The Central Idea
Consider
\[ \begin{cases} y=2x+1,\\ x+y=7. \end{cases} \]The first equation says that wherever the system is true,
\[ y \]and
\[ 2x+1 \]represent the same quantity.
So replace \(y\) in the second equation:
\[ x+(2x+1)=7. \]Then
\[ 3x+1=7, \] \[ 3x=6, \] \[ x=2. \]Back-substitute:
\[ y=2(2)+1=5. \]Therefore,
\[ \boxed{(x,y)=(2,5)}. \]Substitution works because equal quantities may replace one another.
Three Methods, One Goal
The three methods in this lesson are:
- substitution: replace one variable with an equivalent expression;
- comparison: set two expressions equal because both represent the same variable;
- elimination: combine equations so one variable cancels.
All three methods seek the same solution set.
Equivalent Systems
When we solve a system, we often replace it with a simpler system having exactly the same solutions.
Such systems are called equivalent systems.
For example,
\[ \begin{cases} x+y=7,\\ x-y=1 \end{cases} \]and
\[ \begin{cases} x=4,\\ x+y=7 \end{cases} \]have the same unique solution:
\[ \boxed{(4,3)}. \]Why Preserving the Solution Set Matters
A solving step is valid only if it preserves which values satisfy the system.
The equations may look different after substitution or elimination, but the set of simultaneous solutions must remain unchanged.
This idea is more important than any particular solving shortcut.
Substitution
Substitution is especially natural when one equation already isolates a variable.
For example,
\[ \begin{cases} y=3x-2,\\ 2x+y=13. \end{cases} \]The first equation immediately gives an expression for \(y\).
Substitution Example
Replace \(y\) in
\[ 2x+y=13 \]with
\[ 3x-2: \] \[ 2x+(3x-2)=13. \]Then
\[ 5x-2=13, \] \[ 5x=15, \] \[ x=3. \]Now
\[ y=3(3)-2=7. \]So
\[ \boxed{(3,7)}. \]Substitute for \(x\) Instead
Consider
\[ \begin{cases} x=8-2y,\\ 3x+y=14. \end{cases} \]Substitute
\[ 8-2y \]for \(x\):
\[ 3(8-2y)+y=14. \]Then
\[ 24-6y+y=14, \] \[ -5y=-10, \] \[ y=2. \]So
\[ x=8-2(2)=4. \]Therefore,
\[ \boxed{(4,2)}. \]Substitution When Nothing Is Isolated
Consider
\[ \begin{cases} 2x+y=9,\\ x-3y=-5. \end{cases} \]The first equation can be solved quickly for \(y\):
\[ y=9-2x. \]Then substitute into the second:
\[ x-3(9-2x)=-5. \]So
\[ x-27+6x=-5, \] \[ 7x=22, \] \[ x=\frac{22}{7}. \]Finish the Substitution Example
Use
\[ y=9-2x. \]Then
\[ y= 9-2\left(\frac{22}{7}\right) = \frac{63}{7}-\frac{44}{7} = \frac{19}{7}. \]Thus,
\[ \boxed{ \left(\frac{22}{7},\frac{19}{7}\right) }. \]Choose the Easier Variable to Isolate
For
\[ 4x+y=11, \]solving for \(y\) gives
\[ y=11-4x \]without introducing fractions.
Solving for \(x\) gives
\[ x=\frac{11-y}{4}, \]which is also valid but may make the next step less convenient.
Efficiency often comes from choosing the cleanest isolated variable.
Comparison
Comparison is especially natural when both equations are solved for the same variable.
For example,
\[ \begin{cases} y=2x+5,\\ y=-x+11. \end{cases} \]Both right-hand sides equal \(y\).
