Systems of Equations

A system of equations is a set of two or more equations with the same variables. The solution is the set of values that makes all the equations true at the same time. For two linear equations in two variables, the solution is the point where the two lines intersect.

What the Solution Looks Like Graphically

When you graph two linear equations on the same coordinate plane, three things can happen:

Most systems you encounter will have exactly one solution — one intersection point.

Method 1: Substitution

Use substitution when one equation already has a variable isolated (or can easily be solved for one variable).

Steps:

1. Solve one equation for one variable
2. Substitute that expression into the other equation
3. Solve the resulting one-variable equation
4. Substitute back to find the other variable

Example 1

Solve the system:

$y = 3x - 1$

$2x + y = 9$

Step 1 — The first equation already gives $y$ in terms of $x$.

Step 2 — Substitute $3x - 1$ for $y$ in the second equation:

$2x + (3x - 1) = 9$

Step 3 — Solve for $x$:

$5x - 1 = 9$

$5x = 10$

$x = 2$

Step 4 — Substitute $x = 2$ back into the first equation:

$y = 3(2) - 1 = 6 - 1 = 5$

Answer: $(x, y) = (2, 5)$

Check: Equation 1: $5 = 3(2) - 1 = 5$ . Equation 2: $2(2) + 5 = 4 + 5 = 9$ . Both true.

Example 2

Solve the system:

$x + 4y = 14$

$3x - 2y = 0$

Step 1 — Solve the first equation for $x$:

$x = 14 - 4y$

Step 2 — Substitute into the second equation:

$3(14 - 4y) - 2y = 0$

$42 - 12y - 2y = 0$

$42 - 14y = 0$

Step 3 — Solve for $y$:

$14y = 42$

$y = 3$

Step 4 — Substitute back:

$x = 14 - 4(3) = 14 - 12 = 2$

Answer: $(x, y) = (2, 3)$

Method 2: Elimination

Use elimination when both equations are in standard form and neither variable is easily isolated. The idea is to add or subtract the equations to eliminate one variable.

Steps:

1. Multiply one or both equations so that one variable has opposite coefficients
2. Add the equations to eliminate that variable
3. Solve for the remaining variable
4. Substitute back to find the other variable

Example 3

Solve the system:

$3x + 2y = 16$

$x - 2y = -4$

Step 1 — The $y$-coefficients are already opposites ($+2y$ and $-2y$).

Step 2 — Add the equations:

$(3x + 2y) + (x - 2y) = 16 + (-4)$

$4x = 12$

Step 3 — Solve for $x$:

$x = 3$

Step 4 — Substitute into the first equation:

$3(3) + 2y = 16$

$9 + 2y = 16$

$2y = 7$

$y = 3.5$

Answer: $(x, y) = (3, 3.5)$

Example 4: Requiring Multiplication First

Solve the system:

$2x + 3y = 12$

$5x + 2y = 13$

Step 1 — Multiply to create opposite coefficients for $y$:

Multiply the first equation by 2 and the second by $-3$:

$4x + 6y = 24$

$-15x - 6y = -39$

Step 2 — Add:

$4x + (-15x) + 6y + (-6y) = 24 + (-39)$

$-11x = -15$

$x = \frac{15}{11}$

This gives a fraction, which is fine. Now find $y$ by eliminating $x$ instead. Multiply the first equation by 5 and the second by $-2$:

$10x + 15y = 60$

$-10x - 4y = -26$

Add:

$11y = 34$

$y = \frac{34}{11}$

Using elimination twice (once for each variable) avoids messy fraction substitution entirely.

Answer: $(x, y) = \left(\frac{15}{11}, \frac{34}{11}\right)$

Not every system produces whole-number answers, and that is perfectly normal.

Special Cases

No Solution (Parallel Lines)

$2x + y = 5$

$2x + y = 8$

Subtracting the first from the second gives $0 = 3$, which is a contradiction. The lines are parallel and never intersect. There is no solution.

Infinitely Many Solutions (Same Line)

$x + 2y = 6$

$2x + 4y = 12$

The second equation is just the first multiplied by 2. They represent the same line, so every point on that line is a solution.

Real-World Application: Retail — Comparing Two Pricing Models

A store manager is choosing between two suppliers for a product:

• Supplier A: $3 per unit plus a $150 monthly fee: $y = 3x + 150$
• Supplier B: $5 per unit with no monthly fee: $y = 5x$

At what quantity do both suppliers cost the same?

Set the equations equal:

$3x + 150 = 5x$

$150 = 2x$

$x = 75$

Find the cost:

$y = 5(75) = 375$

Answer: At 75 units, both suppliers cost $375. Below 75 units, Supplier B is cheaper (no monthly fee). Above 75 units, Supplier A is cheaper (lower per-unit cost). The manager should choose the supplier based on expected monthly volume.

Common Mistakes to Avoid

1. Forgetting to multiply the entire equation. When multiplying to create opposite coefficients, every term on both sides must be multiplied — including the constant.
2. Adding when you should subtract (or vice versa). If coefficients are the same sign, subtract the equations. If opposite signs, add.
3. Stopping after finding one variable. The solution is an ordered pair — you need both $x$ and $y$.
4. Assuming no fraction answers. Real-world problems often have non-integer solutions.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• A system of equations finds where two lines intersect — the values that satisfy both equations simultaneously
• Substitution works best when one variable is already isolated
• Elimination works best when equations are in standard form — multiply to create opposite coefficients, then add
• Systems can have one solution (intersecting lines), no solution (parallel lines), or infinitely many solutions (same line)
• Always check your solution in both original equations

Return to Algebra for more topics in this section.