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Systems of Equations

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Systems of Equations - Complete Interactive Lesson

Part 1: What Is a System & What Does a Solution Mean?

🔗 Systems of Equations

Part 1 of 7 — What Is a System & What Does a Solution Mean?

Topics in This Part

Section
What Is a System of Equations?
Solutions as Intersection Points
Checking a Candidate Solution
Counting Solutions Graphically

🔑 Key Concept: A system is two or more equations considered at the same time. A solution is an ordered pair (or triple) that makes every equation true at once — geometrically, a point where all the graphs cross.

What Is a System of Equations?

A system of equations is a set of equations sharing the same variables. For two variables, it usually looks like:

$\begin{cases} a_1 x + b_1 y = c_1 \\ a_2 x + b_2 y = c_2 \end{cases}$

We want the values of $x$ and $y$ that satisfy both equations simultaneously.

A first example

$\begin{cases} x + y = 5 \\ x - y = 1 \end{cases}$

The pair $(3, 2)$ works because:

$3 + 2 = 5$ ✓
$3 - 2 = 1$ ✓

No other pair makes both true, so $(3, 2)$ is the solution.

💡 One equation in two variables has infinitely many solutions (a whole line). Pairing it with a second equation usually pins the answer down to a single point.

Solutions Are Intersection Points

Each linear equation graphs as a line. The solution of the system is exactly where those lines cross.

System graph	What it means
Lines cross at one point	One solution
Lines are parallel (never meet)	No solution
Lines are identical (lie on top of each other)	Infinitely many solutions

For $\begin{cases} x + y = 5 \\ x - y = 1 \end{cases}$ , the two lines intersect at — a single crossing, so exactly one solution.

Concept Check 🎯

Verify a Candidate 🧮

To check whether a pair solves a system, substitute it into both equations. Use the system $\begin{cases} 2x + y = 8 \\ x - y = 1 \end{cases}$

1) Substitute into . What value do you get? Substitute into . What value do you get? Is a solution of the system? Enter for yes, for no.

How Many Solutions? 🔽

Match each graphical situation to its number of solutions.

Where We're Headed

You now know what a system is and what its solution means. Graphing works but is imprecise — reading $(3, 2)$ off a sketch is easy, but $(\tfrac{7}{3}, -\tfrac{1}{6})$ is not.

Part 2: The Substitution Method

🔗 Systems of Equations

Part 2 of 7 — The Substitution Method

🔑 The Idea: Solve one equation for one variable, then substitute that expression into the other equation. This collapses two equations in two unknowns into one equation in one unknown.

The Substitution Steps

Isolate one variable in one equation (pick the easiest — a variable with coefficient $1$ is ideal).
Substitute that expression into the other equation.
Solve the resulting one-variable equation.
Back-substitute to find the second variable.
Check in both original equations.

Worked Example

$\begin{cases} y = 2x - 1 \\ 3x + y = 14 \end{cases}$

Part 3: The Elimination Method

🔗 Systems of Equations

Part 3 of 7 — The Elimination Method

🔑 The Idea: Add or subtract the equations so that one variable cancels. If the coefficients don't already match, multiply an equation by a constant first to make them match.

The Elimination Steps

Line up the equations so like terms are in columns.
Match the magnitude of one variable's coefficients (multiply one or both equations if needed).
Add (if the matched coefficients have opposite signs) or subtract (if they have the same sign) to eliminate that variable.
Solve for the remaining variable, then back-substitute.

Worked Example — opposite signs, just add

$\begin{cases} 2x + y = 7 \\ x - y = 2 \end{cases}$

Part 4: No Solution & Infinitely Many Solutions

🔗 Systems of Equations

Part 4 of 7 — No Solution & Infinitely Many Solutions

🔑 The Idea: Not every system has a unique answer. When the variables all cancel, the leftover number-statement tells you which special case you're in.

The Three Outcomes

Outcome	Variables cancel and you're left with…	Graph	Name
One solution	a value, e.g. $x = 3$	lines cross once	consistent, independent
No solution	a false statement, e.g. $0 = 5$

Part 5: Nonlinear Systems

🔗 Systems of Equations

Part 5 of 7 — Nonlinear Systems

🔑 The Idea: When one equation is curved (a parabola, circle, or exponential), substitution still works — but a line can cross a curve twice, once, or not at all, so watch the solution count.

