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:

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:

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 isolatedSubstitution
Both equations solved for the same variableComparison
Opposite or matching coefficientsElimination
Simple graphical formsGraphing 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:

Hybrid Methods Are Allowed

You do not need to remain loyal to one method from start to finish.

You may:

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:

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:

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:

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:

Independent methods increase confidence in the result.

A Reliable Substitution Workflow

  1. Choose an equation with an isolated or easily isolated variable.
  2. Solve for that variable if necessary.
  3. Substitute the equivalent expression into the other equation.
  4. Solve the resulting one-variable equation.
  5. Back-substitute to find the remaining variable.
  6. Verify in the original system.

A Reliable Comparison Workflow

  1. Solve both equations for the same variable.
  2. Set the two expressions equal.
  3. Solve for the remaining variable.
  4. Substitute to find the shared output.
  5. Interpret the result as an intersection or crossover.
  6. Verify in both original equations.

A Reliable Elimination Workflow

  1. Write equations with like variables aligned.
  2. Choose a variable to eliminate.
  3. Scale equations if necessary.
  4. Add or subtract entire equations.
  5. Solve the resulting one-variable equation.
  6. Back-substitute.
  7. Verify the final solution.

A Reliable Method-Selection Workflow

  1. Inspect the equation forms before doing algebra.
  2. If a variable is isolated, consider substitution.
  3. If both equations isolate the same variable, consider comparison.
  4. If coefficients cancel easily, consider elimination.
  5. Prefer the route with simpler arithmetic and fewer fractions.
  6. Switch methods if the structure changes.

A Reliable Special-Case Workflow

  1. Continue simplifying until a variable equation, identity, or contradiction appears.
  2. A variable equation may lead to a unique solution.
  3. An identity such as \(0=0\) signals dependence.
  4. A contradiction such as \(0=5\) signals inconsistency.
  5. 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

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.