Systems of Equations Practice Quiz with a Step-by-Step Interactive Lesson
Use the quiz at the top of the page to practice systems of equations and systems of linear equations: solving a two-variable linear system for the ordered pair \((x,y)\), using the graphing method, substitution method, and elimination method (addition/subtraction), checking solutions by substitution, and identifying whether a system has one solution, no solution, or infinitely many solutions. You will also practice common classifications like consistent vs inconsistent and independent vs dependent systems, plus real word problems with systems of equations. If you want a refresher, click Start lesson to open a step-by-step guide with worked examples.
How this systems of equations practice works
1. Take the quiz: answer the systems of equations questions at the top of the page.
2. Open the lesson (optional): review graphing, substitution, elimination, and how to classify linear systems.
3. Retry: return to the quiz and apply the systems strategies immediately.
What you will learn in the systems of equations lesson
Foundations & vocabulary
System of linear equations and what a solution \((x,y)\) means
Standard form \(Ax+By=C\), slope-intercept form, and interpreting lines
Consistent / inconsistent and independent / dependent systems
Graphing method
Graph each equation and find the intersection point
Recognize parallel lines (no solution) and the same line (infinitely many solutions)
Use slope and intercepts to predict the number of solutions quickly
Substitution & elimination methods
Substitution method: solve for a variable, substitute, then back-substitute
Elimination (addition/subtraction): add or subtract equations to eliminate one variable
Multiply equations to create opposite coefficients and reduce errors
Applications & checking solutions
Check solutions by plugging \((x,y)\) into both equations
Solve word problems (tickets, ages, mixtures, geometry) using systems
Interpret answers in context and spot impossible results early
Back to the quiz
When you are ready, return to the quiz at the top of the page and keep practicing solving systems of equations.
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Systems of Equations
Step-by-step guide
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Systems of Equations Lesson
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Lesson Overview
Lesson overview
Purpose: Build a clear understanding of systems of equations so you can solve systems of linear equations in two variables using graphing, substitution, or elimination, and interpret whether a system has one solution, no solution, or infinitely many solutions.
Success criteria
Explain what a system of equations is and what it means to be a solution \((x,y)\).
Solve a system by graphing and identify the intersection point.
Solve a system by the substitution method.
Solve a system by the elimination method (addition/subtraction), including multiplying equations when needed.
Check a solution by substituting it into both equations.
Decide whether a system has one solution, no solution, or infinitely many solutions.
Use words like consistent / inconsistent and independent / dependent correctly.
Model and solve word problems using systems of equations.
Key vocabulary
System of equations: two (or more) equations with the same variables.
Solution: an ordered pair \((x,y)\) that makes both equations true.
Linear equation: an equation whose graph is a straight line.
Consistent: the system has at least one solution.
Inconsistent: the system has no solution.
Independent: the system has exactly one solution (the lines intersect once).
Dependent: the system has infinitely many solutions (the equations represent the same line).
Substitution / elimination: common algebraic methods to solve systems.
Quick pre-check
Pre-check 1: Which ordered pair \((x,y)\) solves the system \(x + y = 5\) and \(x - y = 1\)?
Hint: Add the two equations to eliminate \(y\), then substitute back.
Pre-check 2: How many solutions does the system \(\begin{cases}2x - 3y = 1\\4x - 6y = 2\end{cases}\) have?
Hint: The second equation is a multiple of the first, so they describe the same line.
Graphing
Graphing systems of linear equations
Learning goal: Graph two linear equations and identify the solution as their intersection point.
Key idea
Each linear equation in two variables graphs as a line. A solution to the system is a point \((x,y)\) that lies on both lines. So, the solution is the intersection point.
Intersect once: one solution (independent system).
Parallel lines: no solution (inconsistent system).
Same line: infinitely many solutions (dependent system).
Worked example
Example: Solve the system by graphing: \(x + 2y = 8\) and \(x - y = 2\).
Rewrite each equation in slope-intercept form. From \(x + 2y = 8\): \(2y = -x + 8 \Rightarrow y = -\tfrac12 x + 4\). From \(x - y = 2\): \(-y = 2 - x \Rightarrow y = x - 2\).
Graph both lines. They intersect at \((4,2)\). Check: \(4 + 2(2) = 8\) and \(4 - 2 = 2\). So the solution is \((4,2)\).
