Solving a formula or an equation for one of its variables: one-step and two-step rearrangements, rewriting a linear equation for y, and factoring out the target variable when it appears in more than one term.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 1 — Equations and Inequalities
Rewrite Formulas and Equations
Objectives
Five outcomes. The fifth is the one that separates this lesson from Lesson 1.3.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.4 Rewrite Formulas and Equations §1.4, pp. 26-31 — the lesson these objectives are drawn from
Warm-up
Solving for a variable uses the same four properties of equality as Lesson 1.3. The only difference is what counts as the answer.
Discussion prompt
Solve 3x = 12, then solve ax = b for x. What is different about the second one, and what extra condition did you have to notice?
Hint: Ask what you divided by, and whether you are certain you were allowed to.
Answer:
\[ 3x = 12 \;\Longrightarrow\; x = 4 \qquad ax = b \;\Longrightarrow\; x = \tfrac{b}{a}, \; a \neq 0 \]
The steps are identical — divide both sides by the coefficient. What changes is that the coefficient is now a letter, so you have to say that it is not zero, because you cannot see whether it is. That is the only new habit in this lesson.
Concept
A formula relates several quantities. Solving it for one variable rewrites it so that variable stands alone, ready to be computed from the others. It is the same work as solving an equation, done symbolically so that it never has to be repeated.
solve for a variable — To rewrite an equation as an equivalent equation in which that variable is on one side by itself and does not appear on the other side at all.
This is why a formula sheet lists the circumference formula once. Everything else you might want to know about a circle comes out of rearranging it.
Figure (svg): A bar chart style comparison showing substituting first repeated five times against rearranging once and substituting five times
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.4 Rewrite Formulas and Equations §1.4, pp. 26-26
Section
Section 1
Concept
Both halves of the definition matter. An expression that still contains the target variable on the right has not been solved for it, no matter how tidy the left side looks.
formula — An equation relating two or more quantities, usually written with letters standing for those quantities.
The textbook flags this with an Avoid Errors note, and it is worth taking seriously: the failed form looks solved at a glance.
Figure (svg): A diagram contrasting an equation solved for r with the same equation not yet solved, because r still appears on both sides
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.4 Rewrite Formulas and Equations §1.4, pp. 26-28 — the Avoid Errors note on rewriting for y
Picture it
Both have the target variable alone on the left.
Figure (svg): A diagram contrasting an equation solved for r with the same equation not yet solved, because r still appears on both sides
The right-hand one is a true statement, and it is useless: to compute y you would already need y. A solved form must let you compute the variable from things you know.
Worked example
Before rearranging anything, be clear on what each letter measures.
\[ d = rt, \quad A = lw, \quad P = 2l + 2w, \quad C = 2\pi r \]
Name each letter and its unit in the distance formula
Why: Distance in miles, rate in miles per hour, time in hours — and the units confirm the multiplication.
\[ d = r t \]
Name each letter in the two rectangle formulas
Why: Area is length times width; perimeter is twice the length plus twice the width. Both use l and w, which is why they are so often confused.
\[ A =\text{ lw and } P = 2 l + 2 w \]
Notice which formulas are products and which are sums
Why: A product can be rearranged in one division; a sum needs a subtraction first, then a division.
Predict which will be harder to rearrange
Why: The trapezoid area is the hardest, because its target variable can sit inside a bracket that is itself multiplied by a half.
\[ A = (\frac{1}{2}) (b 1 + b 2) h \]
Figure (svg): The solution to Worked example the eight standard formulas, read as instructions shown as a ladder of expressions, one row per algebraic move
\[ \text{product formulas: one division.} \quad \text{sum formulas: subtract, then divide.} \]
Verify: test the prediction on the simplest pair
Why: The area formula A equals lw rearranges to w equals A over l in a single division, while the perimeter formula P equals 2l plus 2w needs 2l subtracted before the division by 2 — two moves, exactly as predicted.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.4 Rewrite Formulas and Equations §1.4, pp. 26-26
Sorting
Each of these claims to be solved for y. Two of them are not.
Sort into buckets
Sort each form.
The check takes two seconds: cover the left side and ask whether you could compute the right from x alone.
Worked example
The textbook's Avoid Errors case, worked in both directions.
\[ \text{Is } y = \frac{6 - 2y}{x} \text{ solved for } y? \]
Check the first condition: is y alone on the left?
Why: It is, so a quick glance suggests the work is done.
Check the second condition: does y appear on the right?
Why: It does, inside the numerator. That fails the definition.
