The Central Idea

Consider the two equations

\[ y=2x+1 \]

and

\[ y=-x+7. \]

Each equation describes a line.

A solution to the system must satisfy both equations at once.

Because both expressions equal \(y\), set them equal:

\[ 2x+1=-x+7. \]

Then

\[ 3x=6, \] \[ x=2. \]

Substitute:

\[ y=2(2)+1=5. \]

So the system has solution

\[ \boxed{(2,5)}. \]

Graphically, \((2,5)\) is exactly where the two lines intersect.

What Is a System of Equations?

A system of equations is a collection of two or more equations considered together.

For example,

\[ \begin{cases} x+y=7,\\ 2x-y=5 \end{cases} \]

is a system of two equations in two variables.

The goal is to find values of \(x\) and \(y\) that make both equations true simultaneously.

Simultaneous Conditions

Solving one equation tells us which values satisfy one condition.

Solving a system asks for the overlap among several conditions.

In set language, if

\[ S_1 \]

is the solution set of the first equation and

\[ S_2 \]

is the solution set of the second, then the system solution set is

\[ \boxed{S_1\cap S_2}. \]

Systems therefore connect algebra directly with intersection of sets.

Ordered Pairs as Solutions

For a system in two variables, a solution is usually an ordered pair

\[ (x,y). \]

For example, test

\[ (3,4) \]

in

\[ \begin{cases} x+y=7,\\ 2x-y=2. \end{cases} \]

First equation:

\[ 3+4=7. \]

Second equation:

\[ 2(3)-4=2. \]

Both are true.

Therefore,

\[ \boxed{(3,4)} \]

is a solution of the system.

A Point Must Satisfy Every Equation

Suppose

\[ (3,4) \]

satisfies the first equation of a system but not the second.

Then it is not a solution of the system.

A system uses logical “and”:

\[ \boxed{ \text{equation 1 is true} \quad\text{and}\quad \text{equation 2 is true} }. \]

Graphical Meaning

For equations in two variables, each equation has a graph.

The system solutions are the points common to all the graphs.

For two lines, this means:

One Solution

Consider

\[ \begin{cases} y=2x+1,\\ y=-x+7. \end{cases} \]

The slopes are different:

\[ 2 \]

and

\[ -1. \]

Distinct nonparallel lines intersect exactly once.

So the system has one solution.

No Solution

Consider

\[ \begin{cases} y=3x+2,\\ y=3x-5. \end{cases} \]

Both lines have slope

\[ 3, \]

but different intercepts.

They are distinct parallel lines.

Therefore, they never intersect and the system has

\[ \boxed{\text{no solution}}. \]

The solution set is

\[ \boxed{\varnothing}. \]

Infinitely Many Solutions

Consider

\[ \begin{cases} y=2x+4,\\ 2y=4x+8. \end{cases} \]

The second equation simplifies to

\[ y=2x+4. \]

Both equations describe the same line.

Every point on that line satisfies both equations.

So the system has

\[ \boxed{\text{infinitely many solutions}}. \]

Consistent and Inconsistent Systems

A system is consistent if it has at least one solution.

A system is inconsistent if it has no solution.

Thus:

Independent and Dependent Equations

In a two-equation linear system:

For example,

\[ x+y=5 \]

and

\[ 2x+2y=10 \]

are dependent because one equation is a scalar multiple of the other.

Three Geometric Cases for Two Linear Equations

Geometry Solutions System Type
Lines intersect onceOneConsistent, independent
Distinct parallel linesNoneInconsistent
Same lineInfinitely manyConsistent, dependent

Why Two Nonparallel Lines Have One Solution

Two distinct nonparallel lines in the plane have different directions.

Because neither direction is parallel to the other, the lines meet at exactly one point.

That unique intersection is the unique ordered pair satisfying both equations.

Systems as Intersections of Functions

Suppose

\[ f(x)=2x+5 \]

and

\[ g(x)=-x+11. \]

To find where the graphs intersect, solve

\[ f(x)=g(x). \]

That is,

\[ 2x+5=-x+11. \]

Then

\[ 3x=6, \] \[ x=2. \]

The common output is

\[ 9. \]

So the intersection is

\[ \boxed{(2,9)}. \]

Method 1: Graphing

One way to solve a system is to graph every equation and identify their common point or points.

