➕ Algebra 1 · Systems of Equations
Solving Systems by Substitution
Solve a system by isolating a variable, substituting it into the other equation, finding both values and checking. Read the cases of no solution and infinitely many.
About standards and checking
Written by AI, then independently re-solved and checked by separate AI reviewers and automated answer checks. Written by Claude (Sonnet 5.5) from the concept template. Automated gate: schema, notation and every math answer verified by evaluated expressions. Three blind re-solves by separate AI solvers with no access to the answer key: every keyed answer matched. Two prose audits by a separate, stronger AI reviewer (Claude Opus 5.5): every finding fixed, no wrong answer keys found. Final small edits re-checked by the automated gate (no human teacher review yet). Last checked 2026-11-07.
- A-REI.C.6: Solve systems of linear equations exactly and approximately (e.g., with graphs), focusing on pairs of linear equations in two variables.
- 8.EE.C.8b: Solve systems of two linear equations in two variables algebraically, and estimate solutions by graphing the equations.
By the end you can…
- Solve a system of two linear equations by substitution: isolate a variable, substitute, solve, back-substitute and check.
- Substitute an expression correctly, using parentheses and keeping every negative sign.
- Tell when substitution is the least work, and handle fractions and negative or fractional answers.
- Read a leftover true statement (infinitely many solutions) or false statement (no solution) when the variables cancel.
Lesson (school style)
In the last lesson you found the solution of a system by graphing. That works well when the lines cross at a clear grid point, but a graph can only estimate other crossings. Substitution is an algebra method that gives exact answers. The idea: if one equation tells you what a variable equals, such as y = 3x − 4, write that expression in place of the variable in the other equation. Then only one variable is left, and you already know how to solve one-variable equations. Find that variable, use it to get the other one, and check the pair in both equations. If the variables cancel completely, what is left tells you whether the system has no solution or infinitely many solutions.
The big idea: swap in what a variable equals
Take the system y = 3x − 4 and 2x + y = 11. The first equation says that y and 3x − 4 are the same number. At a solution of the system both equations are true, so there y really is equal to 3x − 4. That lets you write (3x − 4) in place of y in the second equation. Swapping a quantity for something equal to it does not change what an expression is worth, so any solution of the system must also satisfy the new second equation:
2x + (3x − 4) = 11
which has only x in it. Combine like terms: 5x − 4 = 11. Add 4: 5x = 15. Divide by 5: x = 3. Then use y = 3x − 4 to get y = 5. Because you keep the first equation to find y, the pair you get makes both equations true. The solution is (3, 5), and worked example 1 checks it in both equations.
When substitution is convenient
Substitution works on any system of two linear equations. It is the least work when one variable is already alone or has a coefficient of 1 or −1, so it is easy to isolate. (A coefficient is the number multiplying a variable. An unwritten coefficient is 1.)
- Already alone: y = 3x − 4 or x = 5 − 2y. Substitute right away.
- Coefficient 1 or −1: in x + 3y = 7 the x has coefficient 1, so x = 7 − 3y needs no division. In −3x + y = −5 the y has coefficient 1.
Look at both equations before you choose, because the easy variable may be in either one. If every variable has a coefficient like 2, 3 or −5, isolating one usually gives fractions. For example, 2x + 3y = 6 gives y = 2 − (2/3)x. That still works, but the next lesson, elimination, is often easier in that case.
The five steps
- Isolate. Solve one equation for one variable. (Skip this step if a variable is already alone.)
- Substitute. Replace that variable in the other equation with the expression you found. Put the expression in parentheses.
- Solve the new equation. It has only one variable.
- Back-substitute. Put the value you found into the equation from step 1 (or any equation that is easy to use) to get the other variable.
- Check the ordered pair in both original equations. Write the solution as (x, y).
In step 2, use the other equation. If you put the expression back into the equation it came from, you only get a statement that is always true, like 3x − 4 = 3x − 4, and you learn nothing.
Parentheses and negative signs
The expression you substitute usually has more than one term, so put it in parentheses. The parentheses keep it together as one block and show that the number in front multiplies the whole block. They are not always needed (as in 2x + (3x − 4)), but they are never wrong, so make them a habit.
Take y = x − 5 and 3x − 2y = 14.
- With parentheses: 3x − 2(x − 5) = 14. The −2 multiplies both terms: 3x − 2x + 10 = 14, so x = 4.
- Without parentheses: 3x − 2x − 5 = 14. Now only the x got multiplied by −2. This gives x = 19 and y = 14, and the check fails: 3(19) − 2(14) = 29, not 14.
