Equations with variables on both sides
The unknown turns up on both sides of the equals sign. How do you get it onto one side so the equation can be solved?
Where you are 11 in Linear equations
When you reach for it
A two-step equation such as 3x + 3 = 18 has the unknown in one place, and you solve it by undoing what was done to x, last step first. A multi-step equation may have several x terms, but they all sit on the same side, so combining them gets you back to two steps. Now look at
The unknown is on both sides. Undoing does not work directly: subtracting 3 and dividing by 5 gives \displaystyle x = \frac{2x + 15}{5}, which still has x on the right and so says nothing new. Something has to bring the two x terms together first.
Equations like this turn up whenever two quantities that depend on the same unknown are compared. Two phone plans each cost a fixed amount plus a charge per minute: for how many minutes do they cost the same? Each side of the equation is one of the two quantities, and the question “when are they equal?” is the equation itself.
This lesson runs over two days. Day 1 is the move itself: collect the unknown on one side, finish as a two-step equation, and check by substitution. Day 2 adds brackets and fractions, and builds equations from two quantities that become equal.
The procedure
Day 1: collecting the unknown on one side
Subtract a variable term from both sides. In 5x + 3 = 2x + 18, take 2x away from each side:
which simplifies to
The unknown now appears once, and the rest is a two-step equation: subtract 3 to get 3x = 15, then divide by 3 to get x = 5.
Why is it allowed to subtract 2x, when 2x is not a number you know? Because whatever number x is, 2x is a number. If x is a solution, the two sides are the same number; take the same number 2x away from both and they are still the same number. That is the subtraction property of equality from Inverse operations and the properties of equality, used with a quantity that happens to contain the unknown. In the balance picture, each bag holds the same unknown weight x, and taking two bags off each pan cannot tip it.
The step loses no solution and adds none. Adding 2x back to both sides of 3x + 3 = 18 returns the equation you started with. So a number that makes one equation true makes the other true as well: the two equations have exactly the same solutions, and are called equivalent. Every move in this lesson (adding or subtracting the same term on both sides, multiplying or dividing both sides by the same number that is not zero) can be undone this way, which is why the answer at the end is the answer to the equation at the start.
Which term to move is a choice, not a rule. Either x term can be moved, and both choices give the same answer. Moving the term with the smaller coefficient leaves a positive coefficient behind, which avoids dividing by a negative at the end: above, 2 < 5, so 2x moved. Moving 5x instead gives 3 = -3x + 18, then -15 = -3x, and x = 5 again, with one more minus sign to carry. The unknown may finish on the right: 5 = x and x = 5 say the same thing (the symmetric property of equality).
“Smaller” means smaller as a signed number. In 4 - 3w = 9 - 7w the coefficients are -3 and -7, and -7 < -3, so the term to move is -7w. Moving it means adding 7w to both sides, which leaves 4 + 4w = 9, again with a positive coefficient.
Check by substitution in the original equation. Put the answer back into the equation as it was first written and work out each side separately. With x = 5:
The two sides agree, so x = 5 is the solution. Check in the original rather than a later line: a slip in an early step is copied into every line after it, and only the original equation can catch it. Finding a solution and checking one are separate jobs.
Day 2: brackets, fractions and equations from words
Brackets: tidy each side first. When either side has brackets, use the distributive property and combine like terms on each side separately, as in Distribute, then combine. Only then collect the unknown. For example, in
the left side becomes 4x + 8, so 4x + 8 = 7x - 4. Subtract 4x (the smaller term) to get 8 = 3x - 4, add 4 to get 12 = 3x, and divide by 3: x = 4. Check: 4(6) = 24 and 7(4) - 4 = 24.
A minus sign in front of a bracket multiplies every term inside it: -2(a - 4) = -2a + 8, because (-2)(-4) = 8. That is the distributive property with a negative multiplier, not a separate rule.
Fractions. With fraction coefficients you can collect the unknown exactly as before, working with the fractions, or first clear them by multiplying both sides by a common denominator, as Multi-step equations does. Both are legal moves and give the same solution.