Therefore, they equal each other:
\[ 2x+5=-x+11. \]Comparison Example
Continue:
\[ 2x+5=-x+11. \]Add \(x\):
\[ 3x+5=11. \]Subtract \(5\):
\[ 3x=6. \]So
\[ x=2. \]Then
\[ y=2(2)+5=9. \]Therefore,
\[ \boxed{(2,9)}. \]Comparison Is a Form of Substitution
If
\[ y=f(x) \]and
\[ y=g(x), \]then
\[ f(x)=g(x). \]Comparison simply substitutes equal expressions into the same role.
It is the algebraic form of asking where two function graphs have the same output.
Comparison and Graphical Intersection
For
\[ y=2x+5 \]and
\[ y=-x+11, \]the comparison equation
\[ 2x+5=-x+11 \]finds the \(x\)-coordinate where the lines intersect.
So comparison is the symbolic version of graphical intersection.
Comparison with \(x\) Isolated
Comparison also works when both equations are solved for \(x\):
\[ \begin{cases} x=3y-1,\\ x=10-2y. \end{cases} \]Set the expressions equal:
\[ 3y-1=10-2y. \]Then
\[ 5y=11, \] \[ y=\frac{11}{5}. \]Substitute to find \(x\).
Finish the \(x\)-Comparison Example
Using
\[ x=3y-1, \]we get
\[ x= 3\left(\frac{11}{5}\right)-1 = \frac{33}{5}-\frac{5}{5} = \frac{28}{5}. \]Thus,
\[ \boxed{ \left(\frac{28}{5},\frac{11}{5}\right) }. \]Elimination
Elimination combines equations so that one variable disappears.
Consider
\[ \begin{cases} x+y=7,\\ x-y=1. \end{cases} \]Add the equations:
\[ 2x=8. \]Then
\[ x=4. \]Back-substitute:
\[ y=3. \]So
\[ \boxed{(4,3)}. \]Why Elimination Works
If
\[ L_1=R_1 \]and
\[ L_2=R_2, \]then adding the equations gives
\[ L_1+L_2=R_1+R_2. \]If opposite variable terms appear, they cancel.
This produces a simpler equivalent condition.
Elimination by Addition
Consider
\[ \begin{cases} 3x+2y=14,\\ x-2y=2. \end{cases} \]Add:
\[ 4x=16. \]So
\[ x=4. \]Substitute into
\[ x-2y=2: \] \[ 4-2y=2, \] \[ y=1. \]Thus,
\[ \boxed{(4,1)}. \]Elimination by Subtraction
Consider
\[ \begin{cases} 5x+3y=19,\\ 5x-y=7. \end{cases} \]Subtract the second equation from the first:
\[ (5x+3y)-(5x-y)=19-7. \]Then
\[ 4y=12, \] \[ y=3. \]Substitute:
\[ 5x-3=7, \] \[ 5x=10, \] \[ x=2. \]So
\[ \boxed{(2,3)}. \]Scaling Before Elimination
Sometimes no coefficients cancel immediately.
Consider
\[ \begin{cases} 2x+y=7,\\ 3x-2y=4. \end{cases} \]Multiply the first equation by \(2\):
\[ 4x+2y=14. \]Add to the second:
\[ 7x=18. \]So
\[ x=\frac{18}{7}. \]Finish the Scaled Example
Use
\[ 2x+y=7. \]Then
\[ 2\left(\frac{18}{7}\right)+y=7, \] \[ \frac{36}{7}+y=\frac{49}{7}, \] \[ y=\frac{13}{7}. \]Therefore,
\[ \boxed{ \left(\frac{18}{7},\frac{13}{7}\right) }. \]Scale Either Equation
In elimination, you may multiply either equation by any nonzero constant.
The goal is to create opposite coefficients.
For example, from
\[ \begin{cases} 3x+2y=10,\\ 5x+3y=16, \end{cases} \]you could eliminate \(x\) or \(y\).
Choose the route requiring simpler arithmetic.
