Solving Systems of Linear Equations

Quick answer

A system of equations is two equations that must both be true at once, and its solution is the point where the graphs cross. Solve it by graphing, by substitution, or by elimination. Parallel lines never cross so the system has no solution, and identical lines cross everywhere so it has infinitely many.

What you'll learn

Solve a system of two linear equations by graphing, substitution or elimination
Explain why the solution is the point where the graphs intersect
Recognize systems with no solution or infinitely many solutions

What a system is

A system of equations is two or more equations that must hold at the same time:

\begin{cases} y = 2x + 1 \\[2pt] y = -x + 7 \end{cases}

Solving it means finding the values of $x$ and $y$ that satisfy both. One equation alone has infinitely many solutions; demanding both narrows it, usually to a single pair.

Why the answer is where the graphs cross

This follows from something established in graphing linear equations: a graph is the picture of every solution of its equation.

So a point on the first line satisfies the first equation. A point on the second line satisfies the second. A point on both lines satisfies both equations at once — which is the definition of solving the system.

Two different straight lines can share at most one point, so a system of two lines has at most one solution. There is nowhere else to look.

y = 2x + 1
y = -x + 7

The system y = 2x + 1 and y = -x + 7

Three methods

Method	Best when	Gives
Graphing	you want to see what is happening	exact answers only if the crossing lands on grid lines
Substitution	one equation has a variable alone	exact answers
Elimination	coefficients match or are opposites	exact answers

Graphing is the one that explains; the other two are the ones you rely on.

Why elimination cannot change the answer

Elimination replaces one equation with the sum of the two. That produces a different system, and it is fair to ask why the new system has the same solutions as the old one. The answer takes two short arguments.

Call the equations $A$ and $B$ , and replace $A$ with $A + kB$ for some number $k$ , keeping $B$ :

\begin{cases} A \\[2pt] B \end{cases} \quad\longrightarrow\quad \begin{cases} A + kB \\[2pt] B \end{cases}

No solution is lost. Suppose $(x, y)$ makes both $A$ and $B$ true. Then the two sides of $A$ are equal numbers, and so are the two sides of $B$ . Multiplying equal numbers by $k$ keeps them equal, and adding equal numbers to equal numbers keeps them equal. So $(x, y)$ makes $A + kB$ true as well.

No solution is gained. Suppose $(x, y)$ makes both $A + kB$ and $B$ true. Subtract $k$ copies of $B$ from $A + kB$ , by the same reasoning, and what is left is $A$ . So $(x, y)$ makes $A$ true too.

The two systems are satisfied by exactly the same pairs. Every elimination step swaps a system for an equivalent one, which is why the answer at the end is the answer to the system at the start.

Worked examples

Example 3 — By elimination

Solve

\begin{cases} 3x + 2y = 16 \\[2pt] 3x - 2y = 4 \end{cases}

The $y$ -coefficients are $+2$ and $-2$ , which are opposites. Adding the two equations therefore removes $y$ entirely:

6x = 20 \quad\longrightarrow\quad x = \frac{10}{3}

A fraction is a perfectly good answer. The arithmetic checks: $16 + 4 = 20$ and $3x + 3x = 6x$ .

Substitute into the first equation:

3\left(\frac{10}{3}\right) + 2y = 16 \quad\longrightarrow\quad 10 + 2y = 16 \quad\longrightarrow\quad y = 3

The solution is $\left(\tfrac{10}{3}, 3\right)$ .

Check in the second equation. $3\left(\tfrac{10}{3}\right) - 2(3) = 10 - 6 = 4$ ✓

Common mistakes

Practice problems

Solve by substitution: $y = x + 3$ and $y = 2x + 1$ .

Hint

Both give $y$ , so set the two expressions equal.

Answer

$(2, 5)$

Full solution

$x + 3 = 2x + 1$ , so $2 = x$ .

Then $y = 2 + 3 = 5$ . Check: $2(2) + 1 = 5$ ✓
Solve by elimination: $x + y = 10$ and $x - y = 4$ .

Answer

$(7, 3)$

Full solution

The $y$ -coefficients are opposites, so add: $2x = 14$ , giving $x = 7$ .

Then $7 + y = 10$ , so $y = 3$ . Check: $7 - 3 = 4$ ✓
Solve: $y = 4x$ and $y = x + 6$ .

Answer

$(2, 8)$

Full solution

$4x = x + 6$ , so $3x = 6$ and $x = 2$ . Then $y = 4(2) = 8$ .

Check: $2 + 6 = 8$ ✓
Solve: $2x + y = 7$ and $x - y = 2$ .

Hint

The $y$ -coefficients are already opposites.

Answer

$(3, 1)$

Full solution

Add the equations: $3x = 9$ , so $x = 3$ .

