduces the system to two equations with two variables, which can then be solved recursively Turns out it matters..
Steps:
- Choose an equation and solve for one variable.
- Substitute this expression into the remaining two equations.
- Solve the resulting two-variable system using substitution or elimination.
- Back-substitute to find the third variable.
Example:
Solve the system:
[
\begin{cases}
2x + y - z = 5 \quad \text{(1)} \
x - y + 2z = 3 \quad \text{(2)} \
3x + 2y + z = 10 \quad \text{(3)}
\end{cases}
]
Step 1: Solve equation (1) for ( y ):
[
y = 5 - 2x + z
]
Step 2: Substitute ( y ) into equations (2) and (3):
- Equation (2): ( x - (5 - 2x + z) + 2z = 3 )
Simplify: ( x - 5 + 2x - z + 2z = 3 ) → ( 3x + z = 8 \quad \text{(2a)} ) - Equation (3): ( 3x + 2(5 - 2x + z) + z = 10 )
Simplify: ( 3x + 10 - 4x + 2z + z = 10 ) → ( -x + 3z = 0 \quad \text{(3a)} )
Step 3: Solve the system (2a) and (3a):
[
\begin{cases}
3x + z = 8 \
-x + 3z = 0
\end{cases}
]
From (3a): ( x = 3z ). Substitute into (2a):
( 3(3z) + z = 8 ) → ( 10z = 8 ) → ( z = 0.8 ).
Then, ( x = 3(0.8) = 2.4 ) Still holds up..
Step 4: Find ( y ) using ( y = 5 - 2x + z ):
( y = 5 - 2(2.4) + 0.8 = 5 - 4.8 + 0.8 = 1 ).
Solution: ( x = 2.4 ), ( y = 1 ), ( z = 0.8 ).
Method 2: Elimination
The elimination method systematically removes variables by adding or subtracting equations. This method is particularly useful when coefficients are integers or easily manipulated.
Steps:
- Align equations to eliminate one variable.
- Solve the resulting two-variable system.
- Back-substitute to find the third variable.
Example:
Using the same system as before:
[
\begin{cases}
2x + y - z = 5 \quad \text{(1)} \
x - y + 2z = 3 \quad \text{(2)} \
3x + 2y + z = 10 \quad \text{(3)}
\end{cases}
]
Step 1: Eliminate ( y ) from equations (1) and (2). Multiply equation (2) by 1 and add to equation (1):
Equation (1) + Equation (2):
( (2x + y - z) + (x - y + 2z) = 5 + 3 )
Simplify: ( 3x + z = 8 \quad \text{(4)} )
Step 2: Eliminate ( y ) from equations (2) and (3). Multiply equation (2) by 2:
Equation (3) - 2 × Equation (2):
( (3x + 2
