Equations with variable terms on both sides: why a variable term may be added to or subtracted from both sides, choosing the side with the greater coefficient so the result stays positive, combining like terms on each side first, and modelling situations where two changing quantities become equal.
Subject: Algebra 1 · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Algebra 1 · Chapter 3 — Solving Linear Equations
Solving Equations with Variables on Both Sides
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.4 Solving Equations with Variables on Both Sides §3.4, pp. 151-156 — the lesson these objectives are drawn from
Warm-up
Every move you need is already in your hands. What is new is that one of them gets used on a variable term.
Discussion prompt
In Lesson 3.1 you subtracted 6 from both sides of x plus 6 equals 10. Could you subtract 2x from both sides of an equation in the same way? Why or why not?
Hint: Ask what kind of thing 2x is.
Answer:
\[ 7x - 19 = 2x + 55 \;\rightarrow\; 5x - 19 = 55 \]
Yes, and for a simple reason: x stands for a number, so 2x is a number too. The subtraction property of equality permits any number to be taken from both sides, and it does not care whether that number is written as a digit or as a variable term. Nothing new is being allowed — an old permission is being used in a new place.
Concept
When variables appear on both sides, one extra move comes first: add or subtract a variable term so that all the variables end up on one side. After that the equation is an ordinary two-step equation.
variable term — A term containing a variable, such as 2x. Because a variable represents a number, a variable term may be added to or subtracted from both sides of an equation.
Collecting on the side with the greater coefficient leaves a positive coefficient, which is easier to work with.
Figure (svg): A balance scale with variable terms on both pans, and the smaller group being removed from both
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.4 Solving Equations with Variables on Both Sides §3.4, pp. 151-151
Section
Section 1
Concept
Since variables represent numbers, you can transform an equation by adding and subtracting variable terms exactly as you would numbers. The properties of equality make no distinction between the two.
The move is licensed by the same addition and subtraction properties of equality used in Lesson 3.1.
Figure (svg): A reminder that a variable term can be moved because a variable stands for a number
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.4 Solving Equations with Variables on Both Sides §3.4, pp. 151-151 — the Study Tip on transforming an equation with variable terms
Picture it
One x removed from each pan, and the scale stays level.
Figure (svg): A balance scale with variable terms on both pans, and the smaller group being removed from both
The picture is the same one from Lesson 3.1, with an x-block being removed instead of a unit block. Whatever x weighs, removing one of them from each side keeps the balance.
Worked example
This is Example 1 from the textbook, adapted. Seven is greater than two, so the left side is chosen.
\[ \text{Solve } \; 7x - 19 = 2x + 55. \]
Compare the coefficients of the variable terms
Why: Seven on the left, two on the right. Seven is greater, so collect on the left.
Subtract 2x from each side
Why: The right side loses its variable term entirely; the left becomes five x.
\[ 5 x - 19 = 55 \]
Add 19 to each side
Why: Now it is an ordinary two-step equation, and the constant comes off first.
\[ 5 x = 74 \]
Divide each side by 5
Why: Seventy-four over five, which is 14.8.
\[ x = \frac{74}{5} \]
Figure (svg): The equation 7x minus 19 equals 2x plus 55 solved in four steps
\[ 7x - 19 = 2x + 55 \;\Longrightarrow\; x = \tfrac{74}{5} \]
Verify: substitute into both sides separately
Why: At x equal to 74 over 5, the left side is seven times 14.8 minus 19, which is 103.6 minus 19, or 84.6. The right side is two times 14.8 plus 55, which is 29.6 plus 55, also 84.6. Both sides agree, which is what an equation being satisfied means.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.4 Solving Equations with Variables on Both Sides §3.4, pp. 151-151
Sorting
For the equation 5x plus 2 equals 3x plus 10, judge each proposed move.
Sort into buckets
Sort each move by whether it produces an equivalent equation.
Four of six moves are balanced and only two of those four make progress. Writing every operation on both sides explicitly is what makes the difference visible.
Worked example
Example 2 from the textbook. Here the right side has the greater coefficient.
\[ \text{Solve } \; 80 - 9y = 6y. \]
Read the coefficients carefully
Why: The left has negative nine y, since 80 minus 9y is 80 plus negative 9y. The right has positive six.
\[ -9\text{ and } 6 \]
Decide which is greater
Why: Six is greater than negative nine, so collect on the right.