Go back and find the move that avoids it
Why: Starting from 2y + xy = 6, dividing by x too early strands one y on the right. Factoring y out of both terms first is what prevents it.
\[ (2 + x) y = 6 \]
Divide by the whole bracket, not by one term
Why: The bracket is a single quantity multiplying y, so dividing by all of it isolates y properly.
\[ y = \frac{6}{2 + x} \]
Figure (svg): The solution to Worked example recognise a form that is not solved shown as a ladder of expressions, one row per algebraic move
\[ y = \frac{6}{2 + x} \]
Verify: substitute a value of x into both forms
Why: Take x equal to -3. The correct form gives 6 over -1, which is -6. Checking that in the original equation: 2 times -6 plus -3 times -6 is -12 plus 18, which is 6 — correct. The unsolved form would have required knowing y in order to compute y, so it could not have been checked at all.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.4 Rewrite Formulas and Equations §1.4, pp. 28-28
Trap
\[ 4y - xy = 28 \quad \text{solve for } y \]
Divide both sides by 4 to free the first y
Why: The division is aimed at one term rather than at the whole side.
\[ y - \tfrac{xy}{4} = 7 \]
\[ y = 7 + \tfrac{xy}{4} \quad \text{(y is still on the right)} \]
\[ 4y - xy = 28 \quad \text{solve for } y \]
Factor y out of both terms first
Why: Both terms contain y, so it is a common factor and can be pulled out by the distributive property.
\[ (4 - x)y = 28 \]
\[ y = \frac{28}{4 - x}, \quad x \neq 4 \]
Now y appears exactly once and can actually be computed. At x equal to 2 this gives y equal to 14.
Explain it to yourself
A statement can be perfectly true and still useless as a solution.
\[ y = \frac{6 - 2y}{x} \]
Discussion prompt
This equation is true for every pair of x and y that satisfies the original. Explain why that is not enough, and describe what you would actually be able to do with it if someone handed you x equal to 4.
Hint: Try it. Set x to 4 and see how far you get.
Answer:
Setting x to 4 gives y equal to the quantity 6 minus 2y, all over 4 — an equation in y that you would then have to solve. So the form has not removed the work, it has only relocated it.
\[ y = \frac{6 - 2y}{4} \;\Longrightarrow\; 4y = 6 - 2y \;\Longrightarrow\; y = 1 \]
The properly solved form gives it directly: 6 over 6, which is 1. Same answer, no second solve. That saving is exactly what solved for a variable is supposed to buy you.
Matching
Getting the meanings right is what makes the rearrangement worth doing.
Match the pairs
Why: Two of these have subscripted variables or fractional coefficients, which is the visual clue that rearranging them will take more than one step. Note that b sub one and b sub two are two different quantities, not b multiplied by anything — the small lowered numbers are labels, which is why they never enter the arithmetic.
Prediction
Do not rearrange these. Count the moves each would take.
Predict first
Which of these takes the MOST moves to solve for the named variable?
Correct: The trapezoid, solved for b sub two — it takes three moves.
\[ A = \tfrac{1}{2}(b_1 + b_2)h \;\Longrightarrow\; 2A = (b_1 + b_2)h \]
\[ \tfrac{2A}{h} = b_1 + b_2 \;\Longrightarrow\; b_2 = \tfrac{2A}{h} - b_1 \]
Why: The other three are single divisions, because the target variable is one factor of a pure product. The trapezoid needs the half undone by multiplying by 2, then the h undone by dividing, and only then can b sub one be subtracted to leave b sub two alone. Counting moves before starting tells you whether you are on the easy road or the long one.
Section
Section 2
Concept
When the formula is a pure product, the target variable is one of its factors. Dividing both sides by everything else isolates it in a single move.
\[ C = 2\pi r \;\Longrightarrow\; r = \frac{C}{2\pi} \]
Divide by the whole of the rest, not by one piece of it. In the circumference formula that means dividing by two pi as a single quantity.
Figure (svg): A circle with its circumference and radius labelled, and the rearranged formula beside it
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.4 Rewrite Formulas and Equations §1.4, pp. 26-26
Picture it
Example 1: a circle with a circumference of 44 inches.
Figure (svg): A circle with its circumference and radius labelled, and the rearranged formula beside it
The rearranged formula answers the question for any circumference. Substituting 44 into the original would have answered it for exactly one circle.
Worked example
Example 1, both steps. The rearranging and the substituting are deliberately kept apart.
\[ \text{Solve } C = 2\pi r \text{ for } r, \text{ then find } r \text{ when } C = 44 \text{ inches.} \]
Divide each side by 2 pi
Why: Two pi is a single nonzero number multiplying r, so dividing by all of it leaves r alone.
\[ \frac{C}{2 \pi} = r \]
Write the rearranged formula with r on the left
Why: Convention only, but it makes the formula easier to read as an instruction.
\[ r = \frac{C}{2 \pi} \]
Substitute 44 for C
Why: Only now does a specific circle enter the problem.
\[ r = \frac{44}{2 \pi} \]
Evaluate and round
Why: Two pi is about 6.283, and 44 divided by that is about 7.003.
\[ r = 7\text{ approx} \]
Figure (svg): The solution to Worked example solve the circumference formula for r shown as a ladder of expressions, one row per algebraic move
\[ r = \frac{C}{2\pi} \qquad \text{and at } C = 44, \; r \approx 7 \text{ in} \]
Verify: put the radius back into the original formula
Why: Two pi times 7 is about 43.98 inches, which rounds to the 44 inches given. Working forward through the original formula, rather than backward through the rearranged one, is what makes this a genuine check.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.4 Rewrite Formulas and Equations §1.4, pp. 26-26
Fill the middle
Guided Practice 5 and 6, from the triangle area formula.