Eliminate \(x\)
For
\[ \begin{cases} 3x+2y=10,\\ 5x+3y=16, \end{cases} \]multiply the first equation by \(5\):
\[ 15x+10y=50. \]Multiply the second by \(-3\):
\[ -15x-9y=-48. \]Add:
\[ y=2. \]Then
\[ 3x+4=10, \] \[ x=2. \]So
\[ \boxed{(2,2)}. \]Eliminate \(y\) Instead
Using the same system, multiply the first equation by \(3\):
\[ 9x+6y=30. \]Multiply the second by \(-2\):
\[ -10x-6y=-32. \]Add:
\[ -x=-2, \] \[ x=2. \]Then \(y=2\).
Different elimination paths lead to the same solution.
Choose the Easier Elimination
If one variable has coefficients
\[ 2 \]and
\[ -2, \]eliminate that variable immediately.
If coefficients are
\[ 3 \]and
\[ 5, \]you may need scaling.
Method choice is a matter of reducing arithmetic while preserving structure.
Clear Fractions Before Elimination
Consider
\[ \begin{cases} \frac{x}{2}+\frac{y}{3}=5,\\ x-y=1. \end{cases} \]Multiply the first equation by \(6\):
\[ 3x+2y=30. \]The equivalent system
\[ \begin{cases} 3x+2y=30,\\ x-y=1 \end{cases} \]is easier to eliminate.
Finish the Fraction Example
Multiply
\[ x-y=1 \]by \(2\):
\[ 2x-2y=2. \]Add:
\[ 5x=32. \]So
\[ x=\frac{32}{5}. \]Then
\[ y=x-1 = \frac{27}{5}. \]Thus,
\[ \boxed{ \left(\frac{32}{5},\frac{27}{5}\right) }. \]Clear Decimals When Helpful
Consider
\[ \begin{cases} 0.2x+0.5y=4,\\ x+y=10. \end{cases} \]Multiply the first equation by \(10\):
\[ 2x+5y=40. \]The system is now easier to manipulate exactly.
Elimination Can Reveal Classification
Elimination does more than find numerical solutions.
It can reveal:
- one solution;
- no solution;
- infinitely many solutions.
Elimination Producing a Contradiction
Consider
\[ \begin{cases} 2x+y=5,\\ 4x+2y=14. \end{cases} \]Multiply the first equation by \(-2\):
\[ -4x-2y=-10. \]Add:
\[ 0=4. \]This contradiction proves
\[ \boxed{\text{no solution}}. \]Elimination Producing an Identity
Consider
\[ \begin{cases} 2x+y=5,\\ 4x+2y=10. \end{cases} \]Multiply the first by \(-2\):
\[ -4x-2y=-10. \]Add:
\[ 0=0. \]The equations are dependent.
Infinitely many solutions remain.
Do Not Stop at \(0=0\)
The identity
\[ 0=0 \]does not describe the solution set by itself.
Use one remaining equation, such as
\[ 2x+y=5, \]to parameterize the solutions:
\[ y=5-2x. \]Let
\[ x=t. \]Then
\[ \boxed{ (x,y)=(t,5-2t) }. \]Substitution Can Also Reveal No Solution
Consider
\[ \begin{cases} y=2x+1,\\ y=2x+5. \end{cases} \]Comparison gives
\[ 2x+1=2x+5. \]Subtract \(2x\):
\[ 1=5. \]This contradiction shows the system is inconsistent.
Comparison Can Reveal Infinite Solutions
Consider
\[ \begin{cases} y=3x-4,\\ 2y=6x-8. \end{cases} \]The second equation simplifies to
\[ y=3x-4. \]Comparison becomes
\[ 3x-4=3x-4, \]an identity.
Therefore, every point on the line is a solution.
Method Choice from Equation Form
| System Structure | Often Convenient |
|---|---|
| One variable already isolated | Substitution |
| Both equations solved for the same variable | Comparison |
| Opposite or matching coefficients | Elimination |
| Simple graphical forms | Graphing as a check |
No Method Is Mathematically Superior
If performed correctly, substitution, comparison, and elimination produce the same solution set.
