Systems of Equations: Comprehensive Guide
Table of Contents
Introduction to Systems of Equations
A system of equations consists of two or more equations with the same variables. The solution to a system of equations is the set of values for the variables that satisfies all equations in the system simultaneously.
For example, in a system of two linear equations with two unknowns:
2x + y = 5
x - y = 1
The solution is the ordered pair (x, y) that makes both equations true, which in this case is (2, 1).
Types of Systems
Linear Systems
All equations in the system are linear (degree 1 in all variables).
3x + 2y = 8
x - y = 3
Non-linear Systems
At least one equation in the system is non-linear (contains variables with degree higher than 1).
x² + y² = 25
y = x + 1
Number of Equations/Variables
- 2×2 systems: Two equations with two unknowns
- 3×3 systems: Three equations with three unknowns
- m×n systems: m equations with n unknowns
Solution Methods
Substitution Method
The substitution method involves these steps:
- Solve one equation for one variable in terms of the other(s)
- Substitute this expression into the other equation(s)
- Solve the resulting equation
- Back-substitute to find the values of the other variables
Example:
Given the system:
x + 2y = 10
3x - y = 5
Step 1: Solve the first equation for x: x = 10 - 2y
Step 2: Substitute into the second equation: 3(10 - 2y) - y = 5
Step 3: Simplify and solve: 30 - 6y - y = 5
30 - 7y = 5
-7y = -25
y = 25/7
Step 4: Back-substitute to find x: x = 10 - 2(25/7) = 10 - 50/7 = 70/7 - 50/7 = 20/7
Solution: (x, y) = (20/7, 25/7)
Elimination Method
The elimination method (also called addition method) involves these steps:
- Multiply one or both equations by constants to make the coefficients of one variable the same but opposite in sign
- Add the equations to eliminate one variable
- Solve for the remaining variable
- Substitute back to find the other variable(s)
Example:
Given the system:
4x + 3y = 10
2x - 5y = 12
Step 1: Multiply the second equation by 2: 4x - 10y = 24
Step 2: Add the equations: 4x + 3y + 4x - 10y = 10 + 24
8x - 7y = 34
Step 3: Solve for x in terms of y: 8x = 34 + 7y, so x = (34 + 7y)/8
Step 4: Substitute into original equation: 4((34 + 7y)/8) + 3y = 10
Simplify: 4(34 + 7y)/8 + 3y = 10
(34 + 7y)/2 + 3y = 10
17 + 3.5y + 3y = 10
17 + 6.5y = 10
6.5y = -7
y = -7/6.5 = -14/13
Step 5: Find x by substituting y: x = (34 + 7(-14/13))/8 = (34 - 98/13)/8 = (442/13 - 98/13)/8 = 344/104 = 43/13
Solution: (x, y) = (43/13, -14/13)
Graphical Method
The graphical method involves:
- Graph each equation in the system
- Find the point(s) of intersection, if any
- These points are the solutions to the system
For a system of linear equations in two variables, there are three possible outcomes:
- One solution: The lines intersect at exactly one point
- No solution: The lines are parallel (inconsistent system)
- Infinitely many solutions: The lines are the same (dependent system)
Example:
Given the system:
y = 2x - 4
y = -x + 5
When we graph these equations, we see they intersect at the point (3, 2), which is the solution.
Matrix Method
For a system of linear equations, we can use matrices and row operations (Gaussian elimination):
- Write the system in augmented matrix form
- Use elementary row operations to convert to row echelon form
- Back-substitute to find the variables
Example:
Given the system:
2x + y - z = 8
-3x + y + 2z = -11
x + 2y + 3z = 7
Step 1: Write as augmented matrix:
[ 2 1 -1 | 8 ] [ -3 1 2 | -11 ] [ 1 2 3 | 7 ]
Step 2: Convert to row echelon form using row operations (details omitted):
[ 1 0 0 | 2 ] [ 0 1 0 | 1 ] [ 0 0 1 | 1 ]
Solution: (x, y, z) = (2, 1, 1)
Cramer's Rule
Cramer's Rule uses determinants to solve systems of linear equations:
- Create the coefficient matrix D and the matrices Dx, Dy, etc. by replacing columns with the constants
- Calculate the determinants
- The solution is x = |Dx|/|D|, y = |Dy|/|D|, etc.
Example:
Given the system:
3x + 2y = 7
x - y = 2
Step 1: Calculate the determinant of the coefficient matrix:
|D| = |3 2| = 3(-1) - 2(1) = -3 - 2 = -5
| 1 -1|
Step 2: Calculate the determinant of Dx:
|Dx| = |7 2| = 7(-1) - 2(2) = -7 - 4 = -11
| 2 -1|
Step 3: Calculate the determinant of Dy:
|Dy| = |3 7| = 3(2) - 7(1) = 6 - 7 = -1
| 1 2|
Step 4: Apply Cramer's Rule:
x = |Dx|/|D| = (-11)/(-5) = 11/5 = 2.2
y = |Dy|/|D| = (-1)/(-5) = 1/5 = 0.2
Solution: (x, y) = (11/5, 1/5)
Special Cases
Consistent and Inconsistent Systems
A system of equations can be:
- Consistent: Has at least one solution
- Inconsistent: Has no solution
Independent and Dependent Systems
A consistent system can be:
- Independent: Has exactly one solution
- Dependent: Has infinitely many solutions (some equations are linear combinations of others)
Examples:
Independent system (one solution):
x + y = 5
2x - y = 4
Solution: (x, y) = (3, 2)
Dependent system (infinitely many solutions):
2x + 3y = 6
4x + 6y = 12
Solution: All points on the line 2x + 3y = 6
Inconsistent system (no solution):
x + y = 7
x + y = 10
No solution (contradictory equations)
Real-world Applications
Example 1: Mixture Problems
A chemist needs to mix a 30% acid solution with a 50% acid solution to get 10 liters of a 45% acid solution. How much of each solution should be used?
Solution:
Let x = amount of 30% solution and y = amount of 50% solution
We need:
x + y = 10 (total volume)
0.3x + 0.5y = 0.45 × 10 = 4.5 (acid content)
Solving this system:
y = 10 - x
0.3x + 0.5(10 - x) = 4.5
0.3x + 5 - 0.5x = 4.5
-0.2x = -0.5
x = 2.5
y = 10 - 2.5 = 7.5
Therefore, the chemist needs 2.5 liters of the 30% solution and 7.5 liters of the 50% solution.
Example 2: Cost-Revenue Problems
A company's cost function is C(x) = 50x + 2000 and its revenue function is R(x) = 80x, where x is the number of units produced and sold. Find the break-even point.
Solution:
At the break-even point, cost equals revenue:
C(x) = R(x)
50x + 2000 = 80x
2000 = 30x
x = 66.67
Since fractional units don't make sense here, the company breaks even after selling 67 units.