Fill in the blanks
A = \tfrac\frac{2A}{h}___bh \;\Longrightarrow\; 2A = bh \;\Longrightarrow\; b = ___
Why: Multiplying both sides by 2 clears the fraction, giving 2A equal to bh. Then dividing by h isolates b as 2A over h. Note that the 2 ends up in the numerator, not the denominator — a half on one side becomes a doubling on the other, and reversing that is the usual slip here.
Worked example
Guided Practice 4. One division again, but now the divisor is a letter, which brings a condition with it.
\[ \text{Solve } A = lw \text{ for } w, \text{ then find } w \text{ when } A = 40 \text{ m}^2 \text{ and } l = 16 \text{ m}. \]
Divide each side by l
Why: l is a length, and a rectangle with zero length has no area to speak of, so l is not zero and the division is legal.
\[ \frac{A}{l} = w \]
State the condition
Why: Writing l not equal to zero is not pedantry when the divisor is a letter; it is the only place the restriction can be recorded.
\[ w = \frac{A}{l}, l\text{ not } 0 \]
Substitute 40 for A and 16 for l
Why: Square metres divided by metres gives metres, which is the right unit for a width.
\[ w = \frac{40}{16} \]
Simplify
Why: Forty sixteenths reduces to five halves.
\[ w = 2.5 \]
Figure (svg): The solution to Worked example solve the area formula for w shown as a ladder of expressions, one row per algebraic move
\[ w = \frac{A}{l} \qquad \text{and at } A = 40, \, l = 16, \; w = 2.5 \text{ m} \]
Verify: multiply the two dimensions
Why: Sixteen metres times 2.5 metres is 40 square metres, matching the given area exactly. The unit also confirms it: metres times metres gives square metres.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.4 Rewrite Formulas and Equations §1.4, pp. 27-27
Trap
\[ C = 2\pi r \quad \text{solve for } r \]
Divide both sides by 2
Why: Only the visible number is treated as the coefficient; pi is left behind because it looks like part of the variable.
\[ \tfrac{C}{2} = \pi r \quad \text{(pi is still attached to r)} \]
\[ r = \tfrac{C}{2} \quad \text{(wrong)} \]
\[ C = 2\pi r \quad \text{solve for } r \]
Divide both sides by the entire coefficient, 2 pi
Why: Everything multiplying r is part of the coefficient, whether it is a digit or a Greek letter.
\[ r = \frac{C}{2\pi} \]
At a circumference of 44 the wrong version gives 22 and the right one gives about 7 — a factor of pi apart, which is the signature of this error.
Elimination
The formula is the area of a circle. Solve it for r.
Eliminate the wrong options
Which is A = pi r squared, correctly solved for r?
Survives elimination: A
Why: Dividing both sides by pi gives A over pi equal to r squared, and taking the positive square root of both sides gives r. The positive root is the right one because a radius is a length, which is a place where the situation, not the algebra, chooses between the two roots.
Estimation
Guided Practice 1: a circle with a circumference of 25 feet.
Predict first
Roughly what is the radius?
Correct: About 4 feet.
\[ r = \frac{25}{2\pi} \approx 3.98 \text{ ft} \]
Why: Two pi is a bit more than 6, so the radius is 25 divided by a bit more than 6, which is just under 4. The exact value is about 3.98 feet. Getting this bracket first means a slipped keystroke on the calculator — dividing by pi instead of by two pi, say — cannot survive, because that would give almost 8.
Explain it
A classmate insists it is faster to substitute the number first every time.
Discussion prompt
They have to find the radius for six different circumferences. Explain, in three sentences, when their method is fine and when yours wins, and be fair about it.
Hint: Count the number of solves each approach performs.
Answer:
For a single circle they are right — substituting first and solving once is no more work, and arguably less. For six circles their method solves six equations while yours solves one and then evaluates six times.
The real argument is not speed but error rate: one rearrangement can be checked once and then trusted, whereas six separate solves each carry their own chance of a slip.