A minus sign in front of parentheses flips every sign inside: x − (2y − 5) = x − 2y + 5. Also put parentheses around a negative value when you substitute it, so the signs stay right: if x = −2, then 3x + 5 is 3(−2) + 5 = −1.
Isolating a variable first
If no variable is alone, make one alone. From x + 3y = 7, subtract 3y from both sides to get x = 7 − 3y. From −3x + y = −5, add 3x to both sides to get y = 3x − 5. These are the usual balance moves: the same step on both sides.
If you must divide, divide every term. For example, from 2y = 10 − 4x you get y = 5 − 2x, not y = 10 − 2x.
Fractions
Fractions can show up in the expression you substitute, like y = (1/2)x + 1, or in the answer. Handle them carefully:
- Multiply the number outside the parentheses by every term inside. Often the fractions cancel: −4((1/2)x + 1) = −2x − 4.
- If a fraction is in the way, you can multiply both sides of an equation by a common denominator to clear it. Multiply every term.
- Answers do not have to be whole numbers. x = 4/3 is a perfectly good exact answer (see worked example 5). Keep it as a fraction rather than rounding it, because a rounded value is only an estimate.
When the variables vanish
Sometimes after you substitute and simplify, every variable disappears. Read what is left, just as in the lesson on variables on both sides.
- True, like 12 = 12: the two equations describe the same line, so the system has infinitely many solutions. Every pair on that line is a solution.
- False, like 4 = 7: the lines are parallel and never meet, so the system has no solution.
- If a variable is left, you get a value for it and exactly one solution: the lines cross once.
This matches the graphing lesson: parallel lines give no solution, and the same line gives infinitely many. And a leftover like 0 = 0 does not mean x = 0 or y = 0. It means the equation is true for every value of the variable.
Check in both original equations
Check in the original equations, not in a rewritten version where you might have made a slip. When the system has exactly one solution, each check catches different mistakes. If you slipped while isolating a variable (for example, a sign error), the pair can still fit your rewritten equation and the other equation, and only the original version of the equation you isolated shows the error. If you slipped while substituting, solving or back-substituting, at least one of the two original equations will show it. So test the pair in both original equations every time. A pair that fits only one equation is not a solution of the system.
One limit: a check tests the pair, not how many solutions there are. If both equations describe the same line, every pair on that line fits both, so a slip that leaves you with a single pair can pass both checks. For example, in x = 6 − 2y and 3x + 6y = 18, writing 3(6 − 2y) as 18 + 6y gives y = 0 and the pair (6, 0), which fits both equations. Done correctly, the variables vanish and leave 18 = 18: infinitely many solutions. So recheck the substitution step as well, especially the signs.
Key words
- Substitution method
- Solving a system by writing an expression for one variable, taken from one equation, in place of that variable in the other equation.
- Isolate a variable
- Rewrite an equation so that one variable is alone on one side, like y = 3x − 4 or x = 7 − 3y.
- Back-substitute
- Put the value you found into an equation to find the other variable.
- Solution of a system
- An ordered pair (x, y) that makes both equations true at the same time.
- No solution
- No ordered pair makes both equations true. After substituting, the variables cancel and the leftover statement is false, like 4 = 7. The lines are parallel.
- Infinitely many solutions
- Every pair on the shared line is a solution. After substituting, the variables cancel and the leftover statement is true, like 12 = 12. The two equations describe the same line.
Worked examples
Example 1. Solve y = 3x − 4 and 2x + y = 11.
- Isolate: the first equation already says y = 3x − 4.
- Substitute into the other equation, in parentheses: 2x + (3x − 4) = 11.
- Solve: combine like terms to get 5x − 4 = 11. Add 4: 5x = 15. Divide by 5: x = 3.
- Back-substitute: y = 3x − 4 = 3(3) − 4 = 5.
- Check in both equations. y = 3x − 4: 5 = 3(3) − 4, true. 2x + y = 11: 2(3) + 5 = 11, true.
Answer: Answer: (3, 5), that is, x = 3 and y = 5
Example 2. Solve y = x − 5 and 3x − 2y = 14. Watch the parentheses.
- The first equation says y = x − 5. Substitute (x − 5) for y in the other equation, with parentheses: 3x − 2(x − 5) = 14.
- Distribute the −2 to both terms: 3x − 2x + 10 = 14. (−2 times −5 is +10.)
- Combine like terms: x + 10 = 14. Subtract 10: x = 4.
- Back-substitute: y = x − 5 = 4 − 5 = −1.
