# How to solve systems of equations

Canonical: https://duckyhelper.com/learn/algebra-1/systems-of-equations/
Updated: 2026-10-01

A system of equations is two equations that share the same variables. Its solution is the pair \((x, y)\) that makes both equations true, which is where their lines cross. You can solve by graphing, by substitution (solve one equation for a variable and plug it into the other), or by elimination (add or subtract the equations so one variable cancels). Parallel lines mean no solution.

## The key idea

One equation with two variables has endless solutions: every point on its line. A second equation adds a second line. The point where both lines cross is the only pair that works for both, so that is the solution of the system.

$$
\begin{cases} y = 2x - 1 \\ 3x + 2y = 12 \end{cases} \qquad \text{solution: } (2, 3)
$$

**Pick the method that fits the setup**

| Method | Use it when | What you do |
| --- | --- | --- |
| Substitution | one variable is already alone, like \(y = 2x - 1\) | plug that expression into the other equation |
| Elimination | both equations look like \(Ax + By = C\) | add or subtract so x or y cancels, multiplying first if needed |
| Graphing | you want to see the answer or check it | graph both lines and read the crossing point |

**How many solutions?**

| The lines are | Slopes | Solutions | What the algebra shows |
| --- | --- | --- | --- |
| crossing | different | exactly one | a single value for x and for y |
| parallel | same, different intercepts | none | a false statement like \(0 = 6\) |
| the same line | same, same intercept | infinitely many | a true statement like \(0 = 0\) |

## Worked examples

**Example 1: substitution**

Problem: Solve the system \(y = 2x - 1\) and \(3x + 2y = 12\).

1. The first equation says y is the same as \(2x - 1\). Replace y in the second equation with \(2x - 1\), in parentheses.

   $$
   3x + 2(2x - 1) = 12
   $$
2. Distribute and combine like terms.

   $$
   7x - 2 = 12
   $$
3. Add 2, then divide by 7.

   $$
   x = 2
   $$
4. Put \(x = 2\) into the equation that is already solved for y.

   $$
   y = 2(2) - 1 = 3
   $$
5. Check in the other equation: \(3(2) + 2(3) = 12\). It works.

Answer: \((2, 3)\)

**Example 2: elimination by adding**

Problem: Solve the system \(4x + 3y = 6\) and \(2x - 3y = 12\).

1. The y terms are \(3y\) and \(-3y\). Adding the equations makes them cancel.

   $$
   (4x + 3y) + (2x - 3y) = 6 + 12
   $$
2. Simplify.

   $$
   6x = 18
   $$
3. Divide by 6.

   $$
   x = 3
   $$
4. Put \(x = 3\) into either original equation. The first one gives:

   $$
   4(3) + 3y = 6
   $$
5. Subtract 12, then divide by 3.

   $$
   y = -2
   $$

Answer: \((3, -2)\)

**Example 3: elimination after multiplying both equations**

Problem: Solve the system \(3x + 4y = 10\) and \(5x - 6y = 4\).

1. Nothing cancels yet. The y coefficients are 4 and \(-6\), and both go into 12. Multiply the first equation by 3 and the second by 2, every term.

   $$
   9x + 12y = 30
   $$
2. The second equation times 2:

   $$
   10x - 12y = 8
   $$
3. Add the two new equations. The y terms cancel.

   $$
   19x = 38
   $$
4. Divide by 19.

   $$
   x = 2
   $$
5. Put \(x = 2\) into \(3x + 4y = 10\).

   $$
   6 + 4y = 10
   $$
6. Solve for y, then check in the second equation: \(5(2) - 6(1) = 4\).

   $$
   y = 1
   $$

Answer: \((2, 1)\)

**Example 4 (test-hard): a mixture word problem**

Problem: A coffee shop mixes a $9 per pound coffee with a $14 per pound coffee to make 20 pounds of a blend worth $11 per pound. How many pounds of each does it use?

1. Let \(c\) be pounds of the $9 coffee and \(p\) be pounds of the $14 coffee. The weights add to 20, and the values add to \(20 \times 11 = 220\) dollars. Write both equations: \(c + p = 20\) and \(9c + 14p = 220\).
2. Multiply the first equation by 9 so the c terms match: \(9c + 9p = 180\). Subtract it from the second equation.

   $$
   (9c + 14p) - (9c + 9p) = 220 - 180
   $$
3. Simplify.

   $$
   5p = 40
   $$
4. Divide by 5.

   $$
   p = 8
   $$
5. Then \(c = 20 - 8 = 12\). Check the value: \(9(12) + 14(8) = 108 + 112 = 220\).

Answer: 12 pounds of the $9 coffee and 8 pounds of the $14 coffee.