Section
Section 3
Concept
When the formula is a sum, the target variable is buried under both an addition and a multiplication. Undo them in reverse order: remove the added term first, then divide by the coefficient.
\[ P = 2l + 2w \;\Longrightarrow\; l = \frac{P - 2w}{2} \]
The reverse-order rule from Lesson 1.3 is the same rule here. The only change is that the numbers you are undoing are now letters.
Figure (svg): A table of eight standard formulas with the meaning of each variable listed beside it
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.4 Rewrite Formulas and Equations §1.4, pp. 27-27
Picture it
Every sum in this table needs two moves; every pure product needs one.
Figure (svg): A table of eight standard formulas with the meaning of each variable listed beside it
Perimeter, temperature and trapezoid area are the sums. The rest are products, and the trapezoid is a sum inside a product, which is why it is the hardest.
Worked example
Guided Practice 3, with the numerical part afterwards.
\[ \text{Solve } P = 2l + 2w \text{ for } l, \text{ then find } l \text{ when } P = 30 \text{ in and } w = 7 \text{ in.} \]
Subtract 2w from each side
Why: The 2w term is added to the term containing l, so it is the outermost thing wrapped around it.
\[ P - 2 w = 2 l \]
Divide each side by 2
Why: Two is the coefficient of l, and it comes off last.
\[ \frac{P - 2 w}{2} = l \]
Substitute 30 for P and 7 for w
Why: Thirty minus fourteen is sixteen.
\[ l = \frac{30 - 14}{2} \]
Simplify
Why: Sixteen halved is eight.
\[ l = 8 \]
Figure (svg): The solution to Worked example solve the perimeter formula for l shown as a ladder of expressions, one row per algebraic move
\[ l = \frac{P - 2w}{2} \qquad \text{and at } P = 30, \, w = 7, \; l = 8 \text{ in} \]
Verify: compute the perimeter from both dimensions
Why: Twice 8 plus twice 7 is 16 plus 14, which is 30 inches — the given perimeter. Working forward through the original formula is what makes this a check rather than a repetition.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.4 Rewrite Formulas and Equations §1.4, pp. 27-27
Ranking
Solving the trapezoid area formula for h.
Put in order
Why: The half comes off first because it is the outermost factor and clearing it early keeps the rest whole. The bracket is then a single quantity multiplying h, so dividing by all of it isolates h in one move — dividing by b sub one alone would be the classic error. The check is a real step because it is the only one that returns to the original formula.
Worked example
Guided Practice 7. Three moves, and the subscripts are labels rather than factors.
\[ \text{Solve } A = \tfrac{1}{2}(b_1 + b_2)h \text{ for } b_2. \]
Multiply each side by 2
Why: The half is the outermost factor, so clearing it first keeps everything else whole.
\[ 2 A = (b 1 + b 2) h \]
Divide each side by h
Why: The height multiplies the whole bracket, so dividing by it frees the bracket.
\[ 2 A / h = b 1 + b 2 \]
Subtract b sub one from each side
Why: Only now is the target variable exposed, and only a subtraction remains.
\[ 2 A / h - b 1 = b 2 \]
Write it with b sub two on the left
Why: The subscripts never entered the arithmetic; they are names, not numbers.
\[ b 2 = 2 A / h - b 1 \]
Figure (svg): The solution to Worked example the trapezoid, solved for one base shown as a ladder of expressions, one row per algebraic move
\[ b_2 = \frac{2A}{h} - b_1 \]
Verify: test with a trapezoid you can compute by hand
Why: Take b sub one equal to 4, b sub two equal to 6 and h equal to 5. The area is half of 10 times 5, which is 25. Feeding that back in: two times 25 over 5 is 10, minus 4 is 6 — recovering b sub two exactly.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.4 Rewrite Formulas and Equations §1.4, pp. 27-27
Error analysis
A student solves the perimeter formula for the width. The final line is wrong.
Annotate
On: \( P = 2l + 2w \;\Longrightarrow\; P - 2l = 2w \;\Longrightarrow\; w = P - l \)
When you divide a side that is written as a difference, every term in that difference gets divided. This is the same slip as the one in Lesson 1.2, wearing letters instead of numbers.
Comparison
Fill the blanks. The pattern in the right column is the whole point.
Comparison matrix
| Formula | Solve for | Rearranged |
|---|---|---|
| d = rt | t | t = d/r |
| A = lw | w | w = A/l |
| P = 2l + 2w | w | w = (P - 2l)/2 |
| F = (9/5)C + 32 | C | C = (5/9)(F - 32) |
| A = (1/2)bh | h | h = 2A/b |
Every product row has a single division. Every sum row subtracts first and divides second. Knowing which kind you are looking at before you start is most of the work.
Translation
Each rearranged formula is a set of instructions. Match them.
Match the pairs
Why: Reading a formula aloud as instructions is the fastest way to check that you rearranged it correctly, because a wrong rearrangement usually produces an instruction that sounds obviously off. Notice that in every sum case the subtraction is spoken first and the division second, which is the reverse-order rule in words.