Equations from words. When a story compares two quantities, write each one in terms of the same unknown, after saying exactly what the unknown counts and in what unit (“let m be the number of minutes of calls”). The moment the story asks about (“cost the same”, “catches up”) is the moment the two expressions are equal, and setting them equal is the equation. The solution is then checked against the story itself, not only the equation, since the equation could have been written wrongly.
Drawn as graphs, the two sides are two lines, and the solution is where the lines cross. That picture is the start of systems of equations.
A look ahead: when the unknown vanishes. If the two x terms have the same coefficient, subtracting one removes the unknown completely: 3x + 1 = 3x + 4 leaves 1 = 4, false whatever x is, and 3x + 1 = 3x + 1 leaves 1 = 1, true whatever x is. Those two outcomes are the subject of No solution, one solution, or every number. Every equation in this lesson has different coefficients once each side is tidied, so there is exactly one solution.
Worked examples
Day 1: constants on both sides
Solve 7x - 4 = 3x + 24.
SolutionThe x terms are 7x and 3x. Since 3 < 7, subtract 3x from both sides (subtraction property of equality), so the x term that is left has a positive coefficient:
Now it is a two-step equation. Add 4 to both sides (addition property of equality):
Divide both sides by 4, which is allowed because 4 \ne 0:
Check in the original equation: the left side is 7(7) - 4 = 45 and the right side is 3(7) + 24 = 45. They agree, so x = 7.
Day 1: a negative term, moved either way
Solve 6 - 2y = 5y - 22, first by moving the smaller y term and then by moving the larger one.
SolutionThe coefficients are -2 and 5, and -2 is the smaller. Moving -2y means adding 2y to both sides:
Add 22 to both sides to get 28 = 7y, then divide both sides by 7 to get 4 = y, which is y = 4.
The other way: subtract 5y from both sides, so that 6 - 7y = -22. Subtract 6 from both sides: -7y = -28. Divide both sides by -7: y = 4, the same answer with two more minus signs to track.
Check: 6 - 2(4) = -2 and 5(4) - 22 = -2. Both routes give y = 4, as they must, since every step is a legal move on the same equation.
Day 1: the unknown finishes on the right
Solve 4n + 9 = 10n - 5.
SolutionThe smaller coefficient is 4, and it is on the left. Subtract 4n from both sides:
The unknown is now on the right, which is fine. Add 5 to both sides: 14 = 6n. Divide both sides by 6:
The solution need not be a whole number. Check with fractions over 3: the left side is \displaystyle 4 \cdot \frac{7}{3} + 9 = \frac{28}{3} + \frac{27}{3} = \frac{55}{3}, and the right side is \displaystyle 10 \cdot \frac{7}{3} - 5 = \frac{70}{3} - \frac{15}{3} = \frac{55}{3}. They agree, so \displaystyle n = \frac{7}{3}.
Day 2: brackets on both sides, one with a minus in front
Solve 5 - 2(a - 4) = 3(a + 1) - 5.
SolutionTidy each side on its own first. On the left, the distributive property gives -2(a - 4) = -2a + 8, the -2 multiplying both terms in the bracket. Then the constants 5 and 8 are gathered (commutative and associative properties of addition):
On the right, 3(a + 1) = 3a + 3, and 3a + 3 - 5 = 3a - 2. The equation is now
The coefficients are -2 and 3. Moving the smaller, -2a, means adding 2a to both sides: 13 = 5a - 2. Add 2 to both sides: 15 = 5a. Divide by 5: a = 3.
Check in the original: the left side is 5 - 2(3 - 4) = 5 - 2(-1) = 7, and the right side is 3(3 + 1) - 5 = 12 - 5 = 7. So a = 3.
Day 2: fractions on both sides
Solve \displaystyle \frac{2}{3}x + 1 = \frac{1}{2}x + 4.
SolutionCompare the coefficients over a common denominator: \displaystyle \frac{2}{3} = \frac{4}{6} and \displaystyle \frac{1}{2} = \frac{3}{6}, so \displaystyle \frac{1}{2} is the smaller. Subtract \displaystyle \frac{1}{2}x from both sides. Since \displaystyle \frac{2}{3} - \frac{1}{2} = \frac{4}{6} - \frac{3}{6} = \frac{1}{6},
Subtract 1 from both sides: \displaystyle \frac{1}{6}x = 3. Multiply both sides by 6: x = 18.
