Solving Systems of Equations

🔗 Solving Systems of Equations

Part 1 of 5 — What Is a System & Solving by Graphing

Topics in This Part

🔑 Key Concept: A system of equations is two (or more) equations considered together. The solution is the point $(x, y)$ that makes every equation true at the same time — where the lines cross.

What Is a System of Equations?

A system is a set of equations that share the same variables. In Algebra 1 we usually have two linear equations in $x$ and $y$:

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

The brace $\{$ means "all of these at once." We want the single pair $(x, y)$ that satisfies both lines.

A Solution Is a Point

A point $(x, y)$ is a solution only if it works in both equations. Test $(1, 3)$ in the system above:

Both check out, so $(1, 3)$ is the solution.

⚠️ A point that satisfies only one equation is not a solution to the system. It must satisfy all of them.

Concept Check 🎯

Solving by Graphing

Since the solution makes both equations true, it lies on both lines — so it is the point where the graphs intersect.

Steps

1. Graph each line (use slope-intercept form $y = mx + b$, or a table of points).
2. Find the intersection point.
3. Check the point in both equations.

Worked Example

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

• Line 1 has $y$-intercept $1$, slope $2$.
• Line 2 has $y$-intercept $4$, slope $-1$.

Graphing both, they cross at $(1, 3)$.

✅ Check: $3 = 2(1)+1$ ✓ and $3 = -(1)+4$ ✓. Solution: $(1, 3)$.

Read the Intersection 🧮

Each system below is graphed and the lines cross at a single point. Use the equations to find that point. Enter the $x$-value, then the $y$-value.

1) $\begin{cases} y = x \\ y = -x + 6 \end{cases}$ → $x = \,?$, $\;y = \,?$

Find the Crossing Point 🔽

The lines $\begin{cases} y = 3x \\ y = x + 4 \end{cases}$ cross at one point. Since both equal $y$, set the right sides equal and work through it.

When Graphing Is (and Isn't) Enough

Graphing is great for seeing a solution, but it has limits:

• It's slow to draw accurately.
• It struggles with fractions or decimals — is the crossing at $2$ or $2.1$? Hard to tell by eye.

💡 That's why algebra gives us two exact methods: substitution (Part 2) and elimination (Part 3). Graphing builds the picture; algebra nails the numbers.

🔗 Solving Systems of Equations

Part 2 of 5 — The Substitution Method

🔑 The Idea: If one equation already tells you what a variable equals, substitute that expression into the other equation. That collapses two equations into one equation in one variable — which you already know how to solve.

The Substitution Steps

1. Isolate one variable in one equation (if it isn't already).
2. Substitute that expression into the other equation.
3. Solve the resulting one-variable equation.
4. Back-substitute to find the second variable.
5. Check in both original equations.

Worked Example

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

The first equation already gives $y = 2x - 3$. Substitute into the second:

$3x + (2x - 3) = 12$ $5x - 3 = 12 \;\Rightarrow\; 5x = 15 \;\Rightarrow\; x = 3$

Back-substitute $x = 3$ into $y = 2x - 3$:

$y = 2(3) - 3 = 3$

✅ Check: $3(3) + 3 = 12$ ✓. Solution: $(3, 3)$.

Concept Check 🎯

When You Must Isolate First

If no variable is alone yet, isolate the easiest one — usually a variable with a coefficient of $1$.

Worked Example

$\begin{cases} x + 2y = 11 \\ 3x - y = 5 \end{cases}$

Isolate $y$ in the second equation (its $-y$ is easy): $\;y = 3x - 5$.

Substitute into the first:

$x + 2(3x - 5) = 11$ $x + 6x - 10 = 11 \;\Rightarrow\; 7x = 21 \;\Rightarrow\; x = 3$

Then $y = 3(3) - 5 = 4$.

✅ Check: $3 + 2(4) = 11$ ✓ and $3(3) - 4 = 5$ ✓. Solution: $(3, 4)$.

💡 Tip: Watch the distribution — $2(3x - 5) = 6x - 10$, not $6x - 5$. Forgetting to distribute to both terms is the #1 substitution error.

Walk the Steps 🔽

You're solving $\begin{cases} y = x + 1 \\ 2x + y = 7 \end{cases}$ by substitution. Choose what happens at each stage.

Solve by Substitution 🧮

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

1) $\begin{cases} y = 3x \\ x + y = 8 \end{cases}$ → $x = \,?$, $\;y = \,?$ 2) $\begin{cases} y = x - 2 \\ 2x + y = 10 \end{cases}$ → $x = \,?$, $\;y = \,?$

🔗 Solving Systems of Equations

Part 3 of 5 — The Elimination Method

🔑 The Idea: Add or subtract the two equations so that one variable cancels (its coefficients are opposites). One variable disappears, leaving a single equation to solve. This is also called the addition method.