Edge cases
The rearranged formulas carry conditions that the originals hid.
Discussion prompt
Look at t equals d over r and w equals A over l. What does each say when the denominator approaches zero, and what does that mean in the real situation each describes? Which of the two has a denominator that could realistically be near zero?
Hint: Think about a very slow journey and a very thin rectangle.
Answer:
\[ t = \frac{d}{r}, \; r \neq 0 \qquad w = \frac{A}{l}, \; l \neq 0 \]
As the rate approaches zero the time grows without bound, which is exactly right: at zero speed you never arrive. At r equal to zero the formula is undefined, matching the fact that the question has no answer.
A near-zero rate is entirely realistic — a stalled car. A near-zero length is not, for a rectangle you could point at. So the same algebraic condition is a live concern in one case and a formality in the other, which is why conditions are worth reading rather than reciting.
Section
Section 4
Concept
An equation relating x and y can be rewritten with y as the subject. Once it is, you can feed in any x and read off the y, which is exactly what a function does and exactly what graphing needs.
\[ 9x - 4y = 7 \;\Longrightarrow\; y = \tfrac{9}{4}x - \tfrac{7}{4} \]
This is the bridge to Chapter 2, where every line is written in this form so that its slope and intercept can be read straight off.
Figure (svg): A diagram contrasting an equation solved for r with the same equation not yet solved, because r still appears on both sides
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.4 Rewrite Formulas and Equations §1.4, pp. 28-28
Picture it
The same test applies here, and it is where students most often stop early.
Figure (svg): A diagram contrasting an equation solved for r with the same equation not yet solved, because r still appears on both sides
When you divide by a negative coefficient, every term on the other side changes sign. That is the step this section exists to drill.
Worked example
Example 3, both steps, with the negative coefficient handled explicitly.
\[ \text{Solve } 9x - 4y = 7 \text{ for } y, \text{ then find } y \text{ when } x = -5. \]
Subtract 9x from each side
Why: The 9x term does not contain y, so it belongs on the other side.
\[ -4 y = 7 - 9 x \]
Divide each side by -4
Why: Every term on the right is divided, and dividing by a negative flips the sign of each one.
\[ y = -\frac{7}{4} + (\frac{9}{4}) x \]
Substitute -5 for x
Why: The rearranged form takes any x directly.
\[ y = -\frac{7}{4} + (\frac{9}{4}) (-5) \]
Simplify
Why: Nine fourths of negative five is negative forty-five fourths, and negative seven fourths minus that is negative fifty-two fourths.
\[ y = -\frac{52}{4} = -13 \]
Figure (svg): The solution to Worked example solve a linear equation for y shown as a ladder of expressions, one row per algebraic move
\[ y = \tfrac{9}{4}x - \tfrac{7}{4} \qquad \text{and at } x = -5, \; y = -13 \]
Verify: substitute both values into the ORIGINAL equation
Why: Nine times negative five is negative forty-five, and minus four times negative thirteen is positive fifty-two. Negative forty-five plus fifty-two is seven, which matches the right side. Checking in the original rather than the rearranged form is what validates the division by negative four.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.4 Rewrite Formulas and Equations §1.4, pp. 28-28
Discrimination
Do not solve these. Say whether the final division changes signs on the other side.
Sort into buckets
Sort each equation by whether solving for y involves dividing by a negative.
Worked example
Guided Practice 10. Same procedure, and the sign flip does not happen.
\[ \text{Solve } 3x + 2y = 12 \text{ for } y, \text{ then find } y \text{ when } x = 2. \]
Subtract 3x from each side
Why: The x term crosses over; the y term stays.
\[ 2 y = 12 - 3 x \]
Divide each side by 2
Why: Both terms on the right are divided, and since 2 is positive no sign changes.
\[ y = 6 - (\frac{3}{2}) x \]
Substitute 2 for x
Why: Three halves of two is three.
\[ y = 6 - 3 \]
Simplify
Why: Six minus three is three.
\[ y = 3 \]
Figure (svg): The solution to Worked example a positive coefficient, for contrast shown as a ladder of expressions, one row per algebraic move
\[ y = 6 - \tfrac{3}{2}x \qquad \text{and at } x = 2, \; y = 3 \]
Verify: substitute both values into the original
Why: Three times 2 plus 2 times 3 is 6 plus 6, which is 12 — the right side exactly. Comparing with the previous example, the only structural difference is that the coefficient of y was positive here, so no term changed sign.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.4 Rewrite Formulas and Equations §1.4, pp. 28-28
Trap
\[ 9x - 4y = 7 \;\longrightarrow\; -4y = 7 - 9x \]
Divide by -4, flipping the sign of the first term only
Why: The division is applied term by term but the negative is applied only where it was noticed.
\[ y = -\tfrac{7}{4} - \tfrac{9}{4}x \quad \text{(wrong)} \]
At x equal to -5 this gives 38/4, which is 9.5 — nowhere near the correct -13.
\[ 9x - 4y = 7 \;\longrightarrow\; -4y = 7 - 9x \]
Divide EVERY term by -4
Why: Dividing by a negative changes the sign of every term on the other side, without exception.
\[ y = -\tfrac{7}{4} + \tfrac{9}{4}x \]
A one-value check settles it: at x equal to -5 this gives -13, and substituting -5 and -13 into the original gives 7 on both sides.
