➕ Algebra 1 · Equations
Equations with Variables on Both Sides
Solve equations with variables on both sides by gathering the x-terms on one side, and learn to spot one solution, no solution or infinitely many solutions.
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.
- 8.EE.C.7a: Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions.
- 8.EE.C.7b: Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms.
- A-REI.B.3: Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.
- A-CED.A.1: Create equations and inequalities in one variable and use them to solve problems.
By the end you can…
- Solve equations like 5x + 2 = 3x + 10 by gathering the variable terms on one side.
- Combine this with distributing, combining like terms, negatives and fractions.
- Tell whether an equation has one solution, no solution or infinitely many solutions, and explain why.
- Write and solve an equation to find when two quantities are equal.
Lesson (school style)
Sometimes the variable shows up on both sides of the equals sign, like 5x + 2 = 3x + 10. You cannot undo anything yet, because each side has an x. The fix is a balance move: add or subtract a variable term on both sides so all the variable terms end up on one side. Then move the plain numbers to the other side the same way, and finish like a two-step equation. And sometimes the variables vanish completely, which tells you something surprising: either every number works, or no number does.
Why this is different
In 5x + 2 = 3x + 10 there is an x-term on each side, so no single operation frees x. First, get the variable terms together on one side.
Subtract 3x from both sides. The 3x disappears from the right, and the left becomes 5x − 3x + 2 = 2x + 2. Now the equation is 2x + 2 = 10, a two-step equation you already know how to solve.
Why is this allowed? It is the same balance rule as before: subtracting the same amount from both sides keeps the sides equal.
The plan
- Simplify each side. Distribute, then combine like terms.
- Gather the variable terms on one side by adding or subtracting a variable term on both sides.
- Gather the plain numbers on the other side by adding or subtracting on both sides.
- Divide both sides by the coefficient of the variable (it is not zero).
- Check by substituting into the original equation.
Picture a river at the equals sign: variable terms on one bank, plain numbers on the other. Every move that sends a term across is an add or subtract done to both sides.
Which side should keep the variable?
Either side works. Many people keep the variable on the side with the larger coefficient, so the coefficient stays positive.
In 5x + 2 = 3x + 10, that is the left side. Subtract 3x from both sides to get 2x + 2 = 10, so x = 4.
If you subtract 5x instead, you get 2 = −2x + 10, then −8 = −2x, so x = 4 again. Same answer, but you had to divide by a negative number.
Parentheses, negatives and fractions
Simplify first. If there are parentheses, distribute (watch the signs, and remember a minus in front of parentheses flips every sign inside). If there are fractions, multiply every term on both sides by a common denominator to clear them. After that, the plan is the same.
Answers do not have to be whole numbers, and they can be negative.
When the variables cancel
Sometimes gathering the variable terms removes every variable. What is left is a plain statement about numbers. Read it.
- True, like 10 = 10: the equation is true for every value of x. It has infinitely many solutions. For example, 2(x + 5) = 2x + 10: the two sides are the same expression written two ways.
- False, like −1 = 1: no value of x can make it true. It has no solution. For example, x − 2 = x + 7: subtracting 2 from a number can never give the same result as adding 7 to it.
- If an x-term is left, you get exactly one solution, like x = 4.
The variables vanish exactly when both sides have the same coefficient on x after simplifying. Then only the plain numbers matter: equal means always equal, different means never equal.
Be careful: 0 = 0 does not mean x = 0. It means the equation is true for every x. And 0 = 5 does not mean x = 0 either. It means no number works. “Infinitely many” is not a number to plug in for x. It is a way of saying that every number works.
Writing an equation for “when are they equal?”
Many word problems ask when two quantities are the same. Write an expression for each quantity, set the two expressions equal, and solve. The variable lands on both sides on its own. See the saving example in the worked examples.
Check, and sanity-check
Substitute your answer into the original equation and work out each side separately. For a one-solution equation, both sides must match.
For the other two cases, try a couple of numbers. If the leftover was true, every number you try works. If it was false, every number you try fails.
Key words
- Variable term
- A term that contains the variable, such as 3x or −2x.
- Constant
- A plain number with no variable, like the 2 in 5x + 2.
- Solution
- A value of the variable that makes the equation true.
- Infinitely many solutions
- Every number is a solution. This happens when both sides simplify to the same expression, so the leftover statement is true, like 10 = 10.