Add 9y to each side
Why: The left loses its variable term; the right becomes fifteen y.
\[ 80 = 15 y \]
Divide each side by 15
Why: Eighty over fifteen simplifies to sixteen thirds.
\[ y = \frac{16}{3} \]
Figure (svg): The solution to Worked example collect on the right shown as a ladder of expressions, one row per algebraic move
\[ 80 - 9y = 6y \;\Longrightarrow\; y = \tfrac{16}{3} \]
Verify: check both sides at the answer
Why: Sixteen thirds is about 5.33. The left side is 80 minus nine times 5.33, which is 80 minus 48, or 32. The right side is six times 5.33, which is also 32. The two agree, and collecting on the right kept the coefficient positive throughout.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.4 Solving Equations with Variables on Both Sides §3.4, pp. 152-152
Trap
\[ 7x - 19 = 2x + 55 \]
Move the 2x across to the left and write it as plus 2x
Why: Terms are often described as moving across, and moving something usually leaves it unchanged.
\[ 9x - 19 = 55 \quad \text{(wrong)} \]
Nothing moved. Two x was subtracted from the right and must therefore be subtracted from the left as well, giving five x rather than nine x.
\[ 7x - 19 - 2x = 2x + 55 - 2x \;\Longrightarrow\; 5x - 19 = 55 \]
Write the subtraction on both sides rather than describing a term as moving
Why: The property of equality is about applying an operation twice, once per side, and writing it that way makes the sign automatic.
A term that appears to change sign when it crosses the equal sign is really being subtracted from both sides. Writing the operation explicitly removes the guesswork about which sign it should end up with.
Faded example
The move is chosen. Apply it to both sides.
Fill in the blanks
7x - 19 = 2x + 55 \;\rightarrow\; 7x - 19 - 2x = 2x + 55 - 2x \;\rightarrow\; 5x - 19 = 55
Why: Subtracting two x from both sides removes the variable from the right and reduces the left coefficient from seven to five. The two blanks are the same operation seen from each side, which is exactly what a property of equality requires.
Elimination
The equation is 6x plus 1 equals 2x plus 9.
Eliminate the wrong options
Which move puts all the variable terms on one side?
Survives elimination: A
Why: Subtracting two x removes the variable term from the right entirely and leaves four x on the left. Option D is worth noticing: it keeps the balance and moves in the wrong direction, which shows again that a legal move is not automatically a useful one.
Socratic
It looks like a bigger step than subtracting a number, and it is not.
Discussion prompt
Explain why subtracting 2x from both sides is permitted by the same property that permits subtracting 2. Then say what would have to be true for the move to be unsafe.
Hint: Ask what kind of object 2x is once x has a value.
Answer:
The subtraction property of equality says the same number may be taken from both sides. Since x stands for a number, 2x is a number as well — a different one for each value of x, but a number in every case. So the property applies without modification.
The move would be unsafe only if it were not reversible, and subtraction always is: adding 2x back to both sides returns the original equation. Contrast that with dividing by 2x, which is unsafe precisely because 2x might be zero and division by zero is undefined.
Section
Section 2
Concept
Both sides are legal choices, and one of them is easier. Collecting the variable terms on the side whose coefficient is greater leaves a positive coefficient, so the final division is by a positive number.
\[ 7x - 19 = 2x + 55 \;\rightarrow\; \text{collect left, since } 7 > 2 \]
Read the coefficients with their signs: in 80 minus 9y the coefficient is negative nine, not nine.
Figure (svg): Two coefficients compared, with the larger one showing which side to collect on
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.4 Solving Equations with Variables on Both Sides §3.4, pp. 151-151 — the paragraph on collecting on the side with the greater coefficient
Picture it
The larger one marks the side to collect on.
Figure (svg): Two coefficients compared, with the larger one showing which side to collect on
Seven against two, so the left side wins and the surviving coefficient is five — positive, and the final division is straightforward.
Worked example
Collecting on the smaller side is legal. Comparing the two shows why the rule exists.
\[ \text{Solve } \; 7x - 19 = 2x + 55 \; \text{ twice, collecting on each side in turn.} \]
Route one: collect on the left by subtracting 2x
Why: Five x minus nineteen equals fifty-five, then five x equals seventy-four.
\[ x = \frac{74}{5} \]
Route two: collect on the right by subtracting 7x
Why: Negative nineteen equals negative five x plus fifty-five.
\[ -19 = -5 x + 55 \]
Finish route two
Why: Subtract fifty-five from both sides to get negative seventy-four equals negative five x, then divide by negative five.
\[ x = \frac{74}{5} \]
Compare the two
Why: Same answer, but the second route required dividing by a negative number.