For linear systems, graphing is especially useful for understanding:

Graphing May Give Approximate Solutions

If two lines intersect at

\[ \left(\frac{7}{3},\frac{11}{5}\right), \]

a hand-drawn graph may show only an approximate intersection.

Graphing is excellent for structure and checking, but symbolic methods may be better when exact values are required.

Method 2: Substitution

Substitution replaces one variable with an equivalent expression from another equation.

Consider

\[ \begin{cases} y=2x+1,\\ x+y=7. \end{cases} \]

Because the first equation already isolates \(y\), substitute

\[ 2x+1 \]

for \(y\) in the second equation.

Substitution Example

Start with

\[ x+y=7. \]

Substitute

\[ y=2x+1: \] \[ x+(2x+1)=7. \]

Then

\[ 3x+1=7, \] \[ 3x=6, \] \[ x=2. \]

Now find \(y\):

\[ y=2(2)+1=5. \]

So

\[ \boxed{(2,5)}. \]

Why Substitution Works

If

\[ y=2x+1, \]

then \(y\) and \(2x+1\) represent the same quantity wherever the first equation is true.

Replacing one with the other preserves the meaning of the system.

Substitution is therefore an application of equality and equivalence.

Substitution When \(x\) Is Isolated

Consider

\[ \begin{cases} x=4-y,\\ 3x+2y=10. \end{cases} \]

Substitute

\[ x=4-y \]

into the second equation:

\[ 3(4-y)+2y=10. \]

Then

\[ 12-3y+2y=10, \] \[ -y=-2, \] \[ y=2. \]

So

\[ x=4-2=2. \]

The solution is

\[ \boxed{(2,2)}. \]

Method 3: Elimination

Elimination combines equations so that one variable cancels.

Consider

\[ \begin{cases} x+y=7,\\ x-y=1. \end{cases} \]

Add the equations:

\[ (x+y)+(x-y)=7+1. \]

Then

\[ 2x=8, \] \[ x=4. \]

Substitute into either original equation:

\[ 4+y=7, \] \[ y=3. \]

So

\[ \boxed{(4,3)}. \]

Why Elimination Works

If two equalities are true, we may add their left sides and add their right sides.

For

\[ L_1=R_1 \]

and

\[ L_2=R_2, \]

we obtain

\[ L_1+L_2=R_1+R_2. \]

Elimination chooses combinations that make one variable disappear.

Elimination by Subtraction

Consider

\[ \begin{cases} 3x+2y=14,\\ 3x-y=5. \end{cases} \]

Subtract the second equation from the first:

\[ (3x+2y)-(3x-y)=14-5. \]

Then

\[ 3y=9, \] \[ y=3. \]

Substitute:

\[ 3x-3=5, \] \[ 3x=8, \] \[ x=\frac83. \]

So

\[ \boxed{\left(\frac83,3\right)}. \]

Elimination by Scaling an Equation

Sometimes coefficients do not 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:

\[ (4x+2y)+(3x-2y)=14+4. \]

So

\[ 7x=18, \] \[ x=\frac{18}{7}. \]

Finish the Scaled-Elimination Example

Use

\[ 2x+y=7. \]

Substitute

\[ x=\frac{18}{7}. \]

Then

\[ 2\left(\frac{18}{7}\right)+y=7. \]

So

\[ \frac{36}{7}+y=\frac{49}{7}, \] \[ y=\frac{13}{7}. \]

Therefore,

\[ \boxed{ \left(\frac{18}{7},\frac{13}{7}\right) }. \]

Choosing a Method

No single solving method is always best.

A useful rule of thumb:

Different Methods Should Agree

If a system has one exact solution, graphing, substitution, elimination, and later matrix methods all describe the same mathematical solution set.

The methods differ in representation and efficiency, not in the underlying answer.

Verify the Final Ordered Pair

Suppose a system solution is claimed to be

\[ (4,3). \]

For

\[ \begin{cases} x+y=7,\\ x-y=1, \end{cases} \]

check:

\[ 4+3=7, \]

and

\[ 4-3=1. \]

Both equations are true.

The ordered pair is verified.