Fill the middle
Guided Practice 11. The middle line has been erased.
Fill in the blanks
2x + 5y = -1 \;\longrightarrow\; 5y = -1 - 2x \;\longrightarrow\; y = \tfrac______
Why: Subtracting 2x from each side leaves 5y on the left and negative one minus 2x on the right. Dividing by 5 then gives the final form. At x equal to 2 this gives negative five over five, which is -1, and checking in the original: 4 plus 5 times -1 is -1, matching the right side.
Anomaly
A student solves 5y minus x equals 13 for y and reports the value at x equal to 2.
Predict first
The student reports y equals 11. What went wrong?
Correct: They forgot to divide the 13 by 5.
\[ 5y - x = 13 \;\longrightarrow\; 5y = 13 + x \;\longrightarrow\; y = \tfrac{13 + x}{5} \]
\[ \text{at } x = 2: \quad y = \tfrac{15}{5} = 3 \]
Why: Adding x to both sides gives 5y equal to 13 plus x, and dividing by 5 gives y equal to 13 plus x, all over 5. At x equal to 2 that is 15 over 5, which is 3. Reporting 11 means the division was applied to the x term but not to the 13 — dividing 13 plus 2 by 5 as 13 plus two fifths would give 13.4, but taking 13 minus 2 gives 11, which is the likely route. Either way, the whole numerator must be divided.
Real world
You will spend all of Chapter 2 writing equations of lines.
Discussion prompt
The equation 3x + 2y = 12 describes a line. Rewrite it with y as the subject, then say what the two numbers in your answer tell you about the line without drawing anything.
Hint: One of them says where the line crosses the vertical axis; the other says how steep it is.
Answer:
\[ 3x + 2y = 12 \;\Longrightarrow\; y = 6 - \tfrac{3}{2}x \]
The 6 is the value of y when x is zero, so the line crosses the vertical axis at 6. The coefficient of negative three halves says y falls by three for every two that x rises — the line slopes downward.
Neither fact is visible in the original form, which is the practical reason Chapter 2 insists on solving for y before doing anything else.
Section
Section 5
Concept
If the target variable sits in more than one term, no amount of moving things across will isolate it. Pull it out as a common factor, and it becomes a single quantity multiplied by a bracket — which one division removes.
\[ 2y + xy = 6 \;\Longrightarrow\; (2 + x)y = 6 \;\Longrightarrow\; y = \frac{6}{2 + x} \]
This is the distributive property used backwards, which is exactly what combining like terms was in Lesson 1.2 — the same tool, a different job.
Figure (svg): The expression two y plus x y being rewritten as the quantity two plus x times y, showing the shared factor pulled out
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.4 Rewrite Formulas and Equations §1.4, pp. 28-28
Picture it
The whole move, in three lines.
Figure (svg): The expression two y plus x y being rewritten as the quantity two plus x times y, showing the shared factor pulled out
Dividing by the bracket requires the bracket to be nonzero, so the answer carries the condition that x is not negative two. Reading that condition off the denominator is part of the answer.
Worked example
Example 4, including the value at a specific x.
\[ \text{Solve } 2y + xy = 6 \text{ for } y, \text{ then find } y \text{ when } x = -3. \]
Notice that y is a factor of both terms on the left
Why: Two y and x y both contain y, so it is a common factor and nothing else will isolate it.
\[ 2 y + x y = 6 \]
Factor y out using the distributive property
Why: Two plus x is what is left behind once the y is removed.
\[ (2 + x) y = 6 \]
Divide each side by the whole bracket
Why: The bracket is a single quantity multiplying y, so dividing by all of it leaves y alone — and requires the bracket not to be zero.
\[ y = \frac{6}{2 + x}, x\text{ not } -2 \]
Substitute -3 for x
Why: Two plus negative three is negative one.
\[ y = \frac{6}{-1} \]
Simplify
Why: Six divided by negative one is negative six.
\[ y = -6 \]
Figure (svg): The solution to Worked example the variable in two terms shown as a ladder of expressions, one row per algebraic move
\[ y = \frac{6}{2 + x} \qquad \text{and at } x = -3, \; y = -6 \]
Verify: substitute both values into the original
Why: Two times negative six is negative twelve, and negative three times negative six is positive eighteen. Negative twelve plus eighteen is six, matching the right side. The condition also checks out: x equal to -3 is not the forbidden -2.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.4 Rewrite Formulas and Equations §1.4, pp. 28-28
Definition probe
The target variable is y in every case. Count the terms that contain it.