Figure (svg): Two columns contrasting collecting on the larger-coefficient side with collecting on the smaller one
\[ x = \tfrac{74}{5} \text{ either way} \]
Verify: check that the two routes really agree
Why: Both give seventy-four fifths. The second route's final step divides negative seventy-four by negative five, and the two negatives agree so the quotient is positive — correct, but one more place for a sign error than the first route offers.
Discrimination
Read each coefficient with its sign before comparing.
Sort into buckets
Sort each equation by which side you should collect the variables on.
Worked example
Example 2 from the textbook. The comparison is easy to get wrong if the sign is missed.
\[ \text{In } \; 80 - 9y = 6y, \; \text{ decide which side to collect on.} \]
Rewrite the left side as a sum
Why: Eighty minus nine y is eighty plus negative nine y, so the coefficient is negative nine.
\[ \text{coefficient } -9 \]
Read the right coefficient
Why: Six y has coefficient six.
\[ \text{coefficient } 6 \]
Compare with signs included
Why: Six is greater than negative nine, so the right side has the greater coefficient.
\[ 6 > -9 \]
Collect on the right
Why: Add nine y to each side, giving eighty equals fifteen y.
\[ 80 = 15 y \]
Figure (svg): The solution to Worked example reading a negative coefficient shown as a ladder of expressions, one row per algebraic move
\[ 80 = 15y \;\Longrightarrow\; y = \tfrac{16}{3} \]
Verify: check that the surviving coefficient is positive
Why: Fifteen is positive, which is what collecting on the greater side guarantees. Had the sign on the left been read as nine rather than negative nine, the comparison would have gone the other way and the coefficient would have come out negative.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.4 Solving Equations with Variables on Both Sides §3.4, pp. 152-152
Error analysis
The student decided which side to collect on in three equations. Two decisions are wrong.
Annotate
On: \( \begin{aligned} 7x - 19 &= 2x + 55 \;\rightarrow\; \text{collect left, since } 7 > 2 \\ 80 - 9y &= 6y \;\rightarrow\; \text{collect left, since } 9 > 6 \\ 3x + 4 &= 8x - 1 \;\rightarrow\; \text{collect left, since it comes first} \end{aligned} \)
Both errors come from not reading the coefficients as signed numbers in the right order. Rewriting each side as a sum, as Lesson 2.4 recommended, makes the coefficients unambiguous before any comparison is made.
Prediction
The choice of side determines the sign of what survives.
Predict first
In 3x plus 4 equals 8x minus 1, if you collect on the left instead of the right, what coefficient results?
Correct: Negative 5, so you divide by a negative at the end.
\[ \text{left: } -5x + 4 = -1 \;\rightarrow\; x = 1 \]
\[ \text{right: } 4 = 5x - 1 \;\rightarrow\; x = 1 \]
Why: Subtracting eight x from both sides leaves three minus eight, which is negative five, on the left. The answer is the same either way, but the final division is by negative five rather than positive five — one extra sign to get right. Collecting on the greater side is a convenience rule, not a correctness rule.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Collect on the greater side | Collect on the smaller side | |
|---|---|---|
| Resulting coefficient | positive | negative |
| Final division | by a positive number | by a negative number |
| Answer obtained | correct | correct |
The bottom row is the important one: both routes are correct. The rule about the greater side is about reducing the chance of a sign error, not about validity.
Socratic
The rule is a convenience, and it is worth knowing what it buys.
Discussion prompt
Explain why collecting on the side with the greater coefficient guarantees a positive coefficient afterwards. Then say when you might deliberately ignore the rule.
Hint: Think about what subtraction of the smaller from the larger produces.
Answer:
Collecting on the greater side means subtracting the smaller coefficient from the greater one, and a greater number minus a smaller one is positive. Collecting the other way subtracts the greater from the smaller and gives a negative. The rule is simply a way of arranging for the subtraction to go the favourable direction.
You might ignore it when the smaller side has a coefficient of one or zero, since collecting there can leave a particularly clean equation. In 80 minus 9y equals 6y, for instance, either choice is fine; and if one side were simply y, collecting there would leave the variable with a coefficient you never have to divide by at all.
Section
Section 3
Concept
If either side has like terms, combine them before deciding which side to collect on. The coefficients you compare must be the simplified ones, or the comparison is made on the wrong numbers.