The Elimination Steps

1. Line up the equations so $x$, $y$, and the constant are in columns.
2. Match a variable's coefficients to be opposites (multiply an equation if needed).
3. Add the equations — one variable cancels.
4. Solve for the remaining variable.
5. Back-substitute and check.

Worked Example — Variables Already Opposite

$\begin{cases} 3x + y = 9 \\ 2x - y = 1 \end{cases}$

The $y$-terms are $+y$ and $-y$ — already opposites. Add the equations:

$(3x + 2x) + (y - y) = 9 + 1$ $5x = 10 \;\Rightarrow\; x = 2$

Back-substitute into $3x + y = 9$: $\;3(2) + y = 9 \Rightarrow y = 3$.

✅ Check: $2(2) - 3 = 1$ ✓. Solution: $(2, 3)$.

Concept Check 🎯

When You Must Multiply First

If no variable's coefficients are opposites, multiply one (or both) equations by a constant so they become opposites.

Worked Example

$\begin{cases} 2x + 3y = 13 \\ x - y = -1 \end{cases}$

The $x$-coefficients are $2$ and $1$. Multiply the second equation by $-2$ so its $x$-term becomes $-2x$ (opposite of $2x$):

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

Now add to the first equation:

$\begin{aligned} 2x + 3y &= 13 \\ -2x + 2y &= 2 \\ \hline 5y &= 15 \end{aligned}$

So $y = 3$. Back-substitute into $x - y = -1$: $\;x - 3 = -1 \Rightarrow x = 2$.

✅ Check: $2(2) + 3(3) = 4 + 9 = 13$ ✓. Solution: $(2, 3)$.

⚠️ When you multiply an equation, multiply every term — including the constant on the right side.

Plan the Elimination 🔽

You're solving $\begin{cases} 4x + 2y = 14 \\ 3x - y = 8 \end{cases}$. Plan and execute the elimination of $y$.

Solve by Elimination 🧮

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

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

🔗 Solving Systems of Equations

Part 4 of 5 — Special Cases & Choosing a Method

🔑 Heads up: Not every system has exactly one solution. Some have none (parallel lines), and some have infinitely many (the same line twice). Recognizing these is a key Algebra 1 skill.

Three Possibilities

A system with at least one solution is consistent; with no solution it's inconsistent. Lines that are the same are dependent; distinct lines are independent.

💡 The tell: When solving, if both variables vanish and you're left with a number sentence — a false one means no solution, a true one means infinitely many.

Worked Examples of Special Cases

No Solution (parallel lines)

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

Substitute: $2x + 1 = 2x - 3 \Rightarrow 1 = -3$. False! The $x$ canceled and left a false statement, so there is no solution. (Same slope $2$, different intercepts — parallel lines.)

Infinitely Many (same line)

$\begin{cases} y = 3x - 2 \\ 2y = 6x - 4 \end{cases}$

The second equation is just the first times $2$. Substitute: $2(3x - 2) = 6x - 4 \Rightarrow 6x - 4 = 6x - 4 \Rightarrow 0 = 0$. Always true, so there are infinitely many solutions — every point on $y = 3x - 2$.

Concept Check 🎯

Pick the Best Method 🔽

For each system, choose the most efficient first move. (All can be solved any way — pick the easiest.)

How Many Solutions? 🧮

For each system, enter the number of solutions: type 0 for none, 1 for exactly one, or 2 for infinitely many.

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

🔗 Solving Systems of Equations

Part 5 of 5 — Word Problems & Mastery Check

You can now solve systems by graphing, substitution, and elimination, and recognize the special cases. The biggest payoff is solving real problems with two unknowns.

Setting Up a Word Problem

1. Define two variables (e.g. let $x =$ adult tickets, $y =$ child tickets).
2. Write two equations — usually one for a count/total and one for a value/amount.
3. Solve with whichever method fits.
4. Answer in words and check it makes sense.

Worked Example — Tickets

A theater sells adult tickets for $8 and child tickets for $5. One night they sold 200 tickets for a total of $1,300. How many of each?

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

$\begin{cases} a + c = 200 \\ 8a + 5c = 1300 \end{cases}$

From the first equation, $c = 200 - a$. Substitute:

$8a + 5(200 - a) = 1300$ $8a + 1000 - 5a = 1300 \;\Rightarrow\; 3a = 300 \;\Rightarrow\; a = 100$

Then $c = 200 - 100 = 100$.

✅ Check: $100 + 100 = 200$ ✓ and $8(100) + 5(100) = 1300$ ✓. 100 adult and 100 child tickets.

Set Up & Solve 🧮

Two numbers have a sum of 30 and a difference of 8 (first minus second). Find them.

Let $x =$ the larger number, $y =$ the smaller. The system is $\begin{cases} x + y = 30 \\ x - y = 8 \end{cases}$.

Enter the larger number $x$, then the smaller number $y$.

Translate the Problem 🎯

Quick Reference

⚠️ Always check your point in both original equations, and for word problems, answer in words with units.

Exit Quiz ✅

Answer all three to finish the lesson.