Unit 4 · Lesson 1 of 4
Systems of equations: what a solution is, and substitution
You'll be able to: check whether a point solves a system, and solve a system by substitution.
Step 1 of 5Warm-up
Two quick questions from earlier lessons. Both skills get used today. Answer from memory first.
Rewrite \(3x + y = 5\) in slope-intercept form, \(y = mx + b\).
Get \(y\) alone: subtract \(3x\) from both sides.
Answer: \(y = -3x + 5\).
Solve \(5x - 2(x - 3) = 18\).
Distribute the \(-2\): \(5x - 2x + 6 = 18\).
Combine: \(3x + 6 = 18\), so \(3x = 12\).
Answer: \(x = 4\).
Step 2 of 5The idea
A system of equations is two equations that have to be true at the same time. Here is one:
\[y = x + 1 \qquad \text{and} \qquad y = -2x + 4\]
A solution of the system is an ordered pair \((x, y)\) that makes both equations true. On a graph, that is the point where the two lines cross.
Two lines can meet in only three ways:
- They cross once: one solution.
- They are parallel (same slope, different y-intercepts): no solution.
- They are the same line: infinitely many solutions.
Reading a graph is slow and not exact, so we solve with algebra. The substitution method turns two equations with two letters into one equation with one letter.
To check a point: put its \(x\) and \(y\) into every equation. It is a solution only if all of them come out true.
Substitution, step by step:
- Get one letter alone in one equation (pick a letter with no number in front, if you can).
- Put that expression, in parentheses, in place of that letter in the other equation.
- Solve the one-letter equation.
- Put that value back in to find the other letter.
- Check the pair in both original equations.
Why this shows up on the placement test: solving linear systems is one of the core MATH 1010 skills the placement checks, and substitution is also how you'll solve harder systems later.
Step 3 of 5Worked examples
Example 1: Is \((2, -1)\) a solution of \(x + y = 1\) and \(2x - y = 5\)?
- First equation: \(2 + (-1) = 1\). True.
- Second equation: \(2(2) - (-1) = 4 + 1 = 5\). True.
- Both are true, so yes, \((2, -1)\) is a solution. If even one equation had come out false, the answer would be no.
Example 2: Solve \(x - 2y = 4\) and \(3x + y = 5\)
- The first equation has a plain \(x\). Get it alone: \(x = 2y + 4\).
- Substitute into the second equation, in parentheses: \(3(2y + 4) + y = 5\).
- Distribute and combine: \(6y + 12 + y = 5\), so \(7y + 12 = 5\).
- Solve: \(7y = -7\), so \(y = -1\).
- Back to \(x = 2y + 4\): \(x = 2(-1) + 4 = 2\). Use the equation that already has \(x\) alone. It's the quickest.
- Check: \(2 - 2(-1) = 4\) and \(3(2) + (-1) = 5\). Both true. The solution is \((2, -1)\).
Step 4 of 5Practice
Type your answer and press Check. Write points like \((3, 5)\). If you miss twice, open the worked solution. Then mark how it went.
Is \((3, -2)\) a solution of the system \(x + y = 1\) and \(2x + y = 5\)?
First equation: \(3 + (-2) = 1\). True.
Second equation: \(2(3) + (-2) = 6 - 2 = 4\), not 5. False.
One equation fails, so the answer is no.
Solve by substitution: \(y = x + 2\) and \(2x + y = 11\).
Replace \(y\) with \((x + 2)\): \(2x + (x + 2) = 11\).
\(3x + 2 = 11\), so \(3x = 9\) and \(x = 3\).
Then \(y = 3 + 2 = 5\).
Answer: \((3, 5)\).
Solve by substitution: \(y = 3x - 1\) and \(4x - y = 3\).
Replace \(y\) with \((3x - 1)\): \(4x - (3x - 1) = 3\).
The minus sign flips both signs inside: \(4x - 3x + 1 = 3\), so \(x + 1 = 3\) and \(x = 2\).
Then \(y = 3(2) - 1 = 5\).
Answer: \((2, 5)\).
Solve by substitution: \(x = 2y + 1\) and \(3x - 4y = 7\).
Replace \(x\) with \((2y + 1)\): \(3(2y + 1) - 4y = 7\).
\(6y + 3 - 4y = 7\), so \(2y + 3 = 7\), \(2y = 4\), and \(y = 2\).
Then \(x = 2(2) + 1 = 5\).
Answer: \((5, 2)\). Remember: \(x\) comes first.
Solve by substitution: \(x + 3y = 7\) and \(2x - y = 0\).
Get \(x\) alone in the first equation: \(x = 7 - 3y\).
Substitute: \(2(7 - 3y) - y = 0\), so \(14 - 6y - y = 0\) and \(14 - 7y = 0\).
\(7y = 14\), so \(y = 2\). Then \(x = 7 - 3(2) = 1\).
Answer: \((1, 2)\).
How many solutions does the system \(y = 2x + 1\) and \(y = 2x - 3\) have?
- Exactly one solution
- Exactly two solutions
- No solution exists
- Infinitely many solutions
Both lines have slope 2, but their y-intercepts are different (1 and \(-3\)). So they are parallel and never cross.
With algebra: \(2x + 1 = 2x - 3\) gives \(1 = -3\), which is false.
Answer: (c) no solution exists. (Two different lines can never cross twice.)
Solve by substitution: \(2x + 3y = -1\) and \(y = x - 2\).
Replace \(y\) with \((x - 2)\): \(2x + 3(x - 2) = -1\).
\(2x + 3x - 6 = -1\), so \(5x = 5\) and \(x = 1\).
Then \(y = 1 - 2 = -1\).
Answer: \((1, -1)\).
Line A is \(y = 3x - 5\) and line B is \(x + y = 7\). At what point do the two lines cross?
The crossing point solves both equations. Replace \(y\) in line B with \((3x - 5)\): \(x + (3x - 5) = 7\).
\(4x - 5 = 7\), so \(4x = 12\) and \(x = 3\).
Then \(y = 3(3) - 5 = 4\). Check in B: \(3 + 4 = 7\). True.
Answer: \((3, 4)\).
Step 5 of 5Wrap-up
You can now check a point in a system, tell from the slopes whether two lines cross once, never, or everywhere, and solve a system by substitution.
Read or watch next: OpenStax Intermediate Algebra 2e, 4.1 Solve Systems of Linear Equations with Two Variables (free textbook, with more practice at the end of the section).
Also good: Khan Academy: Introduction to systems of equations (short videos). Quick rules: Lines and systems cheat sheet.
Something unclear? Ask your teacher bot a follow-up question, for example: "How do I decide which letter to get alone first?" It knows this course and can explain things another way.