Figure (svg): An equation with like terms on both sides, simplified on each side before collecting
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.4 Solving Equations with Variables on Both Sides §3.4, pp. 152-152 — Example 3, Combine Like Terms First
Picture it
Five lines, and only the first touches one side alone.
Figure (svg): An equation with like terms on both sides, simplified on each side before collecting
The comparison happens after the combining, on seven against five rather than on three against five. Comparing too early would have sent the collection to the wrong side.
Worked example
This is Example 3 from the textbook, adapted.
\[ \text{Solve } \; 3x + 10 + 4x = 5x - 7. \]
Combine the like terms on the left
Why: Three x and four x make seven x. The right side has nothing to combine.
\[ 7 x + 10 = 5 x - 7 \]
Compare the simplified coefficients
Why: Seven on the left against five on the right, so collect on the left.
Subtract 5x from each side
Why: Seven minus five is two.
\[ 2 x + 10 = -7 \]
Finish the two-step equation
Why: Subtract ten from each side to get 2x equals negative seventeen, then divide by two.
\[ x = -\frac{17}{2} \]
Figure (svg): An equation with like terms on both sides, simplified on each side before collecting
\[ 3x + 10 + 4x = 5x - 7 \;\Longrightarrow\; x = -\tfrac{17}{2} \]
Verify: check both sides at the answer
Why: At x equal to negative 8.5, the left side is three times negative 8.5 plus ten plus four times negative 8.5, which is negative 25.5 plus 10 minus 34, or negative 49.5. The right side is five times negative 8.5 minus seven, which is negative 42.5 minus 7, also negative 49.5. The two agree.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.4 Solving Equations with Variables on Both Sides §3.4, pp. 152-152
Faded example
Simplify each side and supply the two coefficients to compare.
Fill in the blanks
3x + 10 + 4x = 5x - 7 \;\rightarrow\; 7x + 10 = 5x - 7
Why: Combining three x and four x gives seven x on the left, while the right side has no like terms to combine and stays at five x. The comparison is then seven against five, which sends the collection to the left — a different answer from comparing three against five would have given.
Worked example
Guided Practice 4. Both sides need simplifying before anything else happens.
\[ \text{Solve } \; 5x + 3x - 4 = 3x + 8. \]
Combine on the left
Why: Five x and three x make eight x.
\[ 8 x - 4 = 3 x + 8 \]
Check the right side
Why: Three x and eight are not like terms, so nothing combines there.
Compare and collect
Why: Eight is greater than three, so subtract three x from both sides.
\[ 5 x - 4 = 8 \]
Finish
Why: Add four to both sides to get 5x equals twelve, then divide by five.
\[ x = \frac{12}{5} \]
Figure (svg): The solution to Worked example like terms on both sides shown as a ladder of expressions, one row per algebraic move
\[ 5x + 3x - 4 = 3x + 8 \;\Longrightarrow\; x = \tfrac{12}{5} \]
Verify: substitute into the original, before combining
Why: At x equal to 2.4 the left side is 12 plus 7.2 minus 4, which is 15.2, and the right side is 7.2 plus 8, also 15.2. Checking against the uncombined original is what tests the combining step as well as the solving.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.4 Solving Equations with Variables on Both Sides §3.4, pp. 152-152
Trap
\[ 3x + 10 + 4x = 5x - 7 \]
Compare the first coefficient on each side — 3 against 5 — and collect on the right
Why: The leftmost variable term on each side is the one the eye lands on first.
\[ 10 + 4x = 2x - 7 \quad \text{(a mess, and the wrong side chosen)} \]
The left side really has a coefficient of seven, not three. The comparison was made on numbers that do not describe either side.
\[ 7x + 10 = 5x - 7 \;\Longrightarrow\; 2x + 10 = -7 \]
Combine each side completely, then compare the simplified coefficients
Why: A side's coefficient is what is left after combining, not the first number you see.
This is the same principle as Lesson 2.7: you cannot work with terms until the expression is simplified, because until then you do not know what the terms are.
Ranking
Five moves, one correct sequence.
Put in order
Why: Combining comes first, because the coefficients to be compared are the simplified ones. Collecting comes second and reduces the problem to a two-step equation. Then the two familiar moves in their usual order, and the check last. Only the first move touches one side at a time.
Sorting
Scan each side separately for like terms.
Sort into buckets
Sort each equation by which sides have like terms to combine.
Four of the six need combining somewhere, and in every one of those the coefficient to compare is different from the first number visible on that side.
Socratic
The order of these two moves changes which side you collect on.