Line Meets Parabola

$\begin{cases} y = x^2 \\ y = x + 2 \end{cases}$

Part 6: Three Variables & Word Problems

🔗 Systems of Equations

Part 6 of 7 — Three Variables & Word Problems

🔑 The Idea: With three unknowns you need three equations. Eliminate one variable to shrink the problem to a familiar two-variable system, then back-substitute up the chain.

Solving a 3-Variable System

$\begin{cases} x + y + z = 6 \\ x - y + z = 2 \\ 2x + y - z = 1 \end{cases}$

Part 7: Mixed Practice & Mastery Check

🔗 Systems of Equations

Part 7 of 7 — Mixed Practice & Mastery Check

You can now (1) verify and count solutions, (2) solve by substitution, (3) solve by elimination, (4) recognize no-solution and infinite cases, (5) tackle nonlinear systems, and (6) handle three variables and word problems. Time to put it all together.

Quick Reference

Goal	Key move
Verify a solution	substitute into all equations; every one must hold
A variable is isolated	use substitution
Coefficients match/oppose	use elimination (add if opposite signs, subtract if same)
Variables cancel → false ( $0=5$ )	no solution (parallel lines)
Variables cancel → true ( $0 =$ )

🔑 Big Idea: "Solve the system" and "find where the graphs intersect" are the same task stated two ways.

The next four parts build exact algebraic methods:

Part 2 — Substitution
Part 3 — Elimination
Part 4 — Special cases (no solution / infinitely many)
Part 5 — Nonlinear systems

Then we extend to three variables (Part 6) and finish with mixed mastery (Part 7).

The first equation already gives $y$ . Substitute $y = 2x - 1$ into the second:

$3x + (2x - 1) = 14$ $5x - 1 = 14 \;\Rightarrow\; 5x = 15 \;\Rightarrow\; x = 3$

Back-substitute: $y = 2(3) - 1 = 5$ . Solution: $(3, 5)$ .

✅ Check: $3(3) + 5 = 9 + 5 = 14$ ✓ and $y = 2(3)-1 = 5$ ✓.

When You Must Isolate First

If no variable is pre-isolated, solve for one yourself.

$\begin{cases} x + 3y = 9 \\ 2x - y = 4 \end{cases}$

Isolate $x$ in the first equation: $x = 9 - 3y$ . Substitute into the second:

$2(9 - 3y) - y = 4$ $18 - 6y - y = 4 \;\Rightarrow\; 18 - 7y = 4$

Back-substitute: $x = 9 - 3(2) = 9 - 6 = 3$ . Solution: $(3, 2)$ .

✅ Check: $3 + 3(2) = 9$ ✓ and $2(3) - 2 = 4$ ✓.

💡 Choosing the variable with coefficient $1$ ( $x$ here) avoids fractions and keeps the algebra clean.

Concept Check 🎯

Solve by Substitution 🧮

Solve each system. Enter the $x$ -value, then the $y$ -value.

1) $\begin{cases} y = x + 2 \\ 2x + y = 11 \end{cases}$ → $x = \,?$ , $y = \,?$ 2) $\begin{cases} x = 3y - 1 \\ 2x + 4y = 18 \end{cases}$ → $x = \,?$ , $y = \,?$

Trace the Steps 🔽

You are solving $\begin{cases} x + 3y = 9 \\ 2x - y = 4 \end{cases}$ by substitution. Choose each stage.

The $y$ -coefficients are $+1$ and $-1$ . Add the equations:

$(2x + y) + (x - y) = 7 + 2 \;\Rightarrow\; 3x = 9 \;\Rightarrow\; x = 3$

Back-substitute into $x - y = 2$ : $3 - y = 2 \Rightarrow y = 1$ . Solution: $(3, 1)$ .

✅ Check: $2(3) + 1 = 7$ ✓ and $3 - 1 = 2$ ✓.

