Solving Systems of Linear Equations

A single linear equation like $2x + y = 8$ has infinitely many solutions — any point on that line works. But when you add a second equation, you're adding a second constraint. Usually there's just one point that satisfies both at the same time, and finding it is what solving a system of equations is all about.

What Is a System of Equations?

A system of equations is a set of two or more equations that share the same variables. You're looking for a solution that makes all equations true at the same time.

Geometrically, each equation in two variables describes a line. The solution to the system is the point where the two lines intersect. That's why:

If the lines cross at one point → one solution
If the lines are parallel (same slope, different intercepts) → no solution
If the lines are identical → infinitely many solutions

Most problems you'll encounter have exactly one solution. There are two main algebraic methods to find it: substitution and elimination.

Solving by Substitution

Substitution works by solving one equation for one variable, then plugging that expression into the other equation. The result is a single equation with one unknown — easy to solve.

Example 1

$$x + y = 10$$ $$3x + 2y = 20$$

Step 1 — Solve the first equation for one variable.

$x + y = 10 \implies y = 10 - x$

Step 2 — Substitute into the second equation.

Replace $y$ with $10 - x$ in the second equation:

$$3x + 2(10 - x) = 20$$ $$3x + 20 - 2x = 20$$ $$x = 0$$

Step 3 — Substitute back to find the other variable.

Plug $x = 0$ into $y = 10 - x$:

$$y = 10 - 0 = 10$$

Solution: $(0, 10)$.

Check: $0 + 10 = 10$ ✓ and $3(0) + 2(10) = 20$ ✓

Example 2

$$y = 2x - 3$$ $$4x - y = 9$$

When one equation is already solved for a variable, substitution is especially quick. Plug $y = 2x - 3$ directly into the second equation:

$$4x - (2x - 3) = 9$$ $$4x - 2x + 3 = 9$$ $$2x = 6 \implies x = 3$$

Then $y = 2(3) - 3 = 3$. Solution: $(3, 3)$.

Solving by Elimination

Elimination works by adding (or subtracting) the two equations to cancel out one variable entirely. The key is to first make the coefficients of one variable equal and opposite.

Example 3

Same system as Example 1:

$$x + y = 10$$ $$3x + 2y = 20$$

To eliminate $x$, match its coefficient in both equations. The first equation has $x$ with coefficient 1, and the second has $3x$. Multiply the first equation by $-3$ throughout:

$$-3(x + y) = -3(10) \implies -3x - 3y = -30$$

Now add this to the second equation:

$$\begin{aligned}-3x - 3y &= -30 \ +\quad 3x + 2y &= \phantom{-}20 \ \hline -y &= -10\end{aligned}$$

So $y = 10$. Substituting back into the first equation: $x + 10 = 10 \implies x = 0$.

Same answer: $(0, 10)$.

Example 4

$$2x + 3y = 12$$ $$5x - 3y = 9$$

This one is set up perfectly for elimination — the $y$ terms are already equal and opposite. Add the two equations directly:

$$7x = 21 \implies x = 3$$

Plug back in: $2(3) + 3y = 12 \implies 6 + 3y = 12 \implies y = 2$. Solution: $(3, 2)$.

Choosing a Method

Both methods always work. A few guidelines:

Use substitution when one equation is already solved for a variable (like $y = 3x + 1$), or when one variable has a coefficient of 1 and is easy to isolate.
Use elimination when both equations are in standard form ($ax + by = c$) and coefficients line up nicely for cancellation.

When you're done, always check your solution by plugging it into both original equations.

Practice Problems

1. Solve by substitution: $$y = x + 2 \qquad 2x + y = 8$$ Show answerSubstitute $y = x + 2$ into the second: $2x + (x+2) = 8 \implies 3x = 6 \implies x = 2$. Then $y = 4$. Solution: $(2, 4)$.

2. Solve by substitution: $$3x - y = 7 \qquad x + 2y = 0$$ Show answerFrom the second equation: $x = -2y$. Substitute: $3(-2y) - y = 7 \implies -7y = 7 \implies y = -1$. Then $x = 2$. Solution: $(2, -1)$.

3. Solve by elimination: $$4x + y = 14 \qquad 2x - y = 4$$ Show answerAdd the equations: $6x = 18 \implies x = 3$. Then $y = 14 - 4(3) = 2$. Solution: $(3, 2)$.

4. Solve by elimination: $$3x + 2y = 16 \qquad 5x - 4y = 6$$ Show answerMultiply the first equation by 2: $6x + 4y = 32$. Add to the second: $6x + 4y + 5x - 4y = 32 + 6$, giving $11x = 38$, so $x = \frac{38}{11}$. Substitute back: $3 \cdot \frac{38}{11} + 2y = 16$, so $2y = \frac{176 - 114}{11} = \frac{62}{11}$, giving $y = \frac{31}{11}$. Solution: $\left(\frac{38}{11}, \frac{31}{11}\right)$.

5. A movie theater charges $12 for adults and $8 for children. A group of 10 people paid $96 total. How many adults and how many children were in the group? Show answerLet $a$ = adults, $c$ = children. System: $a + c = 10$ and $12a + 8c = 96$. From the first: $a = 10 - c$. Substituting: $12(10-c) + 8c = 96 \implies 120 - 4c = 96 \implies c = 6$. So $a = 4$. There were 4 adults and 6 children.