Discussion prompt
Using 3x plus 10 plus 4x equals 5x minus 7, show what happens if you compare coefficients before combining, and explain why the comparison has to come second.
Hint: Work out which side each order would send you to.
Answer:
Comparing before combining looks at three against five and sends you to the right side. Comparing after combining looks at seven against five and sends you to the left. The two orders give different decisions, so at most one of them can be following the rule as intended.
The rule is about the coefficient of the side, and a side with two variable terms does not have a single coefficient until they are combined. So the comparison is meaningless before the combining — three is the coefficient of one term, not of the side. Both routes still reach the correct answer, but only one of them keeps the coefficient positive.
Section
Section 4
Concept
With variables on both sides, the check evaluates each side independently and compares the two results. A single number appearing on both sides is what confirms the solution.
Because the variable appears on both sides, the check does twice as much work as before — and catches twice as much.
Figure (svg): The equation 7x minus 19 equals 2x plus 55 solved in four steps
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.4 Solving Equations with Variables on Both Sides §3.4, pp. 151-151 — the CHECK step of Example 1
Picture it
The check tests the whole chain at once by returning to the top line.
Figure (svg): The equation 7x minus 19 equals 2x plus 55 solved in four steps
Substituting into the first line tests every move below it. Substituting into any later line tests only the moves after that point.
Worked example
The textbook's Example 1 uses the numbers below, where the answer is a whole number.
\[ \text{Check that } x = 4 \text{ solves } \; 7x + 19 = 2x + 39. \]
Write the original equation
Why: Not any of the lines produced while solving.
\[ 7 x + 19 = 2 x + 39 \]
Evaluate the left side at x equal to 4
Why: Seven times four is twenty-eight, plus nineteen is forty-seven.
\[ 47 \]
Evaluate the right side at x equal to 4
Why: Two times four is eight, plus thirty-nine is forty-seven.
\[ 47 \]
Compare
Why: Both sides are forty-seven, so the statement is true and four is a solution.
\[ 47 = 47 \]
Figure (svg): The solution to Worked example check a both-sides solution shown as a ladder of expressions, one row per algebraic move
\[ 7(4) + 19 = 47 = 2(4) + 39 \;\checkmark \]
Verify: notice what the check would have caught
Why: If the collection had been done with the wrong sign, the answer would have differed and the two sides would have produced different numbers. Evaluating the sides separately means the check cannot accidentally reuse an error, since each side is computed from the original equation alone.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.4 Solving Equations with Variables on Both Sides §3.4, pp. 151-151
Elimination
A student has solved 5x plus 2 equals 3x plus 10 and got x equals 4.
Eliminate the wrong options
Which check actually tests the answer?
Survives elimination: A
Why: The equation claims the two sides are equal, so the check must evaluate both and compare. At four the left is twenty-two and the right is twenty-two, which confirms the solution and tests every step of the working at once.
Worked example
The commonest error here produces an answer that looks reasonable.
\[ \text{A student solves } 5x + 2 = 3x + 10 \text{ and gets } x = 6. \text{ Check it.} \]
Evaluate the left side at 6
Why: Five times six is thirty, plus two is thirty-two.
\[ 32 \]
Evaluate the right side at 6
Why: Three times six is eighteen, plus ten is twenty-eight.
\[ 28 \]
Compare
Why: Thirty-two is not twenty-eight, so six is not a solution.
Solve correctly
Why: Subtract three x from both sides to get 2x plus two equals ten, then 2x equals eight, so x is four.
\[ x = 4 \]
Figure (svg): The solution to Worked example a check that catches a sign error shown as a ladder of expressions, one row per algebraic move
\[ \text{at } x = 4: \; 22 = 22 \;\checkmark \]
Verify: check the corrected answer on both sides
Why: At x equal to four the left side is twenty-two and the right side is also twenty-two. The corrected solution passes the check the wrong one failed, which confirms the diagnosis rather than replacing one guess with another.
Trap
\[ 5x + 2 = 3x + 10 \text{ at } x = 4 \]
Substitute into the left side, get 22, and declare the answer correct
Why: One side has been evaluated and produced a definite number, which feels like a completed check.
Twenty-two on its own says nothing. The check is whether the two sides agree, and the second side has not been computed.
\[ \text{left: } 5(4) + 2 = 22 \qquad \text{right: } 3(4) + 10 = 22 \]
Evaluate both sides separately and compare the two numbers
Why: An equation claims the two sides are equal, so a check must produce both and compare them.
This is different from Lesson 3.3, where the right side was already a number. Here both sides need evaluating, and skipping one halves the check.