- No solution
- No number makes the equation true. This happens when the variables vanish and the leftover statement is false, like −1 = 1.
- Equivalent equations
- Equations with exactly the same solution(s), like 5x + 2 = 3x + 10 and 2x = 8.
Worked examples
Example 1. Solve 5x + 2 = 3x + 10.
- There is an x on both sides, so gather the variable terms first. The left side has the larger coefficient (5 versus 3), so keep the x-terms on the left. Subtract 3x from both sides: 5x + 2 − 3x = 3x + 10 − 3x.
- That leaves 2x + 2 = 10.
- Subtract 2 from both sides: 2x = 8.
- Divide both sides by 2: x = 4.
- Check in the original equation. Left: 5(4) + 2 = 22. Right: 3(4) + 10 = 22. The sides match.
- Optional: keep the x-terms on the right instead. Subtract 5x from both sides: 2 = −2x + 10. Subtract 10: −8 = −2x. Divide by −2: 4 = x. Same answer.
Answer: Answer: x = 4
Example 2. Solve 4 − 3x = 2x − 11.
- The variable terms are −3x and 2x, and each sign stays with its term. To keep a positive coefficient, add 3x to both sides: 4 − 3x + 3x = 2x − 11 + 3x.
- That leaves 4 = 5x − 11.
- Add 11 to both sides: 15 = 5x.
- Divide both sides by 5: 3 = x, which is the same as x = 3.
- Check. Left: 4 − 3(3) = 4 − 9 = −5. Right: 2(3) − 11 = 6 − 11 = −5. The sides match.
Answer: Answer: x = 3
Example 3. Solve 2(x + 4) = 3(x − 1) − 5.
- Simplify each side first. Left: 2(x + 4) = 2x + 8. Right: 3(x − 1) − 5 = 3x − 3 − 5 = 3x − 8.
- The equation is now 2x + 8 = 3x − 8. The right side has the larger coefficient, so keep the x-terms there. Subtract 2x from both sides: 8 = x − 8.
- Add 8 to both sides: 16 = x.
- Check. Left: 2(16 + 4) = 2(20) = 40. Right: 3(16 − 1) − 5 = 3(15) − 5 = 45 − 5 = 40. The sides match.
Answer: Answer: x = 16
Example 4. Solve 10 − 2(x − 3) = 4x + 1.
- Distribute −2 with its sign: −2(x − 3) = −2x + 6. The left side is 10 − 2x + 6.
- Combine the plain numbers: 16 − 2x = 4x + 1.
- Add 2x to both sides to gather the x-terms on the right: 16 = 6x + 1.
- Subtract 1 from both sides: 15 = 6x.
- Divide both sides by 6: x = 15/6 = 5/2. The answer is not a whole number, and that is fine.
- Check. Left: 10 − 2(5/2 − 3) = 10 − 2(−1/2) = 10 + 1 = 11. Right: 4(5/2) + 1 = 10 + 1 = 11. The sides match.
Answer: Answer: x = 5/2
Example 5. Solve 2x/3 − 1 = x/2 + 2.
- The denominators are 3 and 2, so the least common denominator is 6. Multiply every term on both sides by 6: 6 · (2x/3) − 6 · 1 = 6 · (x/2) + 6 · 2.
- Simplify each term: 4x − 6 = 3x + 12.
- Subtract 3x from both sides: x − 6 = 12.
- Add 6 to both sides: x = 18.
- Check. Left: 2(18)/3 − 1 = 12 − 1 = 11. Right: 18/2 + 2 = 9 + 2 = 11. The sides match.
Answer: Answer: x = 18
Example 6. Solve 4(x − 1) + 3 = 4x + 1.
- Simplify the left side: 4(x − 1) + 3 = 4x − 4 + 3 = 4x − 1. The equation is 4x − 1 = 4x + 1.
- Subtract 4x from both sides: −1 = 1.
- The x-terms vanished and what is left is false. No value of x can make −1 = 1 true, so the equation has no solution.
- Sanity check with a number. Try x = 2. Left: 4(2 − 1) + 3 = 7. Right: 4(2) + 1 = 9. In fact the right side is always 2 more than the left side, for every x.
Answer: Answer: No solution
Example 7. Solve 5(x + 2) − 3 = 2x + 3x + 7.
- Simplify the left side: 5(x + 2) − 3 = 5x + 10 − 3 = 5x + 7.