Verification Is Especially Important with Fractions

When elimination or substitution produces fractions, arithmetic mistakes are easy to make.

Substituting the final values into every original equation is often the fastest way to catch an error.

Detecting No Solution Algebraically

Consider

\[ \begin{cases} 2x+y=5,\\ 4x+2y=14. \end{cases} \]

Multiply the first equation by \(2\):

\[ 4x+2y=10. \]

But the second equation says

\[ 4x+2y=14. \]

Subtracting gives

\[ 0=4. \]

This contradiction means

\[ \boxed{\text{no solution}}. \]

Detecting Infinitely Many Solutions Algebraically

Consider

\[ \begin{cases} 2x+y=5,\\ 4x+2y=10. \end{cases} \]

Multiply the first equation by \(2\):

\[ 4x+2y=10. \]

This is exactly the second equation.

Elimination produces

\[ 0=0. \]

This identity means the equations describe the same line.

There are infinitely many solutions.

A Free Variable in a Dependent System

For

\[ 2x+y=5, \]

we may solve for

\[ y=5-2x. \]

Because the second equation adds no new condition, \(x\) remains free.

Every point of the form

\[ \boxed{(x,5-2x)} \]

satisfies the dependent system.

Parameterizing Infinitely Many Solutions

We may introduce a parameter

\[ t\in\mathbb{R}. \]

Let

\[ x=t. \]

Then

\[ y=5-2t. \]

So the entire solution set can be written

\[ \boxed{ (x,y)=(t,5-2t), \qquad t\in\mathbb{R} }. \]

Systems in Standard Form

Linear systems are often written

\[ \begin{cases} A_1x+B_1y=C_1,\\ A_2x+B_2y=C_2. \end{cases} \]

Standard form aligns like variables, which makes elimination especially natural.

It also prepares the equations for matrix representation later.

Coefficient Structure Predicts Geometry

Consider

\[ A_1x+B_1y=C_1 \]

and

\[ A_2x+B_2y=C_2. \]

If the coefficient pairs

\[ (A_1,B_1) \]

and

\[ (A_2,B_2) \]

are proportional, the lines have the same direction.

Then they are either parallel or identical.

Example: Proportional Left Sides

Compare

\[ 2x+3y=12 \]

and

\[ 4x+6y=30. \]

The left-side coefficients are proportional:

\[ (4,6)=2(2,3). \]

But doubling the first equation gives

\[ 4x+6y=24, \]

not \(30\).

So the lines are distinct parallel lines and the system has no solution.

Example: Entire Equations Proportional

Compare

\[ 2x+3y=12 \]

and

\[ 4x+6y=24. \]

Every coefficient, including the constant, is doubled.

So the equations are equivalent and the system has infinitely many solutions.

Systems with Horizontal and Vertical Lines

Consider

\[ \begin{cases} x=4,\\ y=7. \end{cases} \]

The vertical line

\[ x=4 \]

and horizontal line

\[ y=7 \]

intersect at

\[ \boxed{(4,7)}. \]

This is one of the simplest linear systems.

A Vertical Line and an Affine Function

Consider

\[ \begin{cases} x=3,\\ y=2x-1. \end{cases} \]

The first equation immediately gives

\[ x=3. \]

Substitute:

\[ y=2(3)-1=5. \]

So

\[ \boxed{(3,5)}. \]

Systems Need Not Be Linear

The general idea of a system is broader than straight lines.

For example,

\[ \begin{cases} y=x^2,\\ y=2x+3 \end{cases} \]

asks for intersections of a parabola and a line.

Such a system may have zero, one, two, or more solutions depending on the equations.

This section begins with linear systems because their structure is especially clear and reusable.

Linear Systems as Simultaneous Models

Systems become powerful when two models describe the same situation.

For example, two cost plans may be

\[ A(x)=5x+20 \]

and

\[ B(x)=3x+40. \]

Solving the system identifies where the plans have the same cost.

Example: Compare Two Plans

Set

\[ 5x+20=3x+40. \]

Then

\[ 2x=20, \] \[ x=10. \]

The common cost is

\[ 5(10)+20=70. \]

So the graphs intersect at

\[ \boxed{(10,70)}. \]

The system identifies the crossover point.