Sorting
Evaluate each side at the claimed value and compare.
Sort into buckets
Sort each claim by whether the solution is correct.
The last item has a fractional solution and checks out perfectly. A fraction is no less a solution than a whole number, and the check works on it identically.
Faded example
Evaluate each side separately.
Fill in the blanks
\text47 x = 4: \quad \text47 7(4) + 19 = ___, \quad \text___ 2(4) + 39 = ___
Why: Both sides come to forty-seven, which is what confirms that four is a solution. The two blanks are computed independently from the original equation, so an error in the solving cannot make them agree by accident — that independence is the point of the check.
Socratic
In Lesson 3.3 one side was already a number. Here neither is.
Discussion prompt
Explain what extra work the check requires when variables appear on both sides, and what extra kind of error that extra work can catch.
Hint: Count how many evaluations each kind of check needs.
Answer:
With a number on one side, the check evaluates one expression and compares it with a number already given. With variables on both sides, it evaluates two expressions and compares them with each other — twice the arithmetic, and neither side is given to you.
The extra work catches errors in the collection step specifically. If a variable term was moved with the wrong sign, the two sides will differ by exactly twice that term's value at the answer, which shows up immediately as two different numbers. A one-sided check could not detect that at all.
Section
Section 5
Concept
A situation with variables on both sides almost always comes from asking when two changing quantities are the same. Each quantity becomes an expression, and the question becomes an equation.
Figure (svg): Two animals starting at different points and running at different speeds, meeting where the distances are equal
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.4 Solving Equations with Variables on Both Sides §3.4, pp. 155-155 — the cheetah-and-gazelle exercises the lesson opens with
Picture it
One starts behind and runs faster; the equation asks when they meet.
Figure (svg): Two animals starting at different points and running at different speeds, meeting where the distances are equal
The head start is a fixed amount and each speed is a rate, so both sides have the fixed-plus-rate shape from Lesson 3.3 — one on each side of the equal sign.
Worked example
A cheetah runs at 70 feet per second; a gazelle running at 50 feet per second has a 40 foot head start.
\[ \text{Solve } \; 70t = 40 + 50t \; \text{ for the time } t \text{ in seconds.} \]
Write an expression for each distance
Why: The cheetah covers 70t feet; the gazelle covers 50t feet from a starting point 40 feet ahead.
\[ 70 t\text{ and } 40 + 50 t \]
Set them equal
Why: The cheetah catches the gazelle when the two distances agree.
\[ 70 t = 40 + 50 t \]
Collect the variables on the greater side
Why: Seventy is greater than fifty, so subtract 50t from both sides.
\[ 20 t = 40 \]
Divide and interpret
Why: Divide by twenty to get t equals two seconds.
\[ 2\text{ seconds} \]
Figure (svg): Two animals starting at different points and running at different speeds, meeting where the distances are equal
\[ 70t = 40 + 50t \;\Longrightarrow\; t = 2 \text{ seconds} \]
Verify: compute both distances at the answer
Why: In two seconds the cheetah covers 140 feet and the gazelle covers 100 feet from a point 40 feet ahead, which is also 140 feet from the cheetah's start. The two positions agree, which is what catching up means.
Translation
Four situations where two quantities become equal. Let t or v be the unknown.
Match the pairs
Why: Every one has the fixed-plus-rate shape on both sides, and the equation asks when the two agree. The third is worth noticing: one rate is negative because the tank is draining, and writing it as minus 8t rather than plus 8t is what makes the model correct.
Worked example
Plan A charges 30 dollars plus 5 dollars a visit; plan B charges 10 dollars plus 9 dollars a visit.
\[ \text{Find the number of visits } v \text{ at which the two plans cost the same.} \]
Write an expression for each plan
Why: Plan A costs 30 plus 5v; plan B costs 10 plus 9v.
\[ 30 + 5 v\text{ and } 10 + 9 v \]
Set them equal
Why: The costs agree at the break-even point.
\[ 30 + 5 v = 10 + 9 v \]
Collect on the greater side
Why: Nine is greater than five, so subtract 5v from both sides.
\[ 30 = 10 + 4 v \]
Finish and interpret
Why: Subtract ten to get 20 equals 4v, then divide by four: five visits.
\[ 5\text{ visits} \]
Figure (svg): The solution to Worked example comparing two payment plans shown as a ladder of expressions, one row per algebraic move
\[ 30 + 5v = 10 + 9v \;\Longrightarrow\; v = 5 \text{ visits} \]
Verify: compute both costs at five visits
Why: Plan A costs 30 plus 25, which is 55. Plan B costs 10 plus 45, also 55. The two agree at five visits, and comparing at six visits shows plan A becoming cheaper — which is the practical conclusion the equation was asked for.