Clearing the fractions first works too: multiplying both sides by 6 gives 4x + 6 = 3x + 24, and subtracting 3x and then 6 gives the same x = 18. Check: \displaystyle \frac{2}{3}(18) + 1 = 13 and \displaystyle \frac{1}{2}(18) + 4 = 13.
Day 2: two phone plans
Plan A costs 20 dollars a month plus 0.15 dollars for each minute of calls. Plan B costs 8 dollars a month plus 0.35 dollars a minute. For how many minutes of calls in a month do the two plans cost the same, and what is that cost? Which plan is cheaper for someone who calls for 30 minutes a month?
SolutionLet m be the number of minutes of calls in a month. The monthly costs in dollars are
The plans cost the same when these are equal:
The smaller coefficient is 0.15. Subtract 0.15m from both sides: 20 = 8 + 0.20m. Subtract 8 from both sides: 12 = 0.20m. Divide both sides by 0.20:
At 60 minutes, plan A costs 20 + 0.15(60) = 20 + 9 = 29 dollars and plan B costs 8 + 0.35(60) = 8 + 21 = 29 dollars. So the plans cost the same, 29 dollars, at 60 minutes.
At 30 minutes, plan A costs 20 + 4.50 = 24.50 dollars and plan B costs 8 + 10.50 = 18.50 dollars, so plan B is cheaper. The graph shows why: plan B starts lower and climbs faster, so it is the cheaper plan for fewer than 60 minutes, and plan A is cheaper for more than 60.
Day 2: a walker and a cyclist
Yusuf-Ali sets off along a canal path, walking at 5 kilometers an hour. One hour later Zinnia sets off from the same place along the same path, cycling at 15 kilometers an hour. How long after Yusuf-Ali set off does Zinnia catch him, and how far from the start?
SolutionLet t be the number of hours since Yusuf-Ali set off. By then he has walked 5t kilometers. Zinnia started one hour later, so she has been cycling for t - 1 hours and has covered 15(t - 1) kilometers. She catches him when they are the same distance from the start:
Distribute on the right: 5t = 15t - 15. The smaller coefficient is 5; subtract 5t from both sides: 0 = 10t - 15. Add 15: 15 = 10t. Divide by 10: t = 1.5.
So Zinnia catches him 1.5 hours after Yusuf-Ali set off, which is half an hour after she left. He has walked 5(1.5) = 7.5 kilometers, and she has cycled 15(0.5) = 7.5 kilometers, so they meet 7.5 kilometers from the start.
The expression 15(t - 1) only describes Zinnia’s distance once she has started, that is, for t \ge 1. The answer t = 1.5 lies in that range, so it fits the story.
Boundaries
- The last division needs a coefficient that is not zero. After collecting, the equation has the form ax = b, and dividing by a is allowed only when a \ne 0. The coefficient a is the difference of the two original coefficients, so it is zero exactly when the x terms had equal coefficients. Then the unknown vanishes and the equation has no solution or every number as a solution; No solution, one solution, or every number deals with that case.
- The moves apply to whole sides. Subtracting 2x from both sides means subtracting it from each side as a whole. A side with brackets is tidied first, because the terms to move are not visible until it is: the left side of 4(x + 2) = 7x - 4 holds no 4x term until expanded.
- The method is for linear equations. It relies on every x term being a number times x, so that terms combine. If x^2 appears on both sides with the same coefficient it cancels; if not, the equation is quadratic and needs other methods. An unknown in a denominator, as in \displaystyle \frac{3}{x} = \frac{1}{x} + 4, belongs to rational equations, where some values of x are not allowed at all.
- A story’s answer must fit the story. The equation does not know that a number of minutes cannot be negative or that a distance formula only applies after someone has set off. A solution outside the range the story allows answers the equation but not the question.
Where this goes next
Collecting the unknown on one side is the last piece of the general method for linear equations: tidy each side, collect the unknown, collect the constants, divide, check. No solution, one solution, or every number explains what happens when the unknown disappears.