When You Must Multiply First

$\begin{cases} 3x + 2y = 16 \\ 5x - 4y = 1 \end{cases}$

To eliminate $y$ , multiply the first equation by $2$ so the $y$ -coefficients become $+4$ and $-4$ :

$\begin{cases} 6x + 4y = 32 \\ 5x - 4y = 1 \end{cases}$

Add: $11x = 33 \Rightarrow x = 3$ . Back-substitute into $3x + 2y = 16$ :

$3(3) + 2y = 16 \;\Rightarrow\; 9 + 2y = 16 \;\Rightarrow\; 2y = 7 \;\Rightarrow\; y = \tfrac{7}{2}$

Solution: $\left(3, \tfrac{7}{2}\right)$ .

⚠️ Watch the sign rule: add when the matched coefficients are opposite ( $+4$ and $-4$ ); subtract when they are the same ( $+4$ and $+4$ ). Adding same-sign coefficients does not eliminate anything.

Concept Check 🎯

Solve by Elimination 🧮

Solve each system. Enter $x$ , then $y$ . Use fractions like 7/2 when needed.

1) $\begin{cases} x + y = 10 \\ x - y = 4 \end{cases}$ → $x = \,?$ , $y = \,?$ 2) $\begin{cases} 3x + 2y = 16 \\ 5x - 4y = 1 \end{cases}$ → $x = \,?$ , $y = \,?$

Substitution or Elimination? 🔽

Pick the method that's usually fastest for each system. (Both methods always work — this is about efficiency.)

💡 The rule of thumb: a false number-statement ( $0 = 5$ ) means no solution; a true one ( $0 = 0$ ) means infinitely many.

Worked Example — No Solution

$\begin{cases} 2x + y = 4 \\ 4x + 2y = 20 \end{cases}$

Multiply the first equation by $2$ : $4x + 2y = 8$ . Now subtract from the second:

$(4x + 2y) - (4x + 2y) = 20 - 8 \;\Rightarrow\; 0 = 12$

The statement $0 = 12$ is false, so there is no solution. The lines have the same slope ( $-2$ ) but different intercepts — they're parallel.

Worked Example — Infinitely Many

$\begin{cases} x - 3y = 6 \\ 2x - 6y = 12 \end{cases}$

Multiply the first by $2$ : $2x - 6y = 12$ — identical to the second. Subtracting gives $0 = 0$ , which is always true, so there are infinitely many solutions (the two equations are the same line).

Concept Check 🎯

Spotting the Case from Slopes

Rewrite both equations as $y = mx + b$ . Then:

Slopes & intercepts	Result
Different slopes	one solution
Same slope, different intercept	no solution
Same slope and same intercept	infinitely many

Example: $2x + y = 4$ becomes $y = -2x + 4$ ; $4x + 2y = 20$ becomes $y = -2x + 10$ . Same slope $-2$ , different intercepts $4$ vs $10$ → no solution, confirming the algebra above.

🔑 You can predict the outcome before solving just by comparing slopes and intercepts.

Classify Each System 🔽

Use slopes/intercepts or the cancellation rule to classify.

Count the Solutions 🧮

For each system, enter the number of solutions: type 0 for no solution, 1 for exactly one, or 9 to mean "infinitely many."

1) $\begin{cases} 3x - y = 2 \\ 6x - 2y = 4 \end{cases}$ → ? 2) $\begin{cases} x + y = 5 \\ x + y = 9 \end{cases}$ → ? 3) $\begin{cases} y = 4x \\ y = -2x + 6 \end{cases}$ → ?

Both equal $y$ , so set them equal and substitute:

$x^2 = x + 2 \;\Rightarrow\; x^2 - x - 2 = 0 \;\Rightarrow\; (x - 2)(x + 1) = 0$

So $x = 2$ or $x = -1$ . Find each $y$ from $y = x + 2$ :

$x = 2 \Rightarrow y = 4$ , giving $(2, 4)$
$x = -1 \Rightarrow y = 1$ , giving $(-1, 1)$

Two solutions — the line slices through the parabola at two points.

✅ Check $(2, 4)$ : $4 = 2^2$ ✓ and $4 = 2 + 2$ ✓.

Concept Check 🎯

Line Meets Circle

$\begin{cases} x^2 + y^2 = 25 \\ y = x + 1 \end{cases}$

Substitute $y = x + 1$ into the circle:

$x^2 + (x + 1)^2 = 25$ $x^{2} + x^{2} + 2 x + 1 = 25 \Rightarrow 2 x^{}$

Divide by $2$ : $x^2 + x - 12 = 0 \Rightarrow (x + 4)(x - 3) = 0$ , so or .