Break-Even as a System

Suppose revenue and cost are

\[ R(q)=25q \]

and

\[ C(q)=15q+400. \]

Break-even means both models have the same output:

\[ R(q)=C(q). \]

This is a system-intersection problem.

The solution is

\[ q=40, \]

with common value

\[ 1000. \]

So the intersection is

\[ \boxed{(40,1000)}. \]

Systems from Two Unknown Quantities

Suppose a theater sells adult tickets for \$12 and student tickets for \$8.

A total of \(100\) tickets are sold for \$1000.

Let

\[ a=\text{adult tickets}, \] \[ s=\text{student tickets}. \]

The total-ticket condition is

\[ a+s=100. \]

The revenue condition is

\[ 12a+8s=1000. \]

Together:

\[ \begin{cases} a+s=100,\\ 12a+8s=1000. \end{cases} \]

Solve the Ticket System

From

\[ a+s=100, \]

we get

\[ s=100-a. \]

Substitute:

\[ 12a+8(100-a)=1000. \]

Then

\[ 12a+800-8a=1000, \] \[ 4a=200, \] \[ a=50. \]

So

\[ s=50. \]

Therefore,

\[ \boxed{(a,s)=(50,50)}. \]

Interpret the Solution in Context

The ordered pair

\[ (50,50) \]

means \(50\) adult tickets and \(50\) student tickets were sold.

A system solution should be translated back into the quantities represented by the variables.

Mixture Example

Suppose \(x\) liters of a \(20\%\) solution are mixed with \(y\) liters of a \(50\%\) solution to make \(30\) liters of a \(40\%\) solution.

Total volume:

\[ x+y=30. \]

Total amount of solute:

\[ 0.20x+0.50y=0.40(30). \]

So the system is

\[ \begin{cases} x+y=30,\\ 0.20x+0.50y=12. \end{cases} \]

Solve the Mixture System

From

\[ x+y=30, \]

we have

\[ x=30-y. \]

Substitute:

\[ 0.20(30-y)+0.50y=12. \]

Then

\[ 6-0.20y+0.50y=12, \] \[ 0.30y=6, \] \[ y=20. \]

Thus,

\[ x=10. \]

The mixture uses

\[ \boxed{10\text{ L at }20\%\text{ and }20\text{ L at }50\%}. \]

Motion Example

Suppose two travelers' positions are modeled by

\[ d_1(t)=50t \]

and

\[ d_2(t)=200-30t. \]

They meet when

\[ d_1(t)=d_2(t). \]

So

\[ 50t=200-30t. \]

Then

\[ 80t=200, \] \[ t=2.5. \]

The meeting position is

\[ 50(2.5)=125. \]

Thus,

\[ \boxed{(t,d)=(2.5,125)}. \]

System Variables Need Not Be \(x\) and \(y\)

Systems may use whatever variables fit the situation:

\[ \begin{cases} a+s=100,\\ 12a+8s=1000 \end{cases} \]

or

\[ \begin{cases} V_1+V_2=30,\\ 0.2V_1+0.5V_2=12. \end{cases} \]

The algebraic structure matters more than the letter names.

Domain Restrictions in Systems

A system may have an algebraic solution that does not make sense in context.

If variables count objects, we may require

\[ x,y\in\mathbb{Z}_{\ge0}. \]

If they represent physical amounts, we may require

\[ x\ge0,\qquad y\ge0. \]

Always check the solution against the model's domain.

Example of an Invalid Contextual Solution

Suppose a ticket system produced

\[ a=120, \qquad s=-20. \]

The equations might be algebraically satisfied, but negative ticket counts are not physically meaningful.

The mathematical solution would reveal that the original assumptions or data do not describe a feasible ticket scenario.

Systems as Constraint Intersections

Each equation restricts the possible values of the variables.

One equation may leave infinitely many possibilities.

A second independent equation narrows the possibilities further.

Geometrically, each line is a one-dimensional constraint in the plane, and their intersection may reduce the solution to a single point.

More Equations Can Add More Constraints

Consider three equations in two variables:

\[ \begin{cases} x+y=5,\\ x-y=1,\\ 2x=6. \end{cases} \]

The first two give

\[ x=3,\qquad y=2. \]

The third equation also gives

\[ x=3. \]

So the third condition is consistent and redundant.