Try it
Try it 1: What is the solution \((x,y)\) to the system \(3x + y = 8\) and \(x + y = 4\)?
Hint: Subtract the second equation from the first to eliminate \(y\), then substitute back.
Try it 2: If two lines have the same slope but different \(y\)-intercepts, the system has:
Hint: Same slope means parallel; different intercepts means they never meet.
Summary
Graphing turns a system into two lines and the solution is their intersection.
Parallel lines → no solution; same line → infinitely many solutions.
Substitution
Solve systems by substitution
Learning goal: Solve a system by rewriting one equation and substituting into the other.
Key idea
The substitution method works best when one equation is already solved for a variable (or can be solved easily). Steps:
Solve one equation for \(x\) or \(y\).
Substitute that expression into the other equation.
Solve the resulting one-variable equation.
Back-substitute to find the other variable.
Check in both equations.
Worked example
Example: Solve by substitution: \(\begin{cases}y = -x + 4\\2x + y = 6\end{cases}\).
Substitute \(y=-x+4\) into \(2x+y=6\): \(2x+(-x+4)=6 \Rightarrow x+4=6 \Rightarrow x=2\). Now find \(y\): \(y=-2+4=2\). Check: \(2x+y=2(2)+2=6\). So the solution is \((2,2)\).
Try it
Try it 1: Solve by substitution: \(\begin{cases}y = 2x + 1\\x + y = 7\end{cases}\). What is \((x,y)\)?
Hint: Substitute \(y=2x+1\) into \(x+y=7\).
Try it 2: In the system \(\begin{cases}y = -x + 4\\2x + y = 6\end{cases}\), after substituting \(y\) into the second equation you should solve:
Hint: Substitute \(y=-x+4\) into \(2x+y=6\) and simplify.
Summary
Substitution replaces one variable with an equivalent expression.
Always back-substitute and check the ordered pair in both equations.
Elimination
Solve systems by elimination (addition and subtraction)
Learning goal: Solve a system by adding or subtracting equations to eliminate one variable.
Key idea
The elimination method (also called addition/subtraction) is fast when coefficients line up. Steps:
Write both equations in matching form (like \(Ax+By=C\)).
If needed, multiply one or both equations to create opposite coefficients.
Add or subtract to eliminate a variable.
Solve for the remaining variable, then substitute back.
Subtract the second equation from the first to eliminate \(y\): \((3x+2y)-(x+2y)=11-5 \Rightarrow 2x=6 \Rightarrow x=3\). Substitute into \(x+2y=5\): \(3+2y=5 \Rightarrow 2y=2 \Rightarrow y=1\). Check: \(3(3)+2(1)=11\). So the solution is \((3,1)\).
Try it
Try it 1: What is the solution \((x,y)\) to the system \(x + 2y = 8\) and \(x - y = 2\)?
Hint: Subtract \((x-y=2)\) from \((x+2y=8)\) to eliminate \(x\).
Try it 2: What is the solution \((x,y)\) to the system \(2x + 3y = 11\) and \(x + y = 4\)?
Hint: Multiply \(x+y=4\) by \(-2\) and add to \(2x+3y=11\).
Summary
Elimination removes one variable by adding/subtracting equations.
Multiplying an equation is allowed as long as you multiply every term.
Number of Solutions
One solution, no solution, or infinitely many solutions
Learning goal: Classify a system as independent, inconsistent, or dependent using algebra.
Key idea
When you eliminate and solve, one of three things happens:
One solution: you get a single ordered pair \((x,y)\).
No solution: you get a contradiction like \(0=5\). (Lines are parallel.)
Infinitely many solutions: you get an identity like \(0=0\). (Same line.)
Worked example
Example: Determine the number of solutions of \(\begin{cases}x+2y=4\\2x+4y=8\end{cases}\).
Multiply the first equation by \(2\): \(2(x+2y)=2\cdot4 \Rightarrow 2x+4y=8\). This matches the second equation exactly, so both equations represent the same line. The system has infinitely many solutions (dependent system).
Try it
Try it 1: Determine the number of solutions of \(\begin{cases}x - 2y = 3\\2x - 4y = 8\end{cases}\).
Hint: Multiply the first equation by \(2\) and compare it to the second equation.
Try it 2: Determine the number of solutions of \(\begin{cases}2x - 3y = 1\\4x - 6y = 2\end{cases}\).
Hint: The second equation is exactly \(2\) times the first.