Sort into buckets
Sort each equation by the move it needs.
Worked example
Guided Practice 13. Same move, but the factoring leaves a difference in the bracket.
\[ \text{Solve } 4y - xy = 28 \text{ for } y, \text{ then find } y \text{ when } x = 2. \]
Check that y is a factor of both terms
Why: Four y and negative x y both contain y.
\[ 4 y - x y = 28 \]
Factor y out, keeping the minus sign inside the bracket
Why: Removing y from negative x y leaves negative x, so the bracket is four minus x.
\[ (4 - x) y = 28 \]
Divide each side by the bracket
Why: The condition is that four minus x is not zero, so x is not four.
\[ y = \frac{28}{4 - x}, x\text{ not } 4 \]
Substitute 2 for x
Why: Four minus two is two.
\[ y = \frac{28}{2} \]
Simplify
Why: Twenty-eight halved is fourteen.
\[ y = 14 \]
Figure (svg): The solution to Worked example a subtraction between the two terms shown as a ladder of expressions, one row per algebraic move
\[ y = \frac{28}{4 - x} \qquad \text{and at } x = 2, \; y = 14 \]
Verify: substitute both values into the original
Why: Four times fourteen is fifty-six, and two times fourteen is twenty-eight. Fifty-six minus twenty-eight is twenty-eight, matching the right side exactly.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.4 Rewrite Formulas and Equations §1.4, pp. 28-28
Error analysis
A student attempts Guided Practice 12 and produces something that is true but useless.
Annotate
On: \( 3 = 2xy - x \;\Longrightarrow\; 3 + x = 2xy \;\Longrightarrow\; y = \frac{3 + x}{2x} \quad \text{(this one is right)} \)
Count first, then choose the tool. Factoring is for two or more terms containing the target; a single term needs nothing more than Lesson 1.3.
Reverse engineer
Here is a solved form. Something was rearranged to produce it.
Fill in the blanks
y = \frac4 - x___}
Why: Factoring y out of 4y minus xy leaves the bracket four minus x, so the denominator of the solved form is exactly what the bracket contained. Reading a solved form backwards like this is the fastest way to check a factoring step: multiply the denominator by y and you should recover the original left side.
Counterexample
A classmate proposes a faster route.
\[ 2y + xy = 6 \;\Longrightarrow\; y = \frac{6}{2} + \frac{6}{x} \]
Discussion prompt
They claim you can divide 6 by each part of the left side separately. Find a value of x that shows this fails, and then say in one sentence what rule they have confused with what.
Hint: Use the value from the worked example, where you already know the right answer.
Answer:
\[ \text{at } x = -3: \quad \frac{6}{2} + \frac{6}{-3} = 3 - 2 = 1 \quad \text{but the true value is } -6 \]
They have confused splitting a numerator over a common denominator, which is legal, with splitting a denominator, which is not. A sum in the denominator has to be evaluated as one quantity before the division happens.
Commit first
Answer, then rate your confidence.
Predict first
After solving 2y + xy = 6 for y, what value of x must be excluded, and why?
Correct: x = -2, because it makes the denominator 2 plus x equal to zero.
\[ x = -2: \quad 2y + (-2)y = 0 \neq 6 \]
Why: The solved form divides by two plus x, so that quantity must not be zero, which excludes x equal to negative two. Checking against the original equation confirms this is not an artefact of the rearrangement: at x equal to -2 the original becomes 2y minus 2y equals 6, which is 0 equals 6 — false for every y, so there genuinely is no solution there. The condition on the rearranged formula is telling you something real about the original.
Comparison
Fill the blanks. The count of terms containing the target decides everything.
Comparison matrix
| Situation | Diagnosis | The move |
|---|---|---|
| Target is one factor of a product | one term | divide by everything else |
| Target is in a term added to another | one term | subtract, then divide |
| Target is in two terms | two terms | factor it out, divide by the bracket |
| Target is inside a bracket times something | one term | clear the outer factors, then subtract |
| Target has a negative coefficient | one term | divide by the negative; every sign flips |
Only one row calls for factoring, and it is the row you can identify before writing anything down.
Pattern
One routine rearranges any formula or equation in this chapter.
Step five is not decoration. A rearrangement that divides by a letter has changed what the equation is allowed to say, and the condition is where that change gets recorded.
OpenStax Algebra and Trigonometry 2e, §2.2 Linear Equations in One Variable §2.2
Check
A one-step rearrangement. Divide by the whole coefficient.