The same move appears well beyond this lesson. Writing a repeating decimal as a fraction leads to an equation like 10x = 21 + x, as in Turning a repeating decimal into a fraction. Multi-step inequalities use the same steps, with extra care about negatives. Solving a formula for one of its letters collects the wanted letter on one side when it appears more than once. And solving two equations together by substitution produces a single equation with the unknown on both sides, the algebra behind two lines crossing.
Exercises
Day 1: collecting the unknown
- Solve 9x + 2 = 4x + 32.
- Solve 3m - 8 = 7m + 12.
- Solve 11 - 4k = 2k - 13.
- Solve 8p = 3p + 35.
- Solve 2.5y + 3 = 1.3y + 9.
- Solve -3w + 10 = -7w - 6.
- Solve 6x - 1 = 2x + 4.
Day 1: checking and choosing
- Is x = 3 a solution of 5x - 7 = 2x + 2? Decide by substitution.
- Is t = -2 a solution of 4 - 3t = t + 8? If not, solve the equation.
- Solve 2a + 15 = 7a - 10 twice: once by subtracting 2a from both sides, once by subtracting 7a. Show that both routes give the same answer.
Day 2: brackets and fractions
- Solve 4(x + 3) = 2(x + 9).
- Solve 5(2n - 1) = 3(n + 4) + 4.
- Solve 7 - (y - 2) = 3(y - 1).
- Solve 2(3c - 4) - c = 4(c + 1).
- Solve \displaystyle \frac{3}{4}x - 2 = \frac{1}{4}x + 5.
- Solve \displaystyle \frac{x}{3} + 4 = \frac{x}{2} - 1.
- Solve 0.4(5d + 10) = 0.5(2d + 20).
- Solve -3(2 - z) = 2(z + 5) - 4.
Day 2: from words
- Wendeline has 140 dollars and saves 15 dollars a week. Xavi has 60 dollars and saves 25 dollars a week. After how many weeks will they have the same amount, and how much will each have?
- Taxi firm A charges 4.50 dollars plus 1.20 dollars a kilometer. Taxi firm B charges 2.10 dollars plus 1.50 dollars a kilometer. For what distance do they charge the same, and how much is that? Which is cheaper for a 10 kilometer ride?
- A full tank holds 500 liters and drains at 20 liters a minute. At the same moment a second tank holding 80 liters starts filling at 15 liters a minute. After how many minutes do the tanks hold the same amount, and how much is in each?
- Brontë-May sets off walking at 4 kilometers an hour. Half an hour later Anselm sets off from the same place along the same route at 6 kilometers an hour. Let t be the number of hours since Anselm set off. Write an equation for the moment he catches her, solve it, and say how far from the start they meet.
- Odalys is 38 years old and her son Tobiah is 10. In how many years will Odalys be three times as old as Tobiah, and how old will each be then?
Reasoning
- Anselm subtracts 4x from both sides of 4x + 1 = 4x + 6. Explain what he is left with, and what that says about which numbers solve the equation.
- Explain why subtracting 3x from both sides of an equation can neither create a new solution nor lose one.
Answers
- x = 6. Subtract 4x: 5x + 2 = 32, so 5x = 30 and x = 6. Check: 56 = 56.
- m = -5. Subtract 3m: -8 = 4m + 12, so -20 = 4m and m = -5. Check: both sides are -23.
- k = 4. Add 4k: 11 = 6k - 13, so 24 = 6k and k = 4. Check: both sides are -5.
- p = 7. Subtract 3p: 5p = 35, so p = 7. Check: 56 = 21 + 35.
- y = 5. Subtract 1.3y: 1.2y + 3 = 9, so 1.2y = 6 and y = 5. Check: both sides are 15.5.
- w = -4. The smaller coefficient is -7, so add 7w: 4w + 10 = -6, so 4w = -16 and w = -4. Check: 12 + 10 = 22 and 28 - 6 = 22.
- \displaystyle x = \frac{5}{4}. Subtract 2x: 4x - 1 = 4, so 4x = 5 and \displaystyle x = \frac{5}{4}. Check: both sides are 6.5.