$x = -4 \Rightarrow y = -3$ , giving $(-4, -3)$
$x = 3 \Rightarrow y = 4$ , giving

✅ Check $(3, 4)$ : $3^2 + 4^2 = 9 + 16 = 25$ ✓ and ✓.

⚠️ Don't forget to expand $(x+1)^2$ fully — the middle term $2x$ is the one students lose most often.

Solve the Nonlinear System 🧮

Use the system $\begin{cases} y = x^2 - 1 \\ y = 3 \end{cases}$ .

1) Set the right sides equal: $x^2 - 1 = 3$ . Solve for $x^2$ . Enter $x^2 = \,?$ 2) Enter the negative $x$ -solution. 3) Enter the positive $x$ -solution.

Order the Nonlinear Steps 🔽

You're solving $\begin{cases} x^2 + y^2 = 25 \\ y = x + 1 \end{cases}$ .

x+y+z=6x−y+z=22x+y−z=1

Step 1 — eliminate $z$ . Add equations (1) and (3): the $+z$ and $-z$ cancel:

$(x + y + z) + (2x + y - z) = 6 + 1 \;\Rightarrow\; 3x + 2y = 7 \quad \text{(A)}$

Add equations (2) and (3):

$(x - y + z) + (2x + y - z) = 2 + 1 \;\Rightarrow\; 3x = 3 \;\Rightarrow\; x = 1 \quad \text{(B)}$

Step 2 — back up. From (A): $3(1) + 2y = 7 \Rightarrow 2y = 4 \Rightarrow y = 2$ .

Step 3 — find $z$ . From equation (1): $1 + 2 + z = 6 \Rightarrow z = 3$ .

Solution: $(x, y, z) = (1, 2, 3)$ .

✅ Check in equation (2): $1 - 2 + 3 = 2$ ✓.

Concept Check 🎯

Word Problems → Systems

Problem. Adult tickets cost $8 and child tickets cost $5. A family bought $7$ tickets for $47 total. How many of each?

Let $a$ = adult tickets, $c$ = child tickets. Translate:

$\begin{cases} a + c = 7 \\ 8a + 5c = 47 \end{cases}$

Solve the first for $c = 7 - a$ and substitute:

$8a + 5(7 - a) = 47 \;\Rightarrow\; 8a + 35 - 5a = 47 \;\Rightarrow\; 3a = 12 \;\Rightarrow\; a = 4$

So $c = 7 - 4 = 3$ : 4 adult tickets and 3 child tickets.

✅ Check: $8(4) + 5(3) = 32 + 15 = 47$ ✓ and $4 + 3 = 7$ ✓.

💡 Translation tips: "total count" gives the count equation; "total cost/value" gives the value equation. Label your variables before writing equations.

Set Up & Solve a Word Problem 🧮

A coffee shop sells small drinks for $3 and large drinks for $5. One morning it sold $20$ drinks for $78 total. Let $s$ = number of small and $\ell$ = number of large.

1) The count equation is $s + \ell = \,?$ 2) Solve the system. How many small drinks were sold? 3) How many large drinks were sold?

Translate the Story 🔽

A theater sold $x$ adult and $y$ student tickets. Adults are $12 each, students $7 each. They sold $50$ tickets for $465.

⚠️ Always check your answer in the original equations — especially after multiplying, expanding $(x+1)^2$ , or distributing a negative sign.

Mixed Practice 🎯

Mixed Drill 🧮

1) Solve $\begin{cases} 3x + 2y = 12 \\ x = 2y \end{cases}$ . Enter $x$ , then $y$ . 2) The line $y = 2x$ meets the parabola $y = x^2$ at two points. Enter the two $x$ -values (smaller first).

Diagnose Before Solving 🔽

For each system, identify the smartest first move or the outcome.

Exit Quiz ✅

Answer all three to finish the lesson.

-7y = -14 \;\Rightarrow\; y = 2

x^{2} + x^{2} + 2 x + 1 = 25 \Rightarrow 2 x^{2} + 2 x - 24 = 0