Then $3 - y = 2$ , giving $y = 1$ . Check: $2(3) + 1 = 7$ ✓
How many solutions does this system have: $y = 5x + 2$ and $y = 5x - 3$ ?

Answer

None.

Full solution

Setting them equal gives $2 = -3$ , which is false.

Both lines have slope $5$ , so they are parallel and never meet.
How many solutions does this system have: $y = x + 4$ and $3y = 3x + 12$ ?

Answer

Infinitely many.

Full solution

Dividing the second by $3$ gives $y = x + 4$ , the same equation.

The two describe one line, so every point on it is a solution.
Solve: $3x + 2y = 12$ and $x + 2y = 8$ .

Hint

The $y$ -coefficients match, so subtract rather than add.

Answer

$(2, 3)$

Full solution

Subtract the second from the first: $2x = 4$ , so $x = 2$ .

Then $2 + 2y = 8$ , giving $2y = 6$ and $y = 3$ .

Check: $3(2) + 2(3) = 12$ ✓
A store sells pens at $2 and notebooks at $5. A customer buys 12 items for $39. How many of each?

Hint

One equation counts items, the other counts money.

Answer

7 pens and 5 notebooks.

Full solution

Let $p$ be pens and $n$ notebooks: $p + n = 12$ and $2p + 5n = 39$ .

From the first, $p = 12 - n$ . Substituting: $2(12 - n) + 5n = 39$ .

So $24 + 3n = 39$ , giving $3n = 15$ and $n = 5$ . Then $p = 7$ .

Check: $7 + 5 = 12$ ✓ and $2(7) + 5(5) = 14 + 25 = 39$ ✓
The system $x + y = 10$ and $x - y = 4$ has the single solution $(7, 3)$ . Kai replaces both equations with their sum. What happens to the solutions, and why?

Hint

Write out the system Kai ends up with.

Answer

Kai’s system has infinitely many solutions: every pair $(7, y)$ .

Full solution

The sum of the two equations is $2x = 14$ . Kai’s system is that equation written twice, which says only that $x = 7$ .

Any $y$ at all now works, so $(7, 0)$ , $(7, 3)$ and $(7, 100)$ are all solutions. The original system allowed only $(7, 3)$ .

Replacing one equation keeps the other one on the books, and that retained equation is what rules out the extra pairs. Kai threw it away, so solutions were gained.

Frequently asked questions

Which method should I use?

Substitution is easiest when one equation already has a variable alone, like y = 2x + 1. Elimination is easiest when the same variable has matching or opposite coefficients. Graphing shows what is happening but only gives exact answers when the crossing point lands on grid lines.

Why does the solution have to be the intersection?

A point on a line is a solution of that line's equation. A point on both lines satisfies both equations at once, which is exactly what solving a system asks for. So the crossing point is the only candidate.

What if the variables all cancel out?

Then the lines are parallel or identical. If what remains is false, like 0 = 5, the lines never meet and there is no solution. If it is true, like 0 = 0, they are the same line and every point on it is a solution.

What to learn next

Formulas on this page

Standard form of a line

Key terms in this lesson

System of equations: A system of equations is two or more equations that must all be true at once. For two lines, the solution is the point where their graphs cross.

Standards alignment

This lesson covers the following Common Core State Standards for Mathematics.

CCSS.MATH.CONTENT.HSA.REI.C.5Reasoning with Equations and InequalitiesProve that, given a system of two equations in two variables, replacing one equation by the sum of that equation and a multiple of the other produces a system with the same solutions.
CCSS.MATH.CONTENT.HSA.REI.C.6Reasoning with Equations and InequalitiesSolve systems of linear equations exactly and approximately (e.g., with graphs), focusing on pairs of linear equations in two variables.
CCSS.MATH.CONTENT.8.EE.C.8Expressions and EquationsAnalyze and solve pairs of simultaneous linear equations.
CCSS.MATH.CONTENT.8.EE.C.8aExpressions and EquationsUnderstand that solutions to a system of two linear equations in two variables correspond to points of intersection of their graphs, because points of intersection satisfy both equations simultaneously.
CCSS.MATH.CONTENT.8.EE.C.8bExpressions and EquationsSolve systems of two linear equations in two variables algebraically, and estimate solutions by graphing the equations. Solve simple cases by inspection.
CCSS.MATH.CONTENT.8.EE.C.8cExpressions and EquationsSolve real-world and mathematical problems leading to two linear equations in two variables.

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Quick answer

What you'll learn

What a system is

Why the answer is where the graphs cross

Three methods

Why elimination cannot change the answer

Worked examples

Common mistakes

Practice problems

Frequently asked questions

Which method should I use?

Why does the solution have to be the intersection?

What if the variables all cancel out?

What to learn next

Formulas on this page

Key terms in this lesson

Standards alignment

More Algebra 1 lessons