## Common mistakes

- **Plugging back into the equation you just used.** In substitution, putting x back into the rearranged equation you substituted from only gives \(0 = 0\). Fix: use the other equation, or the one already solved for y.
- **Scaling only part of an equation.** When you multiply \(3x + 4y = 10\) by 3, the 10 becomes 30 too.
- **Sign slips when subtracting equations.** Subtracting \(9c + 9p\) means subtracting both terms. Fix: put the whole equation in parentheses, or multiply it by \(-1\) and add instead.
- **Stopping after one variable.** The solution of a system is an ordered pair. Find both values and write \((x, y)\).
- **Mixing up 0 = 0 and 0 = 6.** A true statement means the same line (infinitely many solutions). A false one means parallel lines (no solution).

## Quick methods

> **Tip: Check any answer by graphing**
>
> Type both equations into a graphing calculator such as Desmos and tap the point where the lines cross. If it matches your pair, you are done. It is a great check, and it shows no-solution systems as parallel lines. Teachers often still want the algebra, so show the work too.

> **Note: Spot the number of solutions without solving**
>
> Write both equations in \(y = mx + b\) form. Different slopes: one solution. Same slope, different b: none. Same slope and same b: infinitely many.

## Practice

**5 practice questions**

1. Solve the system \(y = x + 4\) and \(2x + y = 10\).
   A. \((2, 6)\)
   B. \((6, 2)\)
   C. \((3, 7)\)
   D. \((2, 4)\)

   Answer: \((2, 6)\). Substitute: \(2x + (x + 4) = 10\), so \(3x = 6\) and \(x = 2\). Then \(y = 2 + 4 = 6\). The pair \((6, 2)\) has the right numbers in the wrong order. \((3, 7)\) fits the first equation but not the second.

2. Solve the system \(5x + 2y = 1\) and \(3x - 2y = 15\).
   A. \(\left(2, -\frac{9}{2}\right)\)
   B. \(\left(2, \frac{9}{2}\right)\)
   C. \(\left(-2, \frac{11}{2}\right)\)
   D. \(\left(-\frac{9}{2}, 2\right)\)

   Answer: \(\left(2, -\frac{9}{2}\right)\). Add the equations: \(8x = 16\), so \(x = 2\). Then \(10 + 2y = 1\), so \(2y = -9\) and \(y = -\frac{9}{2}\). The choice with \(+\frac{9}{2}\) loses the sign when subtracting 10.

3. How many solutions does the system \(y = 3x + 2\) and \(6x - 2y = 8\) have?
   A. None
   B. Exactly one
   C. Exactly two
   D. Infinitely many

   Answer: None. Solve the second for y: \(y = 3x - 4\). Both lines have slope 3 but different intercepts (2 and \(-4\)), so they are parallel and never cross. Two lines can never cross exactly twice.

4. For what value of \(k\) does the system \(2x + 5y = 7\) and \(4x + ky = 14\) have infinitely many solutions?
   A. \(5\)
   B. \(7\)
   C. \(10\)
   D. \(14\)

   Answer: \(10\). Infinitely many solutions means the second equation is the first one multiplied by a number. \(4x\) and 14 are both 2 times \(2x\) and 7, so \(ky\) must be 2 times \(5y\): \(k = 10\). With \(k = 5\) the lines just cross once.

5. A farm has chickens and goats. Together they have 30 heads and 84 legs. How many goats are there?
   A. 12
   B. 18
   C. 21
   D. 24

   Answer: 12. Let \(c\) be chickens and \(g\) be goats: \(c + g = 30\) and \(2c + 4g = 84\). Double the first: \(2c + 2g = 60\). Subtract: \(2g = 24\), so \(g = 12\). 18 is the number of chickens, and 24 forgets to divide by 2.

## Frequently asked questions

### Which method should I use to solve a system?

Use substitution when one equation already has x or y alone. Use elimination when both equations are in \(Ax + By = C\) form, especially if a variable already has matching or opposite coefficients. Graphing is best for checking, or when the question asks what the system looks like.

### What does the solution of a system mean on a graph?

Each equation is a line. The solution is the point where the two lines cross, because that point is on both lines, so its x and y make both equations true. Parallel lines never cross, so they have no solution.

### How can I tell if a system has no solution or infinitely many?

Solve it and look at the end. If the variables cancel and you get something false, like \(0 = 6\), there is no solution. If you get something true, like \(0 = 0\), the equations are the same line and there are infinitely many solutions.

### Do I always have to find both x and y?

For a full solution, yes, because the answer is a point \((x, y)\). In a word problem, read the question: if it only asks for the number of goats, you can stop once you have that, but finding the other value is a good way to check.

## Related

- [Slope-intercept form and how to write the equation of a line](https://duckyhelper.com/learn/algebra-1/slope-intercept-form/)
- [How to solve multi-step equations](https://duckyhelper.com/learn/algebra-1/solving-equations/)
- [Systems of linear equations on the SAT](https://duckyhelper.com/learn/sat-math/systems-of-equations/)
- [Algebra 1 study guides](https://duckyhelper.com/learn/algebra-1/)

## Try asking Ducky

- "Should I use substitution or elimination for this system?"
- "I got (6, 2) but the answer is (2, 6). Did I just mix up the order?"
- "Help me turn this word problem into two equations."

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