- Check. y = x − 5: −1 = 4 − 5, true. 3x − 2y = 14: 3(4) − 2(−1) = 12 + 2 = 14, true.
- What if you forget the parentheses? Then 3x − 2x − 5 = 14 gives x = 19 and y = 14. But 3(19) − 2(14) = 29, not 14, so the check fails and tells you something went wrong.
Answer: Answer: (4, −1)
Example 3. Solve 2x − 5y = −19 and x + 3y = 7.
- Neither equation is solved for a variable, so look for a coefficient of 1. In the second equation x has coefficient 1. Subtract 3y from both sides: x = 7 − 3y.
- Substitute into the other (first) equation, with parentheses: 2(7 − 3y) − 5y = −19.
- Distribute: 14 − 6y − 5y = −19. Combine: 14 − 11y = −19. Subtract 14: −11y = −33. Divide by −11: y = 3.
- Back-substitute into x = 7 − 3y: x = 7 − 3(3) = −2.
- Check in the original equations. 2x − 5y = −19: 2(−2) − 5(3) = −4 − 15 = −19, true. x + 3y = 7: −2 + 3(3) = 7, true.
Answer: Answer: (−2, 3)
Example 4. Solve y = (1/2)x + 1 and 3x − 4y = 6.
- Substitute into the other equation, with parentheses: 3x − 4((1/2)x + 1) = 6.
- Multiply −4 by both terms inside: −4 · (1/2)x = −2x and −4 · 1 = −4. The equation is 3x − 2x − 4 = 6.
- Combine like terms: x − 4 = 6. Add 4: x = 10.
- Back-substitute: y = (1/2)(10) + 1 = 5 + 1 = 6.
- Check. y = (1/2)x + 1: (1/2)(10) + 1 = 6, true. 3x − 4y = 6: 3(10) − 4(6) = 30 − 24 = 6, true.
Answer: Answer: (10, 6)
Example 5. Solve y = 2x − 1 and y = −x + 3. (In the graphing lesson, the lines of this system crossed between grid lines, so a graph could only estimate the crossing.)
- Both equations say what y equals. Substitute the expression from the first equation for y in the second: 2x − 1 = −x + 3. This is the same as setting the two right sides equal.
- Add x to both sides: 3x − 1 = 3. Add 1: 3x = 4. Divide by 3: x = 4/3.
- Back-substitute into the first equation: y = 2(4/3) − 1 = 8/3 − 3/3 = 5/3.
- Check. y = 2x − 1: 2(4/3) − 1 = 5/3, true. y = −x + 3: −4/3 + 3 = −4/3 + 9/3 = 5/3, true.
- The exact solution is (4/3, 5/3). As decimals it is about (1.33, 1.67), but the fractions are the exact values.
Answer: Answer: (4/3, 5/3)
Example 6. Solve x = 3y + 2 and 2x − 6y = 7.
- The first equation says x = 3y + 2. Substitute into the other equation, with parentheses: 2(3y + 2) − 6y = 7.
- Distribute: 6y + 4 − 6y = 7. Combine like terms: 4 = 7.
- The y-terms vanished and what is left is false. No pair (x, y) can make both equations true, so the system has no solution.
- Graph meaning: write both equations as y = mx + b. The first is y = (1/3)x − 2/3 and the second is y = (1/3)x − 7/6. The slopes are equal and the y-intercepts are different, so the lines are parallel.
Answer: Answer: No solution
Example 7. Solve y = −2x + 6 and 4x + 2y = 12.
- Substitute into the other equation, with parentheses: 4x + 2(−2x + 6) = 12.
- Distribute: 4x − 4x + 12 = 12. Combine like terms: 12 = 12.
- The x-terms vanished and what is left is true for every x. So the system has infinitely many solutions: every pair on the line y = −2x + 6 works.
- Try three of them. (0, 6), (1, 4) and (−3, 12) all satisfy y = −2x + 6. In the second equation: 4(0) + 2(6) = 12, 4(1) + 2(4) = 12 and 4(−3) + 2(12) = −12 + 24 = 12.
- Why: dividing every term of 4x + 2y = 12 by 2 gives 2x + y = 6, which is y = −2x + 6. The two equations describe the same line.
Answer: Answer: Infinitely many solutions (every pair on the line y = −2x + 6)
Example 8. A snack stand sold 40 items in all, pretzels and lemonades. Each pretzel costs 3 dollars and each lemonade costs 2 dollars, and the total sales were 95 dollars. How many of each did it sell?