- Simplify the right side by combining like terms: 2x + 3x + 7 = 5x + 7.
- The equation is 5x + 7 = 5x + 7. Subtract 5x from both sides: 7 = 7.
- The x-terms vanished and what is left is true. Every value of x makes the equation true, so there are infinitely many solutions.
- Sanity check with numbers. Try x = 0: left 5(2) − 3 = 7 and right 0 + 0 + 7 = 7. Try x = −4: left 5(−2) − 3 = −13 and right −8 − 12 + 7 = −13.
Answer: Answer: Infinitely many solutions (every number works)
Example 8. Ava has 40 dollars and saves 6 dollars each week. Ben has 10 dollars and saves 11 dollars each week. Neither spends any money. After how many weeks do they have the same amount?
- Let w be the number of weeks. After w weeks, Ava has 40 + 6w dollars and Ben has 10 + 11w dollars.
- They have the same amount when 40 + 6w = 10 + 11w.
- Ben's coefficient is larger, so keep the w-terms on the right. Subtract 6w from both sides: 40 = 10 + 5w.
- Subtract 10 from both sides: 30 = 5w. Divide both sides by 5: w = 6.
- Check. Ava has 40 + 6(6) = 76 dollars and Ben has 10 + 11(6) = 76 dollars.
Answer: Answer: After 6 weeks (they both have 76 dollars)
In plain English
Here is a puzzle. Five identical mystery bags plus 2 coins balance a scale against three of the same bags plus 10 coins. How much does one bag weigh?
Take 3 bags off each side. The scale still balances. Now 2 bags plus 2 coins balance 10 coins. Take 2 coins off each side. Now 2 bags balance 8 coins, so one bag weighs the same as 4 coins. That is how you solve 5x + 2 = 3x + 10: take the same thing off both sides, then share what is left equally until one bag stands alone.
There are two surprise endings. Say 5 bags plus 2 coins balance 5 bags plus 9 coins. Take 5 bags off each side and you get 2 coins balancing 9 coins. That can never happen, so no bag weight works. Now say 5 bags plus 2 coins balance 5 bags plus 2 coins. They match whatever a bag weighs, so every weight works.
See it
The same idea, the fun way
Pick the one that clicks. Each one explains the same concept from a different angle.
The head-start race
Two runners each run at a steady speed. Runner A gets a 40-meter head start and runs 4 meters per second. Runner B starts at the line and runs 6 meters per second. After t seconds, A is 40 + 4t meters from the line and B is 6t meters. They are side by side when 40 + 4t = 6t, which gives t = 20 seconds. If both ran 4 meters per second, A would stay 40 meters ahead: 40 + 4t = 4t gives 40 = 0, which is false, so they never meet. If both started together at the same speed, they would always be side by side: 4t = 4t is true for every t.
How it connects: When the t-terms match, only the starting points matter, so the answer is no time or every time. The limit of the picture: in a real race only t ≥ 0 makes sense, the speeds are assumed steady, and B only catches A if the race is at least 120 meters long. The matched-speed equations themselves, though, 40 + 4t = 4t and 4t = 4t, are true for no number or for every number, not just for times in a race.
Meeting up
Mia and Sam live 48 kilometers apart on one straight road. At the same moment they start biking toward each other at steady speeds: Mia from her house at 14 kilometers per hour and Sam from Sam's house at 10 kilometers per hour. Measure from Mia's house. After t hours Mia is 14t kilometers from her house. Sam is 48 − 10t kilometers from Mia's house, because Sam gets 10 kilometers closer every hour. They meet when 14t = 48 − 10t. Add 10t to both sides: 24t = 48, so t = 2 hours.
How it connects: “They meet” means “they are at the same place,” so you set the two position expressions equal. Sam is heading toward Mia's house, so Sam's distance shrinks, which is why that side has −10t. They do not meet halfway: Mia has gone 28 kilometers and Sam 20. The limit of the picture: the speeds are assumed steady, and only t ≥ 0 up to the meeting time makes sense.
Playlist lengths
Playlist A has 7 songs plus 3 minutes of talking in total. Playlist B has 4 songs plus 12 minutes of talking in total. Every song is the same length, s minutes. The two playlists run the same total time when 7s + 3 = 4s + 12.
How it connects: Take 4 songs off both playlists: 3s + 3 = 12, so 3s = 9 and s = 3 minutes. Removing the same songs from both is subtracting 4s from both sides. It only keeps the totals equal because you remove the same thing from both playlists, never from just one.