Summary
\(0=0\) after elimination means infinitely many solutions (same line).
\(0=\text{nonzero}\) after elimination means no solution (parallel lines).
Strategy & Practice
Choose a method and solve efficiently
Learning goal: Decide when to use graphing, substitution, or elimination, and solve cleanly.
Key idea
Graphing: good for visualizing and checking, but can be imprecise without a calculator.
Substitution: best when a variable is already isolated (like \(y=2x+1\)).
Elimination: best when coefficients match (or can match after multiplying).
Use elimination: multiply \(x+y=3\) by \(4\) to match the \(4y\): \(4x + 4y = 12\). Subtract from \(5x+4y=13\): \((5x+4y)-(4x+4y)=13-12 \Rightarrow x=1\). Then \(x+y=3 \Rightarrow 1+y=3 \Rightarrow y=2\). So \((x,y)=(1,2)\).
Try it
Try it 1: Solve the system \(\begin{cases}x + 2y = 8\\x - 2y = 4\end{cases}\). What is \((x,y)\)?
Hint: Add the equations to eliminate \(y\), then substitute back.
Try it 2: What is the solution \((x,y)\) to the system \(4x + y = 9\) and \(x + y = 3\)?
Hint: Subtract \(x+y=3\) from \(4x+y=9\) to eliminate \(y\).
Summary
Pick the method that uses the structure of the system (isolated variable → substitution, matching coefficients → elimination).
Always check \((x,y)\) in both equations.
Applications
Translate word problems into systems of equations
Learning goal: Turn a real situation into two equations, solve, and interpret the answer.
Where systems show up
Tickets and money: different prices and totals.
Ages: sum and difference relationships.
Geometry: perimeter or area with relationships between sides.
Mixtures: two components combine to a total amount.
Worked example: ticket problem
Example: Adult tickets cost \$12 and student tickets cost \$8. A total of 25 tickets were sold for \$260. How many adult tickets and student tickets were sold?
Let \(a\) = adult tickets and \(s\) = student tickets. Total tickets: \(a+s=25\). Total cost: \(12a+8s=260\).
Multiply the first equation by \(8\): \(8a+8s=200\). Subtract from the cost equation: \((12a+8s)-(8a+8s)=260-200 \Rightarrow 4a=60 \Rightarrow a=15\). Then \(s=25-15=10\). So there were 15 adult tickets and 10 student tickets.
Try it
Try it 1: Two numbers add to \(11\) and their difference is \(3\). If \(x\) is larger, what is \((x,y)\)?
Hint: Use \(x+y=11\) and \(x-y=3\).
Try it 2: A rectangle has perimeter \(34\). Its length is \(3\) more than its width. If \((w,\ell)\) is (width, length), what is \((w,\ell)\)?
Hint: Perimeter \(34\) means \(2w+2\ell=34\), and \(\ell=w+3\).
Summary
Define variables clearly, then write one equation for each relationship.
After solving, check if the answer makes sense in the context.
Big Picture
Check your solution and connect the ideas
Learning goal: Verify solutions reliably, summarize the main strategies, and finish with a final check.
How to check a solution
To check \((x,y)\), substitute the values into both equations. If both equations are true, the ordered pair is a solution. If even one is false, it is not a solution.
Worked example: verify a solution
Example: Check whether \((2,3)\) is a solution to \(2x+y=7\) and \(x+2y=8\).
First equation: \(2(2)+3=4+3=7\) ✅ Second equation: \(2+2(3)=2+6=8\) ✅ So \((2,3)\) is a solution.
Try it
Try it 1: What is the solution \((x,y)\) to the system \(2x + y = 7\) and \(x + 2y = 8\)?
Hint: Multiply the first equation by \(2\) or use elimination to remove \(y\) or \(x\).
Try it 2: Which method is usually most efficient for the system \(x + y = 7\) and \(x - y = 1\)?
Hint: Adding the equations cancels \(y\) immediately.
Final recap
Graphing: intersection point is the solution.
Substitution: replace one variable with an equivalent expression.
Elimination: add/subtract equations to remove a variable (multiply if needed).
Classification: one solution (independent), no solution (inconsistent), infinitely many (dependent).
Check: substitute \((x,y)\) into both equations every time.
Next step: Close this lesson and try your quiz again. If you miss a question, reopen the book and review the page that matches the systems skill you need.
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