The best method is usually the one that:
- uses the structure already present;
- minimizes unnecessary fractions;
- reduces arithmetic;
- keeps the reasoning easy to verify.
Hybrid Methods Are Allowed
You do not need to remain loyal to one method from start to finish.
You may:
- simplify equations first;
- use elimination to isolate one variable;
- finish with substitution;
- graph afterward as a check.
Good solving adapts to the structure of the problem.
Example of a Hybrid Method
Consider
\[ \begin{cases} 4x+2y=18,\\ y=7-x. \end{cases} \]Substitution is immediate:
\[ 4x+2(7-x)=18. \]Then
\[ 4x+14-2x=18, \] \[ 2x=4, \] \[ x=2. \]Now use the already isolated equation:
\[ y=7-2=5. \]So
\[ \boxed{(2,5)}. \]Preserve Exact Values
If the system produces
\[ x=\frac{17}{6}, \]keep the fraction during symbolic work.
Convert to a decimal only when interpretation requires it.
Premature rounding can introduce avoidable error into the second variable.
Check the Ordered Pair
Suppose the claimed solution is
\[ (2,5) \]for
\[ \begin{cases} 4x+2y=18,\\ y=7-x. \end{cases} \]Check:
\[ 4(2)+2(5)=18, \]and
\[ 5=7-2. \]Both are true.
Verification Uses the Original System
Always verify with the original equations, not only with a transformed version.
This catches mistakes made during:
- distribution;
- sign changes;
- scaling;
- substitution;
- copying equations.
Substitution with Applied Variables
Suppose adult tickets \(a\) and student tickets \(s\) satisfy
\[ \begin{cases} a+s=80,\\ 12a+8s=800. \end{cases} \]From the first equation,
\[ s=80-a. \]Substitute into revenue:
\[ 12a+8(80-a)=800. \]Finish the Ticket Example
Continue:
\[ 12a+640-8a=800, \] \[ 4a=160, \] \[ a=40. \]Then
\[ s=40. \]So
\[ \boxed{40\text{ adult tickets and }40\text{ student tickets}}. \]Elimination in a Mixture Problem
Suppose
\[ \begin{cases} x+y=20,\\ 0.10x+0.40y=5. \end{cases} \]Multiply the second equation by \(10\):
\[ x+4y=50. \]Subtract the first equation:
\[ 3y=30, \] \[ y=10. \]Then
\[ x=10. \]Comparison in Break-Even Problems
Suppose
\[ R(q)=30q \]and
\[ C(q)=18q+480. \]At break-even, both outputs are equal:
\[ 30q=18q+480. \]Comparison directly expresses the economic meaning of break-even.
Finish the Break-Even Example
Continue:
\[ 12q=480, \] \[ q=40. \]The common output is
\[ 1200. \]So the two models intersect at
\[ \boxed{(40,1200)}. \]Elimination with Three Variables Preview
Consider
\[ \begin{cases} x+y+z=6,\\ x-y+z=2,\\ 2x+z=5. \end{cases} \]Elimination can combine equations to reduce the number of variables step by step.
This same idea scales naturally to larger systems.
First Elimination Step
Subtract the second equation from the first:
\[ (x+y+z)-(x-y+z)=6-2. \]Then
\[ 2y=4, \] \[ y=2. \]The three-variable system has already been simplified by eliminating two variables from one relation.
Continue the Three-Variable Example
Use
\[ y=2 \]in the first equation:
\[ x+2+z=6, \] \[ x+z=4. \]The third equation is
\[ 2x+z=5. \]Subtract:
\[ x=1. \]Then
\[ z=3. \]So
\[ \boxed{(x,y,z)=(1,2,3)}. \]Why Elimination Scales Better
Substitution can become cumbersome in large systems because expressions may grow rapidly.
Elimination organizes the work around coefficient cancellation.
This is why Gaussian elimination becomes a standard method for larger systems.