Six of one, half a dozen of the other
People say this when two choices come out the same, so it makes no difference which you pick. A dozen is 12, so half a dozen is 6. Six and half a dozen are exactly the same amount.
How it connects: That is a true statement like 6 = 6. When the x-terms cancel and you are left with a true statement, both sides match for every x: infinitely many solutions. If you are left with a false statement like 6 = 7, there is no solution.
Birds of a feather flock together
The proverb says that similar people tend to be found together.
How it connects: Gather like with like: all the x-terms go to one side and all the plain numbers go to the other. But terms do not fly across the equals sign on their own. Each move is an add or subtract done to both sides.
The party nobody could attend
An equation with no solution threw a party, and no number showed up. Its cousin x + 1 = x + 1 threw one, and every number came.
How it connects: No solution means no number makes the equation true. Infinitely many solutions means every number does.
Memory help
STAND
Simplify each side (distribute, combine like terms). Take the variable terms to one side (same move on both sides). Add or subtract the plain numbers on both sides. Now divide both sides by the coefficient (not zero). Double-check in the original equation. Stand your ground: S-T-A-N-D.
The gather rhyme
Gather the x’s, then the plain; do each move on both sides, again and again. If the x’s vanish, read what is there: true means all numbers, false means none anywhere.
True: Tons. False: Found none.
When the x-terms vanish, look at the leftover statement. True means Tons of solutions (every number works). False means Found none (no number works).
Draw the river
Draw a vertical line through the equals sign. Circle every x-term in one color and box every plain number in another. Your job is to get all the circles on one bank and all the boxes on the other, using the same move on both sides each time.
0 = 0 is not x = 0
When the x-terms vanish, do not write x = 0. A leftover like 0 = 0 is true for every x, so every number works. A leftover like 0 = 5 is false for every x, so no number works. Neither one says x = 0.
Opposite sides do not add
In 5x + 2 = 3x + 10, the 5x and the 3x are on opposite sides, so they do not combine into 8x. To bring them together, subtract 3x from both sides to get 2x.
Study notes
Key ideas
- When x is on both sides, gather the variable terms on one side by adding or subtracting a variable term on both sides.
- Either side can keep the variable. Keeping it on the side with the larger coefficient avoids a negative coefficient.
- If the variables vanish, read what is left. A true statement (like 7 = 7) means infinitely many solutions. A false statement (like 2 = 9) means no solution.
- After simplifying ax + b = cx + d: if a ≠ c there is exactly one solution. If a = c and b = d there are infinitely many. If a = c and b ≠ d there is none.
- Check in the original equation.
Steps
- Simplify each side: distribute, then combine like terms (clear fractions if needed).
- Add or subtract a variable term on both sides so the variable terms are on one side.
- Add or subtract on both sides so the plain numbers are on the other side.
- Divide both sides by the coefficient of the variable (it is not zero).
- Check in the original equation. If the variables vanished, read the leftover statement instead.
Formulas
- One solution: ax + b = cx + d with a ≠ c gives x = (d − b) ÷ (a − c)
- Infinitely many solutions: ax + b = ax + b is true for every x
- No solution: ax + b = ax + d with b ≠ d is never true
Watch out
- Add or subtract the variable term on both sides, not just one.
- Do not add variable terms that are on opposite sides: in 5x + 2 = 3x + 10 you subtract 3x from both sides.
- When the x-terms vanish, do not write x = 0. Read the leftover statement: true means every number works, false means no number works.
- Keep each term's sign with it: in 4 − 3x the term is −3x.
Why it works
- Why can you subtract a variable term from both sides? It is the balance rule: doing the same thing to both sides keeps the equation true.
- Why can the x-terms vanish? When both sides have the same coefficient on x, the gap between the two sides is the same for every x. The plain numbers decide: equal means the sides always match, different means they never match.
Common mistakes
Teach it back
Explain to a friend (or a pet!) how to solve 6x + 5 = 2x + 21, and what it means if the x-terms ever vanish.
Explaining it out loud shows what you really know. Did you…
- I subtracted 2x from both sides to get 4x + 5 = 21, then subtracted 5 to get 4x = 16.
- I divided by 4 to get x = 4 and checked it in the original equation (29 = 29).
- I said that if the x-terms vanish, a true leftover means every number works 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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