Too Many Conditions Can Conflict

Suppose the third equation were instead

\[ 2x=8. \]

The first two require

\[ x=3, \]

while the third requires

\[ x=4. \]

No point satisfies all three equations.

The expanded system is inconsistent.

Three Variables

A linear system may involve three variables:

\[ \begin{cases} x+y+z=6,\\ 2x-y+z=3,\\ x+2y-z=4. \end{cases} \]

A solution is an ordered triple

\[ \boxed{(x,y,z)} \]

that satisfies all three equations simultaneously.

Geometric Meaning in Three Variables

A linear equation in three variables such as

\[ Ax+By+Cz=D \]

typically describes a plane in three-dimensional space.

A system of such equations asks for the common intersection of several planes.

That intersection may be:

Why Systems Lead Naturally to Matrices

As systems become larger, repeatedly writing every variable name becomes cumbersome.

For example,

\[ \begin{cases} 2x+3y-z=7,\\ x-4y+2z=5,\\ 3x+y+5z=12 \end{cases} \]

contains a regular grid of coefficients.

Matrices provide a compact way to organize those coefficients and perform systematic elimination.

Coefficient Pattern Preview

The system

\[ \begin{cases} 2x+3y=7,\\ 4x-y=5 \end{cases} \]

contains coefficient rows

\[ (2,3) \]

and

\[ (4,-1). \]

Later these can be organized into a coefficient matrix, while the constants form a separate column or an augmented matrix.

Systems and Row Operations Preview

Elimination already uses operations that will later become row operations:

These operations preserve the solution set when used correctly.

Why Equivalent Systems Matter

When solving by substitution or elimination, we replace the original system with a simpler system having the same solution set.

Such systems are equivalent systems.

The goal is not to preserve the appearance of the equations, but to preserve exactly which variable values satisfy them.

Example of Equivalent Systems

Start with

\[ \begin{cases} x+y=7,\\ x-y=1. \end{cases} \]

Add the equations to obtain

\[ 2x=8. \]

The system can then be represented as

\[ \begin{cases} x=4,\\ x+y=7. \end{cases} \]

Then

\[ y=3. \]

Each stage preserves the unique solution

\[ (4,3). \]

Do Not Change Only One Equation Arbitrarily

An equation in a system may be replaced only by an equivalent consequence that preserves the common solution set.

Changing a coefficient or constant without a valid operation changes the problem itself.

Approximate Systems from Data

Real models may not intersect perfectly because of measurement noise.

Two fitted lines may produce an approximate intersection rather than an exact physical truth.

In such cases, use approximate notation and interpret the intersection as model-based.

Example: Approximate Crossover

Suppose fitted cost models are

\[ A(x)\approx4.9x+21.2 \]

and

\[ B(x)\approx3.1x+39.8. \]

Set the models equal:

\[ 4.9x+21.2=3.1x+39.8. \]

Then

\[ 1.8x=18.6, \] \[ x\approx10.33. \]

The crossover is itself an estimate based on approximate models.

Graphing as a Check on Symbolic Work

If algebra gives

\[ (10.33,71.8) \]

as an approximate intersection, a graph should show the fitted lines crossing near that point.

If the graph suggests a completely different region, recheck the algebra or model.

Substitution Versus Elimination

Consider

\[ \begin{cases} y=5x-2,\\ 3x+2y=10. \end{cases} \]

Substitution is natural because \(y\) is isolated.

But for

\[ \begin{cases} 3x+2y=10,\\ 5x-2y=6, \end{cases} \]

elimination is natural because the \(y\)-coefficients already cancel.

Method choice is part of mathematical efficiency.

One Equation Can Be Solved Before Combining

Consider

\[ \begin{cases} 2x=8,\\ x+3y=13. \end{cases} \]

The first equation immediately gives

\[ x=4. \]

Then

\[ 4+3y=13, \] \[ 3y=9, \] \[ y=3. \]

There is no need to force a more complicated method.

Systems as Information Structures

Each independent equation contributes information.

In two variables:

Underdetermined Systems

A system has fewer independent constraints than needed when variables remain free.