From Elimination to Row Operations
The familiar elimination moves become matrix row operations:
- swap two equations;
- multiply an equation by a nonzero scalar;
- add a scalar multiple of one equation to another.
Matrices remove repeated variable symbols while preserving the same coefficient structure.
Equation Scaling and Row Scaling
When you multiply
\[ 2x+y=7 \]by \(3\), you obtain
\[ 6x+3y=21. \]In matrix language, this becomes multiplication of an entire row by \(3\).
Equation Addition and Row Replacement
When you replace one equation with the sum of itself and a multiple of another equation, you are performing a row replacement.
For example:
\[ E_2\leftarrow E_2-2E_1. \]This notation becomes useful when many equations are present.
Comparison and Function Equality
Comparison has a broader role than systems alone.
Whenever two functions represent competing quantities, the equation
\[ f(x)=g(x) \]finds their crossover inputs.
This includes:
- break-even points;
- meeting times;
- equal-cost thresholds;
- curve intersections.
Substitution and Composition
Substitution is also connected to function composition.
If
\[ y=g(x) \]and another relation contains \(y\), replacing \(y\) with \(g(x)\) inserts one relationship into another.
This is structurally similar to composition.
Elimination as Information Combination
Elimination combines two constraints to produce a new consequence containing less information about one variable but more direct information about another.
For example, two equations in \(x\) and \(y\) may combine to produce one equation in \(x\) alone.
Do Not Cancel Across Equations Illegally
You cannot simply “cancel” matching terms that occur in separate equations.
Cancellation must arise from a valid operation such as adding or subtracting entire equations.
Every term and constant must be included in that operation.
Do Not Scale Only One Side
If you multiply an equation by \(3\), multiply every term on both sides.
From
\[ 2x+y=7, \]the correct scaled equation is
\[ 6x+3y=21. \]Writing
\[ 6x+3y=7 \]changes the equation and the solution set.
Sign Discipline in Subtraction
When subtracting equations, parentheses help prevent sign errors.
For example,
\[ (5x+3y)-(5x-y) \]becomes
\[ 5x+3y-5x+y = 4y. \]The minus sign applies to every term in the second equation.
Choose Addition When Possible
If one coefficient is \(+4\) and the other is \(-4\), adding equations is often safer than subtracting because fewer sign changes are involved.
Good elimination choices reduce opportunities for error.
Substitution Can Introduce Fractions
Consider
\[ 3x+4y=7. \]Solving for \(x\) gives
\[ x=\frac{7-4y}{3}. \]This is valid, but if another equation has convenient matching coefficients, elimination may be cleaner.
Elimination Can Avoid Fractions
With
\[ \begin{cases} 3x+4y=7,\\ 6x-y=5, \end{cases} \]elimination can remove \(x\) after multiplying only one equation by a small integer.
This may be preferable to substituting a fractional expression.
But Substitution Can Be Best Even with Fractions
If a variable is already isolated, substitution may still be simplest even when the resulting arithmetic contains fractions.
There is no universal rule requiring integer-only work.
Choose the method that keeps the structure clearest.
Graphing as an Independent Check
After solving symbolically, graph the equations if useful.
If your answer is
\[ (3,2), \]the graphs should intersect at that point.
If they do not, recheck the symbolic work.
Method Agreement Is a Powerful Check
You can sometimes solve a system by one method and verify with another.
For example:
- solve by elimination;
- check by substitution;
- confirm graphically.
Independent methods increase confidence in the result.
A Reliable Substitution Workflow
- Choose an equation with an isolated or easily isolated variable.
- Solve for that variable if necessary.
- Substitute the equivalent expression into the other equation.
- Solve the resulting one-variable equation.
- Back-substitute to find the remaining variable.
- Verify in the original system.
A Reliable Comparison Workflow
- Solve both equations for the same variable.
- Set the two expressions equal.
- Solve for the remaining variable.