- Let p be the number of pretzels sold and d be the number of lemonades sold. The item count gives p + d = 40. The money gives 3p + 2d = 95 (in dollars).
- In p + d = 40, d has coefficient 1. Isolate it: d = 40 − p.
- Substitute into the money equation, with parentheses: 3p + 2(40 − p) = 95.
- Distribute: 3p + 80 − 2p = 95. Combine: p + 80 = 95. Subtract 80: p = 15.
- Back-substitute: d = 40 − 15 = 25.
- Check. Items: 15 + 25 = 40. Sales: 3(15) + 2(25) = 45 + 50 = 95 dollars.
Answer: Answer: 15 pretzels and 25 lemonades
In plain English
You have two clues about two mystery numbers, x and y. A system of equations is those two clues together.
Sometimes one clue already says what y is. Maybe it says y is three times x, minus 4. That is great news. You can take that description and write it in place of y in the other clue.
Now the other clue has only x in it. That is a plain one-variable equation, and you can solve it. Then you use x to find y. Last, you test both numbers in both clues.
Be careful with brackets. When you write a whole description in place of y, put it in parentheses. Then the number outside multiplies all of it.
Sometimes all the x's cancel. If you are left with a true statement, like 12 = 12, every point on the line works. If you are left with a false statement, like 4 = 7, no pair works.
See it
The same idea, the fun way
Pick the one that clicks. Each one explains the same concept from a different angle.
Sticks of butter
In the United States, one stick of butter is 8 tablespoons. A recipe asks for 2 sticks plus 4 tablespoons of butter, but you only own a tablespoon measure. Let s be the number of tablespoons in one stick, so s = 8. The recipe needs 2s + 4 tablespoons. Swap in what s equals, in parentheses, and you get 2(8) + 4 = 20 tablespoons.
How it connects: The fact s = 8 lets you write 8 anywhere s appears, just as y = 5x + 1 lets you write (5x + 1) anywhere y appears. Notice 2(8), not 28: the parentheses show that the 2 multiplies the swapped-in amount. The limit of the picture: the swap is allowed only because a stick really equals 8 tablespoons. You cannot swap in something that is merely close. And in a system, you use one equation to get the swap and the other equation to put it into.
Two clues about two ages
Jo is 4 years older than Sam, and together their ages add up to 30. Let j be Jo’s age and s be Sam’s age, both in years. Clue 1 says j = s + 4. Clue 2 says j + s = 30. Use clue 1 to rewrite Jo’s age inside clue 2: (s + 4) + s = 30, so 2s + 4 = 30, then 2s = 26 and s = 13. Then j = 13 + 4 = 17.
How it connects: Clue 1 is the swap and clue 2 is where it goes. Only s is left, so you can solve, and then back-substitute to get j. Check both clues: 17 = 13 + 4 and 17 + 13 = 30. The limit of the picture: both clues must describe the same moment. If one clue said “in 5 years,” you would have to write it in terms of today’s ages first. And you never put clue 1 back into clue 1.
Counting baskets
In one game, a team made 20 baskets worth 47 points in all. Every basket was worth 2 or 3 points (ignore free throws). Let a be the number of 2-point baskets and b be the number of 3-point baskets. The basket count gives a + b = 20, so a = 20 − b. The points give 2a + 3b = 47. Swap in for a: 2(20 − b) + 3b = 47. Distribute: 40 − 2b + 3b = 47, so b = 7 and a = 13.
How it connects: The parentheses show that each of the (20 − b) two-point baskets is worth 2 points: 2(20 − b), not 2 · 20 − b. Check: 13 + 7 = 20 baskets, and 2(13) + 3(7) = 26 + 21 = 47 points. The limit of the picture: counts must be whole numbers that are not negative, and real games also have free throws worth 1 point each, which this example ignores.
The cart before the horse
To “put the cart before the horse” means to do things in the wrong order. The horse is supposed to be in front, pulling the cart.
How it connects: Substitution has an order too: isolate before you substitute, solve for the first variable before you back-substitute, and check last. The limit of the idiom: the order matters for these steps, but which variable you isolate first does not. Either choice leads to the same solution, with more or less work.
The proof of the pudding is in the eating
This proverb means you can only judge how good something really is by trying it.
How it connects: A solution is only proven by trying it: substitute the pair into both original equations. If both are true, the pair really is the solution. The limit of the proverb: the “eating” is one specific test, substituting into both equations. A pair that merely looks right, or fits only one equation, is not proven.