For example,

\[ x+y+z=6 \]

is one equation in three variables.

Many ordered triples satisfy it.

Additional independent equations are needed to determine a unique triple.

Overdetermined Systems

A system may contain more equations than variables.

That does not automatically mean no solution exists.

The extra equations may be:

System Rank Preview

Later matrix methods formalize the idea of how many genuinely independent constraints a system contains.

This concept is connected to rank.

For now, the important intuition is that repeated or dependent equations do not add new information.

Exact Arithmetic Is Often Worth Preserving

If a system produces

\[ x=\frac{18}{7}, \]

keep the fraction when exact values matter.

A decimal such as

\[ 2.571428\ldots \]

may be useful for interpretation, but the fraction preserves exact structure.

Units in Applied Systems

Each equation should be dimensionally meaningful.

For ticket equations:

\[ a+s=100 \]

counts tickets, while

\[ 12a+8s=1000 \]

measures dollars.

The equations represent different conditions but share the same unknown quantities.

Variable Definitions Prevent Confusion

Before solving an applied system, write definitions such as

\[ a=\text{number of adult tickets}, \] \[ s=\text{number of student tickets}. \]

This makes the final ordered pair interpretable and helps prevent switching variable meanings during the calculation.

A Reliable Graphing Workflow

  1. Graph each equation accurately.
  2. Identify all common points.
  3. Estimate coordinates if necessary.
  4. Check the candidate solution in every original equation.
  5. Recognize parallel or coincident graphs when no single intersection appears.

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 the ordered pair in every original equation.

A Reliable Elimination Workflow

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

A Reliable Classification Workflow

  1. Simplify both equations.
  2. Compare slopes or coefficient structure.
  3. If the lines intersect, identify the unique solution.
  4. If the lines are distinct and parallel, report no solution.
  5. If the equations are equivalent, report infinitely many solutions.
  6. Use algebraic identities or contradictions to confirm the classification.

A Reliable Applied-System Workflow

  1. Define the unknown quantities.
  2. Translate each independent condition into an equation.
  3. Check units and domain restrictions.
  4. Choose a solving method.
  5. Solve and verify.
  6. Interpret the ordered pair or tuple in context.
  7. Reject solutions that violate physical or discrete constraints.

Common Errors

Solving Only One Equation

A system solution must satisfy every equation, not just one.

Forgetting to Back-Substitute

Solving for \(x\) alone does not complete a two-variable solution unless \(y\) is also found.

Reporting \(x=2\) Instead of \((2,5)\)

For two variables, the solution should usually be reported as an ordered pair.

Substituting into the Wrong Expression

Replace a variable with an expression known to be equal to that variable.

Scaling Only Part of an Equation

When multiplying an equation, multiply every term on both sides by the same nonzero factor.

Missing the No-Solution Case

A contradiction such as \(0=4\) means the system is inconsistent.

Missing the Infinite-Solution Case

An identity such as \(0=0\) after correct elimination may indicate dependent equations.

Trusting a Graph Without Verification

A hand-drawn intersection may be approximate. Substitute the candidate values into the equations.

Ignoring Domain Restrictions

Negative counts or impossible physical values may invalidate an algebraic solution in context.

Assuming More Equations Always Mean More Information

Dependent equations may repeat information already present.

Practice

1. Check a Solution

Does

\[ (2,3) \]

solve

\[ \begin{cases} x+y=5,\\ 2x-y=1? \end{cases} \]

Check Answer \[ 2+3=5, \]

and

\[ 2(2)-3=1. \]

Both equations are true.

\[ \boxed{\text{Yes, }(2,3)\text{ is a solution.}} \]

2. Solve by Substitution

Solve

\[ \begin{cases} y=x+4,\\ 2x+y=10. \end{cases} \]
Check Answer

Substitute:

\[ 2x+(x+4)=10. \] \[ 3x=6. \] \[ x=2. \]

Then

\[ y=6. \]

So

\[ \boxed{(2,6)}. \]

3. Solve by Elimination

Solve

\[ \begin{cases} x+y=9,\\ x-y=3. \end{cases} \]
Check Answer

Add the equations:

\[ 2x=12, \] \[ x=6. \]