- Substitute to find the shared output.
- Interpret the result as an intersection or crossover.
- Verify in both original equations.
A Reliable Elimination Workflow
- Write equations with like variables aligned.
- Choose a variable to eliminate.
- Scale equations if necessary.
- Add or subtract entire equations.
- Solve the resulting one-variable equation.
- Back-substitute.
- Verify the final solution.
A Reliable Method-Selection Workflow
- Inspect the equation forms before doing algebra.
- If a variable is isolated, consider substitution.
- If both equations isolate the same variable, consider comparison.
- If coefficients cancel easily, consider elimination.
- Prefer the route with simpler arithmetic and fewer fractions.
- Switch methods if the structure changes.
A Reliable Special-Case Workflow
- Continue simplifying until a variable equation, identity, or contradiction appears.
- A variable equation may lead to a unique solution.
- An identity such as \(0=0\) signals dependence.
- A contradiction such as \(0=5\) signals inconsistency.
- Parameterize remaining free variables when infinitely many solutions exist.
Common Errors
Substituting an Expression for the Wrong Variable
If \(y=2x+1\), replace \(y\), not \(x\), with \(2x+1\).
Forgetting Parentheses
Substituting \(y=3-x\) into \(2y\) requires
\[ 2(3-x), \]not
\[ 2(3)-x. \]Scaling Only Part of an Equation
Multiply every term on both sides.
Subtracting Without Distributing the Minus Sign
Use parentheses around the equation being subtracted.
Stopping After Finding One Variable
A two-variable system usually requires an ordered pair.
Stopping at \(0=0\)
Parameterize the remaining solution family.
Treating \(0=5\) as a Numerical Answer
It is a contradiction showing no solution exists.
Rounding Too Early
Keep fractions exact until the end when possible.
Using a Complicated Method When a Simple One Is Available
Inspect the structure before beginning the algebra.
Failing to Verify
Check the final ordered pair in the original equations.
Practice
1. Substitution
Solve
\[ \begin{cases} y=x+3,\\ 2x+y=12. \end{cases} \]Check Answer
Substitute:
\[ 2x+(x+3)=12. \] \[ 3x=9. \] \[ x=3. \]Then
\[ y=6. \]So
\[ \boxed{(3,6)}. \]2. Substitution for \(x\)
Solve
\[ \begin{cases} x=7-y,\\ 2x+3y=11. \end{cases} \]Check Answer
\[ 2(7-y)+3y=11. \] \[ 14-2y+3y=11. \] \[ y=-3. \]Then
\[ x=10. \]So
\[ \boxed{(10,-3)}. \]3. Comparison
Solve
\[ \begin{cases} y=4x-1,\\ y=x+8. \end{cases} \]Check Answer
\[ 4x-1=x+8. \] \[ 3x=9. \] \[ x=3. \]Then
\[ y=11. \]Therefore,
\[ \boxed{(3,11)}. \]4. Elimination by Addition
Solve
\[ \begin{cases} 2x+y=9,\\ x-y=0. \end{cases} \]Check Answer
Add:
\[ 3x=9. \] \[ x=3. \]Then
\[ y=3. \]So
\[ \boxed{(3,3)}. \]5. Elimination by Subtraction
Solve
\[ \begin{cases} 4x+3y=18,\\ 4x-y=10. \end{cases} \]Check Answer
Subtract the second equation from the first:
\[ 4y=8. \] \[ y=2. \]Then
\[ 4x-2=10, \] \[ 4x=12, \] \[ x=3. \]Thus,
\[ \boxed{(3,2)}. \]6. Elimination After Scaling
Solve
\[ \begin{cases} 2x+y=5,\\ 3x-2y=4. \end{cases} \]Check Answer
Multiply the first equation by \(2\):
\[ 4x+2y=10. \]Add:
\[ 7x=14. \] \[ x=2. \]Then
\[ y=1. \]So
\[ \boxed{(2,1)}. \]7. No Solution
Solve or classify
\[ \begin{cases} y=2x+3,\\ y=2x-4. \end{cases} \]Check Answer
Comparison gives
\[ 2x+3=2x-4. \] \[ 3=-4. \]This is impossible.
\[ \boxed{\text{No solution.}} \]8. Infinitely Many Solutions
Solve or classify
\[ \begin{cases} x+2y=6,\\ 2x+4y=12. \end{cases} \]Check Answer
The second equation is twice the first.