Trap
\[ 30 + 5v = 10 + 9v \;\Longrightarrow\; v = 5 \]
Report that the plans cost the same at 5 dollars
Why: The number five appeared and the problem is about money, so the units get attached from the context rather than from the definition.
Five was the number of visits, not a cost. The cost at that point is 55 dollars, which is a different number entirely.
The two plans cost the same at 5 visits, where both cost 55 dollars.
Look up what the letter stood for before attaching a unit
Why: The variable was defined as a number of visits, so the answer is a count and the cost has to be computed separately if it is wanted.
Break-even problems have two natural answers — when, and how much — and the equation gives the first. Reporting the second as well is usually what the situation actually calls for.
Elimination
A cheetah runs at 70 feet per second. A gazelle running at 50 feet per second starts 40 feet ahead. Let t be the time in seconds.
Eliminate the wrong options
Which equation gives the catch-up time?
Survives elimination: A
Why: The cheetah's distance is speed times time, and the gazelle's is its head start plus its own speed times time. Setting them equal gives two seconds. Option B is worth noticing: a negative answer would have signalled the model was wrong, which is the reasonableness check from Lesson 1.6 doing its job.
Prediction
The equation gives the moment of equality. The situation continues past it.
Predict first
Plan A costs 30 plus 5 per visit and plan B costs 10 plus 9 per visit, breaking even at 5 visits. Which plan is cheaper at 8 visits?
Correct: Plan A, since it has the smaller rate.
\[ \text{at } v = 8: \quad 30 + 40 = 70 \quad \text{against} \quad 10 + 72 = 82 \]
Why: At eight visits plan A costs 30 plus 40, which is 70, and plan B costs 10 plus 72, which is 82. Past the crossing point the plan with the smaller rate wins, because the rate is what dominates as the quantity grows. Below the crossing point the smaller fixed cost wins instead, which is why break-even questions have practical answers on both sides.
Socratic
Earlier word problems produced a variable on one side only.
Discussion prompt
Explain what feature of a situation puts the variable on both sides of the equation, and contrast it with the taxi problem from Lesson 3.3, which did not. Then say what kind of question a both-sides equation is answering.
Hint: Count how many quantities in each situation are changing.
Answer:
In the taxi problem only one quantity changed with the number of miles — the fare — and it was compared with a fixed total. In the race, both the cheetah's distance and the gazelle's distance change with time, so both sides of the comparison contain the variable.
A both-sides equation is answering a question of the form when do these two become equal, rather than when does this reach a fixed value. That is why they appear in comparisons, break-even calculations and catch-up problems, and why Chapter 7 will handle the same kind of question with two equations at once.
Comparison
Fill the blanks from memory before you scroll back. Only the first row is new.
Comparison matrix
| Variable on one side | Variable on both sides | |
|---|---|---|
| First move | undo the constant | collect the variable terms on one side |
| Remaining moves | undo the coefficient | the same as a two-step equation |
| Check | evaluate one side | evaluate both sides and compare |
One extra move at the front and one extra evaluation in the check. Everything in between is Lesson 3.3 unchanged.
Pattern
Whether the equation comes from a page or from a comparison, the same five moves cover it.
Step two must come after step one. A side with two variable terms has no single coefficient to compare until they are combined.
OpenStax Elementary Algebra 2e, §2.3 Solve Equations with Variables and Constants on Both Sides §2.3
Check
Collect first. Compare the coefficients before you move anything.
Check your understanding
Solve 6x plus 1 equals 2x plus 9.
Answer: A
Why: Six is greater than two, so subtract 2x from both sides, giving 4x plus one equals nine. Subtracting one gives 4x equals eight, and dividing by four gives two. Both sides check at thirteen.
Check
Combine first. The coefficient to compare is the simplified one.
Check your understanding
Solve 4x plus 2x minus 3 equals 3x plus 9.
Answer: A
Why: Combining four x and two x gives six x, so the equation is six x minus three equals three x plus nine. Subtracting three x gives three x minus three equals nine, then three x equals twelve, so x is four. Both sides check at twenty-one.
Check
A comparison problem. Set the two expressions equal.
Check your understanding
One saver has 60 dollars and adds 15 a week. Another has 200 dollars and adds 5 a week. After how many weeks do they have the same amount?