Check your understanding
Solve A = pi r squared for r.
Answer: A
Why: Dividing both sides by pi gives r squared equal to A over pi, and taking the positive square root of both sides gives r. The positive root is chosen because a radius is a length.
Check
Solving for y with a negative coefficient. Watch every sign.
Check your understanding
Solve 9x - 4y = 7 for y, then find y when x = -5.
Answer: A
Why: Subtracting 9x gives -4y = 7 - 9x, and dividing by -4 gives y = (9/4)x - 7/4. At x = -5 that is -45/4 - 7/4, which is -52/4, or -13.
Check
The variable appears twice. Count the terms first.
Check your understanding
Solve 4y - xy = 28 for y, then find y when x = 2.
Answer: A
Why: Both terms contain y, so factor: (4 - x)y = 28, giving y = 28/(4 - x). At x = 2 the denominator is 2 and y is 14. Checking: 56 minus 28 is 28.
Real world
A phone plan's total cost is C equals f plus rm, where f is a fixed monthly fee, r is the rate per minute, and m is minutes used.
Discussion prompt
Solve this for m, say what condition your answer carries, and then explain what that condition means about the plan in plain English. Then solve the same formula for r and say which of the two rearrangements you would actually want if you were comparing plans.
Hint: One of the two rearrangements answers how many minutes can I afford, and the other answers what am I really paying per minute.
Answer:
\[ C = f + rm \;\Longrightarrow\; m = \frac{C - f}{r}, \quad r \neq 0 \]
\[ C = f + rm \;\Longrightarrow\; r = \frac{C - f}{m}, \quad m \neq 0 \]
The condition on the first says the per-minute rate cannot be zero — and if it were, the question how many minutes can I afford would have no answer, because every number of minutes would fit within any budget.
For comparing plans you want the second: given a bill and the minutes used, it recovers the effective rate, which is the number the advertising does not print.
Commit first
Answer, then rate your confidence.
Predict first
You solve ab = c for b and get b equals c over a. Does the same reasoning let you solve a + b = c for b as b equals c over a?
Correct: No — a sum needs subtraction, so b equals c minus a.
\[ ab = c \;\Longrightarrow\; b = \tfrac{c}{a} \qquad a + b = c \;\Longrightarrow\; b = c - a \]
Why: Division undoes multiplication and subtraction undoes addition; they are not interchangeable. In ab equals c the a is a factor of b, so dividing removes it. In a plus b equals c the a is added to b, so subtracting removes it. Reading which operation joins the target to its neighbours, before choosing a move, is the whole diagnostic step of this lesson.
Explain it
They can solve for x when x appears once but freeze when it appears twice.
Discussion prompt
In four sentences or fewer, explain what to do when the variable you want appears in two terms, why moving things across cannot work, and how they would check their answer without an answer key.
Hint: The reason moving cannot work is worth one whole sentence.
Answer:
Moving terms across only relocates the variable, so with two copies you can never get both onto the same side and out of the way at once. Factoring pulls both copies into one, which is why it is the only route.
Once factored, one division by the whole bracket finishes it. They check by putting their answer back into the original equation and confirming both sides give the same number — and by noticing what value of the other letter would make the denominator zero.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For a fractional coefficient, clear it first by multiplying both sides. For a negative coefficient, write the division out term by term so no sign is missed. For the two-term case, count the terms containing the target before choosing a move. For conditions, read the denominator of your answer and say what would make it zero. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Across the top of a page write the eight formulas from this lesson and mark each one P for pure product or S for sum. Down the left side write the four routes: divide once, subtract then divide, clear the outer factor then subtract, and factor out then divide by the bracket. Beside each route write one formula from the top that needs it. In the middle of the page take the trapezoid area formula and solve it three times, once for h, once for b sub one and once for b sub two, showing every line. At the bottom write one equation in which y appears in two terms, solve it for y, and circle the value of x that your answer forbids, writing beside it what goes wrong in the ORIGINAL equation at that value.
If the circled value does not break the original equation too, the condition was an artefact of your algebra rather than a fact about the problem, and something went wrong. Go back to Section 5.
Recap
Five things, and the third is the one that makes Chapter 2 possible.
| If the target variable is | The move is |
|---|---|
| One factor of a product | Divide by everything else |
| In a term added to others | Subtract first, then divide |
| Inside a bracket times something | Clear the outer factors first |
| Carrying a negative coefficient | Divide, and flip every sign |
| In two or more terms | Factor it out, then divide by the bracket |
Lesson 1.5 stops handing you the equation altogether: you build the model from a described situation, using the formulas you have just learned to rearrange.
McDougal Littell Algebra 2 (Texas Edition), Ch. 1 Equations and Inequalities — Lesson 1.4 Rewrite Formulas and Equations §1.4, pp. 26-31 — everything on these slides traces back here
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