The team player
Why does the equation y = 5x + 1 make a great teammate? Any time someone needs a substitute for y, it volunteers.
How it connects: y = 5x + 1 says that y and 5x + 1 are the same number, so one can stand in for the other. The limit of the joke: a real substitute player is a different person, but here the stand-in is exactly equal to what it replaces. And the swap goes into the other equation. Putting 5x + 1 back into the equation it came from tells you nothing.
Memory help
SWAP
Solve one equation for a variable. Write that expression, in parentheses, in the other equation. Algebra: solve the new equation, which has one variable. Plug the value back in to find the other variable. Then check both original equations.
Parentheses first
When the expression you substitute has more than one term, wrap it in parentheses: 3x − 2(x − 5), not 3x − 2x − 5. A minus sign in front of the parentheses flips every sign inside.
Use the other equation
You get the expression from one equation, so it goes into the other equation. Putting it back into the equation it came from gives a statement that is always true and teaches you nothing.
Vanish and read
“Variables vanish? Read what’s there: true means endless, false means none anywhere.” A true leftover (like 12 = 12) means infinitely many solutions. A false leftover (like 4 = 7) means no solution.
Box it, then drop it
Draw a box around the expression you got from one equation. In the other equation, circle the variable it equals. Copy the boxed expression in place of the circle, with parentheses around it, and cross out the circle.
Study notes
Key ideas
- Substitution: if one equation says what a variable equals, write that expression in place of the variable in the other equation. Then only one variable is left.
- It is the least work when a variable is already alone or has a coefficient of 1 or −1. Look at both equations to find an easy variable.
- Put the substituted expression in parentheses, so the number in front multiplies every term and negative signs are kept.
- Back-substitute to find the other variable, then check the pair in both original equations.
- If the variables vanish, read what is left. True (like 12 = 12) means infinitely many solutions. False (like 4 = 7) means no solution.
Steps
- Isolate one variable in one equation (skip this if one is already alone).
- Substitute that expression, in parentheses, into the other equation.
- Solve the new one-variable equation.
- Back-substitute to find the other variable.
- Check the ordered pair in both original equations and write it as (x, y). If the variables vanished, read the leftover statement instead.
Formulas
- The swap: If y = E (where E is an expression), replace y with (E) in the other equation
- Minus before parentheses: a − (b + c) = a − b − c and a − (b − c) = a − b + c
- True leftover: Variables vanish and the statement is true (like 12 = 12): infinitely many solutions
- False leftover: Variables vanish and the statement is false (like 4 = 7): no solution
Watch out
- Wrap the substituted expression in parentheses. 3x − 2(x − 5) is not 3x − 2x − 5.
- Substitute into the other equation, not the one the expression came from.
- A solution is a pair. After you find one variable, back-substitute to find the other, then write (x, y).
- When the variables vanish, read the leftover. 0 = 0 is true (infinitely many solutions) and does not mean x = 0. 0 = 5 is false (no solution) and does not mean y = 5.
- Check in both original equations. Do not stop after the first one fits.
- When you divide to isolate a variable, divide every term.
Why it works
- Why is substitution allowed? At a solution of the system, both equations are true. So there the variable really equals the expression from the first equation, and swapping something for an equal amount does not change the second equation. You keep the first equation to find the other variable, so the pair you end up with satisfies both.
- Why does a true leftover mean infinitely many solutions? The new equation is true for every value of the variable. Each value gives a matching value of the other variable from the isolated equation, and each such pair makes both equations true.
- Why does a false leftover mean no solution? A solution would have to turn the new equation into a true statement, but the leftover is false for every value of the variable. So no pair works.
Common mistakes
Teach it back
Explain to a friend how to solve the system x = y + 2 and 3x − 2y = 9 by substitution, and what it means if the variables ever cancel.
Explaining it out loud shows what you really know. Did you…
- I substituted (y + 2) for x in the other equation, with parentheses: 3(y + 2) − 2y = 9.
- I solved 3y + 6 − 2y = 9, so y + 6 = 9 and y = 3, then back-substituted to get x = 3 + 2 = 5.
- I checked (5, 3) in both equations: 5 = 3 + 2 and 3(5) − 2(3) = 9.
- I said that if the variables cancel, a true leftover means infinitely many solutions and a false leftover means no solution.
Your 3-week memory plan
- TodayRead the lesson, then try the flashcards.
- TomorrowRecall it from memory first, then check.
- In 3 daysTake the true/false and practice questions.
- In 1 weekMix it with older topics and teach it back.
- In 3 weeksA last quick review before it fades.
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