- Yes. The left side is 5(3) - 7 = 8 and the right side is 2(3) + 2 = 8; they agree.
- No: the left side is 4 - 3(-2) = 10 but the right side is -2 + 8 = 6. Solving, add 3t: 4 = 4t + 8, so -4 = 4t and t = -1. Check: 4 + 3 = 7 and -1 + 8 = 7.
- a = 5. Subtracting 2a: 15 = 5a - 10, so 25 = 5a and a = 5. Subtracting 7a: -5a + 15 = -10, so -5a = -25 and a = 5. Both routes are legal moves on the same equation, so they must agree.
- x = 3. Expand: 4x + 12 = 2x + 18. Subtract 2x: 2x + 12 = 18, so 2x = 6 and x = 3. Check: both sides are 24.
- n = 3. Expand: 10n - 5 = 3n + 16. Subtract 3n: 7n - 5 = 16, so 7n = 21 and n = 3. Check: both sides are 25.
- y = 3. The left side is 7 - y + 2 = 9 - y and the right is 3y - 3. Add y: 9 = 4y - 3, so 12 = 4y and y = 3. Check: both sides are 6.
- c = 12. The left side is 6c - 8 - c = 5c - 8 and the right is 4c + 4. Subtract 4c: c - 8 = 4, so c = 12. Check: both sides are 52.
- x = 14. Subtract \displaystyle \frac{1}{4}x: \displaystyle \frac{1}{2}x - 2 = 5, so \displaystyle \frac{1}{2}x = 7 and x = 14. Check: both sides are 8.5.
- x = 30. The smaller coefficient is \displaystyle \frac{1}{3}; subtract \displaystyle \frac{x}{3}: \displaystyle 4 = \frac{x}{6} - 1, so \displaystyle \frac{x}{6} = 5 and x = 30. Check: both sides are 14.
- d = 6. Expand: 2d + 4 = d + 10. Subtract d: d + 4 = 10, so d = 6. Check: both sides are 16.
- z = 12. Expand: -6 + 3z = 2z + 10 - 4 = 2z + 6. Subtract 2z: z - 6 = 6, so z = 12. Check: both sides are 30.
- After 8 weeks, with 260 dollars each. With w weeks, 140 + 15w = 60 + 25w. Subtract 15w: 140 = 60 + 10w, so 80 = 10w and w = 8. Then 140 + 15(8) = 260 and 60 + 25(8) = 260.
- At 8 kilometers, costing 14.10 dollars. With k kilometers, 4.50 + 1.20k = 2.10 + 1.50k, so 2.40 = 0.30k and k = 8; then 4.50 + 9.60 = 14.10. For 10 kilometers firm A charges 16.50 and firm B charges 17.10, so firm A is cheaper.
- After 12 minutes, with 260 liters in each. With t minutes, 500 - 20t = 80 + 15t. Add 20t: 500 = 80 + 35t, so 420 = 35t and t = 12. Then 500 - 240 = 260 and 80 + 180 = 260.
- 4(t + 0.5) = 6t, so t = 1, and they meet 6 kilometers from the start. Brontë-May has been walking for t + 0.5 hours. Expand: 4t + 2 = 6t; subtract 4t: 2 = 2t, so t = 1. Anselm has gone 6(1) = 6 kilometers and Brontë-May 4(1.5) = 6 kilometers.
- In 4 years, when Odalys is 42 and Tobiah is 14. With y years from now, Odalys will be 38 + y and Tobiah 10 + y, so 38 + y = 3(10 + y). Expand: 38 + y = 30 + 3y. Subtract y: 38 = 30 + 2y, so 8 = 2y and y = 4. Check: 42 = 3(14).
- He is left with 1 = 6, with no x in it. That statement is false, and no choice of x can change it, so no number makes the original equation true: it has no solution. The x vanished because both sides had the same coefficient, 4; No solution, one solution, or every number takes this up.
- If a number makes the equation true, both sides are the same number for it, and taking 3x (the same number on each side) from both leaves them equal, so it solves the new equation too. In the other direction, adding 3x to both sides of the new equation gives back the old one, so any solution of the new equation also solves the old. The two equations have exactly the same solutions.