Answer: A
Why: The equation is 60 plus 15t equals 200 plus 5t. Subtracting 5t gives 60 plus 10t equals 200, then 10t equals 140, so t is fourteen weeks. Checking: 60 plus 210 is 270, and 200 plus 70 is also 270.
Real world
Two mobile plans: one costs 25 dollars a month with unlimited data, the other 10 dollars a month plus 3 dollars per gigabyte.
Discussion prompt
Write an equation for the amount of data at which the two plans cost the same, solve it, and then say which plan is cheaper for someone using 3 gigabytes and which for someone using 8. Explain how the crossing point divides the two cases.
Hint: One side has no variable at all, and the other has both a fixed part and a rate.
Answer:
\[ 25 = 10 + 3g \;\Longrightarrow\; 15 = 3g \;\Longrightarrow\; g = 5 \text{ gigabytes} \]
At five gigabytes both plans cost twenty-five dollars. Below that the metered plan is cheaper: at three gigabytes it costs 10 plus 9, which is 19, against 25. Above it the unlimited plan wins: at eight gigabytes the metered plan costs 10 plus 24, which is 34.
The crossing point is the boundary between the two regimes, which is why break-even calculations are worth doing before choosing. Notice that this equation has a variable on only one side — the unlimited plan does not depend on usage — so it is really a Lesson 3.3 problem wearing a Lesson 3.4 disguise.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
When you subtract 2x from the right side of an equation, what must you do to the left side?
Correct: Subtract 2x, the same operation on both sides.
\[ 7x - 19 - 2x = 2x + 55 - 2x \;\Longrightarrow\; 5x - 19 = 55 \]
Why: The subtraction property of equality requires the identical operation on both sides. The familiar description of a term moving across and changing sign is a shorthand for exactly this: subtracting 2x from the right removes it there, and subtracting the same 2x from the left is what makes the term appear on the left with a minus sign. Writing the operation on both sides rather than describing a movement is what keeps the sign correct without having to remember a rule.
Explain it
They can solve two-step equations and freeze when a letter appears on both sides.
Discussion prompt
In no more than four sentences, explain what to do first when the variable is on both sides and why that move is allowed. Then tell them how to choose which side to collect on and what that choice saves them.
Hint: The reason it is allowed is about what a letter stands for.
Answer:
A usable answer: get all the letters onto one side first by subtracting the smaller variable term from both sides. That is allowed because a letter stands for a number, and you have always been allowed to subtract a number from both sides. Once the letters are together it is an ordinary two-step equation.
Choose the side with the bigger coefficient — the bigger number in front of the letter. Doing so leaves a positive number in front at the end, so the final division is by a positive rather than a negative, which is one less sign to get right.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: The collection sign is fixed by writing the operation on both sides instead of describing a term as moving. Negative coefficients are fixed by rewriting each side as a sum first, so the sign is attached to the number. Combining first is fixed by scanning each side before looking at the equal sign at all. The two-sided check is fixed by computing each side as a separate number and then comparing. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of a page draw a balance scale holding an equation with variable blocks on both pans, then draw it again after removing the smaller group from each pan, writing the equation under each drawing. Underneath, solve one equation with variables on both sides in full, writing one move per line with its reason, and marking which line was the collection step. Beside it, solve the same equation again collecting on the other side, and compare the two routes in one sentence. Near the bottom, show the check as two separate evaluations, one per side, ending in two numbers you compare. Finally, in the margin, write one comparison situation from your own life and the both-sides equation it produces.
Your two routes should reach the same answer, and one of them should have involved dividing by a negative. If both divisions were by positives, check whether you really collected on different sides.
Recap
Five things, and the first two are the only genuinely new ideas in the lesson.
| If the question says | Your first move is |
|---|---|
| 7x - 19 = 2x + 55 | Compare 7 and 2, then subtract 2x from both sides |
| 80 - 9y = 6y | Read the left coefficient as -9, then collect on the right |
| 3x + 10 + 4x = 5x - 7 | Combine the left side before comparing |
| When do the two plans cost the same | Write an expression for each, then set them equal |
| Check your solution | Evaluate both sides and compare the two numbers |
Lesson 3.5 puts everything together: equations with brackets on both sides, needing the distributive property before any collecting can begin, and a formal four-step procedure that covers every linear equation you will meet.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.4 Solving Equations with Variables on Both Sides §3.4, pp. 151-156 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Algebra 1 — $55/session, free consultation.