Let
\[ y=t. \]Then
\[ x=6-2t. \]So
\[ \boxed{ (x,y)=(6-2t,t), \qquad t\in\mathbb{R} }. \]9. Choose the Method
Which method is most natural for
\[ \begin{cases} y=5x+1,\\ y=-2x+8? \end{cases} \]Check Answer
Because both equations are solved for \(y\),
\[ \boxed{\text{comparison}} \]is especially natural.
10. Applied Ticket System
Adult tickets cost \$15 and child tickets cost \$9. A total of \(60\) tickets brings in \$660. Find the number of each.
Check Answer
Let \(a\) be adult tickets and \(c\) child tickets:
\[ \begin{cases} a+c=60,\\ 15a+9c=660. \end{cases} \]Use
\[ c=60-a. \]Then
\[ 15a+9(60-a)=660. \] \[ 15a+540-9a=660. \] \[ 6a=120. \] \[ a=20. \]Then
\[ c=40. \]So
\[ \boxed{20\text{ adult and }40\text{ child tickets}}. \]11. Three Variables
Solve
\[ \begin{cases} x+y+z=6,\\ x-y+z=2,\\ 2x+z=5. \end{cases} \]Check Answer
Subtract the second equation from the first:
\[ 2y=4, \] \[ y=2. \]Then
\[ x+z=4. \]Compare with
\[ 2x+z=5. \]Subtract:
\[ x=1. \]Then
\[ z=3. \]Thus,
\[ \boxed{(1,2,3)}. \]12. Verify a Solution
Verify that
\[ (2,1) \]solves
\[ \begin{cases} 2x+y=5,\\ 3x-2y=4. \end{cases} \]Check Answer
\[ 2(2)+1=5, \]and
\[ 3(2)-2(1)=4. \]Both equations are true.
\[ \boxed{\text{Verified.}} \]Key Ideas
- Substitution, comparison, and elimination are different ways to preserve and simplify the same system solution set.
- Substitution replaces a variable with an equivalent expression.
- Comparison sets two expressions equal because both represent the same variable or output.
- Elimination combines equations so that one variable cancels.
- Equivalent systems may look different while having exactly the same solutions.
- Substitution is especially convenient when a variable is isolated or easy to isolate.
- Comparison is especially convenient when both equations are solved for the same variable.
- Elimination is especially convenient when coefficients already match or can be made to match simply.
- Scaling an equation by a nonzero constant preserves its solution set.
- Adding a multiple of one equation to another is the core operation behind elimination.
- An identity such as \(0=0\) signals dependence; a contradiction such as \(0=5\) signals inconsistency.
- Hybrid methods are often more efficient than forcing one method throughout.
- Exact fractions should usually be preserved until interpretation requires decimals.
- Final solutions should be checked in the original equations.
- Elimination naturally develops into row operations, augmented matrices, and Gaussian elimination.
Where This Leads
Substitution, comparison, and elimination reveal that solving systems is fundamentally about replacing a complicated set of simultaneous conditions with simpler but equivalent conditions. The three methods differ in style, but all preserve the same solution set and all rely on the equality properties developed earlier in Malone Math.
From here, elimination becomes increasingly systematic. The next matrix-oriented lessons can encode coefficients in augmented matrices, formalize row operations, build row-echelon and reduced row-echelon forms, and develop Gaussian and Gauss-Jordan elimination. Those procedures are not new mathematics so much as organized versions of the elimination ideas already used here.