Then

\[ y=3. \]

Therefore,

\[ \boxed{(6,3)}. \]

4. Eliminate After Scaling

Solve

\[ \begin{cases} 2x+y=8,\\ 3x-2y=5. \end{cases} \]
Check Answer

Multiply the first equation by \(2\):

\[ 4x+2y=16. \]

Add:

\[ 7x=21. \] \[ x=3. \]

Then

\[ 2(3)+y=8, \] \[ y=2. \]

So

\[ \boxed{(3,2)}. \]

5. Classify the System

Classify

\[ \begin{cases} y=4x+1,\\ y=4x-6. \end{cases} \]
Check Answer

The slopes are equal and the intercepts differ.

The lines are distinct and parallel.

\[ \boxed{\text{No solution; inconsistent system.}} \]

6. Dependent System

Classify

\[ \begin{cases} x+2y=6,\\ 2x+4y=12. \end{cases} \]
Check Answer

The second equation is twice the first.

They describe the same line.

\[ \boxed{\text{Infinitely many solutions; dependent system.}} \]

7. Horizontal and Vertical Lines

Solve

\[ \begin{cases} x=-3,\\ y=5. \end{cases} \]
Check Answer

The lines intersect at

\[ \boxed{(-3,5)}. \]

8. Compare Two Plans

Plan A costs

\[ A(x)=6x+12 \]

and Plan B costs

\[ B(x)=4x+30. \]

Find the crossover point.

Check Answer \[ 6x+12=4x+30. \] \[ 2x=18. \] \[ x=9. \]

The common cost is

\[ 6(9)+12=66. \]

So

\[ \boxed{(9,66)}. \]

9. Ticket System

Adult tickets cost \$10 and child tickets cost \$6. A total of \(80\) tickets brings in \$640. How many of each were sold?

Check Answer

Let \(a\) be adult tickets and \(c\) child tickets:

\[ \begin{cases} a+c=80,\\ 10a+6c=640. \end{cases} \]

From \(c=80-a\):

\[ 10a+6(80-a)=640. \] \[ 10a+480-6a=640. \] \[ 4a=160. \] \[ a=40. \]

Then

\[ c=40. \]

So

\[ \boxed{40\text{ adult and }40\text{ child tickets}}. \]

10. Mixture System

How many liters of \(10\%\) solution and \(40\%\) solution are needed to make \(20\) liters of \(25\%\) solution?

Check Answer

Let \(x\) be liters of \(10\%\) solution and \(y\) liters of \(40\%\) solution:

\[ \begin{cases} x+y=20,\\ 0.10x+0.40y=5. \end{cases} \]

Use \(x=20-y\):

\[ 0.10(20-y)+0.40y=5. \] \[ 2-0.10y+0.40y=5. \] \[ 0.30y=3. \] \[ y=10. \]

Then

\[ x=10. \]

Therefore,

\[ \boxed{10\text{ L of each solution}}. \]

11. Identify a Contradiction

What does the system

\[ \begin{cases} 2x+3y=7,\\ 4x+6y=20 \end{cases} \]

tell you?

Check Answer

Doubling the first equation gives

\[ 4x+6y=14, \]

but the second says

\[ 4x+6y=20. \]

The equations are contradictory.

\[ \boxed{\text{The system has no solution.}} \]

12. Parameterize a Dependent System

Describe all solutions of

\[ \begin{cases} x+y=4,\\ 2x+2y=8. \end{cases} \]
Check Answer

The equations are equivalent.

Let

\[ x=t. \]

Then

\[ y=4-t. \]

So all solutions are

\[ \boxed{ (x,y)=(t,4-t), \qquad t\in\mathbb{R} }. \]

Key Ideas

Where This Leads

Systems of equations extend algebra from one condition at a time to simultaneous relationships. A solution is no longer merely a zero or intercept; it is a point, tuple, or set of values where several mathematical statements agree at once.

From here, the course can develop substitution and elimination in greater depth, organize larger systems with augmented matrices, formalize row operations and row-echelon forms, study Gaussian and Gauss-Jordan elimination, examine matrix equations and inverses, and connect system structure with determinants, rank, and applications throughout algebra, geometry, modeling, and numerical computation.