Exact and approximate solutions: solving to an exact value and rounding only at the end, using the approximately-equal symbol honestly, checking a rounded answer and knowing how close to expect, clearing decimals by multiplying through, and rounding error in real situations.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 3 — Solving Linear Equations
Solving Decimal Equations
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 3.6 Solving Decimal Equations §3.6, pp. 163-168 — the lesson these objectives are drawn from
Warm-up
Every equation in this lesson is one you can already solve. What is new is what to do with the answer.
Discussion prompt
Solve 3x equals 10 exactly, then write the answer as a decimal. Are the two forms the same number, and is one of them more useful?
Hint: Try writing the decimal down completely.
Answer:
\[ 3x = 10 \;\Longrightarrow\; x = \tfrac{10}{3} = 3.3333\ldots \]
They are the same number, and the decimal cannot be written down completely — it repeats forever. So any decimal you actually write is a rounded version, slightly different from the exact answer. This lesson is about handling that gap honestly.
Concept
Carry the exact value through every step and round only the final answer. Rounding early introduces an error that every later step then multiplies.
rounding error — The difference between an exact value and a rounded one, which can accumulate when a rounded value is used in further calculation.
When you round, say so with the approximately-equal symbol rather than an equal sign.
Figure (svg): A chain showing the solving carried out exactly and only the final answer rounded
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.6 Solving Decimal Equations §3.6, pp. 163-163
Section
Section 1
Concept
Dividing two whole numbers gives an exact fraction. Its decimal form may terminate, in which case the decimal is exact too, or repeat forever, in which case any written decimal is a rounding.
Exact answers are not always practical; sometimes a rounded answer makes more sense.
Figure (svg): Two columns contrasting an exact fractional answer with a rounded decimal one
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.6 Solving Decimal Equations §3.6, pp. 163-163 — the opening paragraph on exact and practical answers
Picture it
One is precise; the other is usable.
Figure (svg): Two columns contrasting an exact fractional answer with a rounded decimal one
Neither form is better in general. A fraction is right when precision matters and a decimal is right when the answer has to be acted on.
Worked example
This is Example 1 from the textbook. Solve exactly, then round to the nearest hundredth.
\[ \text{Solve } \; 38x - 39 = 118 \; \text{ and round to the nearest hundredth.} \]
Add 39 to each side
Why: The constant comes off first, as always.
\[ 38 x = 157 \]
Divide each side by 38
Why: This is the exact answer: one hundred and fifty-seven over thirty-eight.
\[ x = \frac{157}{38} \]
Convert to a decimal
Why: A calculator gives 4.131578947 and continuing.
\[ 4.131578... \]
Round to the nearest hundredth
Why: The third decimal place is one, which is below five, so the second place stays as it is.
\[ x \approx 4.13 \]
Figure (svg): A chain showing the solving carried out exactly and only the final answer rounded
\[ 38x - 39 = 118 \;\Longrightarrow\; x = \tfrac{157}{38} \approx 4.13 \]
Verify: substitute the rounded answer
Why: Thirty-eight times 4.13 is 156.94, minus thirty-nine is 117.94 — close to 118 but not equal, because the answer was rounded. Closeness rather than equality is exactly what a rounded answer should produce.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.6 Solving Decimal Equations §3.6, pp. 163-163
Sorting
A decimal is exact only when it terminates.
Sort into buckets
Sort each value by whether it is an exact form of its fraction.
Three of the six lost nothing and three lost a little. Knowing which case you are in tells you whether to write an equal sign or the wavy one.
Worked example
Guided Practice 1 to 3. Solve each exactly before rounding.
\[ \text{Solve to the nearest hundredth: } \; 24x + 43 = 66, \quad 42x - 28 = 87, \quad 22x = 39x - 19. \]
Solve the first exactly
Why: Subtract forty-three to get 24x equals 23, then divide.
\[ x = \frac{23}{24} \]
Round the first
Why: Twenty-three over twenty-four is 0.9583 and continuing.
\[ x \approx 0.96 \]
Solve the second exactly
Why: Add twenty-eight to get 42x equals 115, then divide.
\[ x = \frac{115}{42} \approx 2.74 \]
Solve the third exactly
Why: Subtract 39x from both sides to get negative 17x equals negative 19, then divide.
\[ x = \frac{19}{17} \approx 1.12 \]
Figure (svg): The solution to Worked example three from guided practice shown as a ladder of expressions, one row per algebraic move
\[ x \approx 0.96, \quad x \approx 2.74, \quad x \approx 1.12 \]
Verify: check that each rounded answer is close
Why: Substituting 0.96 into the first gives 66.04 against 66; substituting 2.74 into the second gives 87.08 against 87. Each is off by less than a tenth, which is what rounding to two decimal places should cost.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.6 Solving Decimal Equations §3.6, pp. 163-163
Trap
\[ 38x = 157 \;\rightarrow\; x \approx 4.1 \;\rightarrow\; \text{use } 4.1 \text{ in the next step} \]
Round as soon as a decimal appears, to keep the numbers manageable
Why: Long decimals are awkward to write, so trimming them feels like tidying.
Every later step then works with a value that is already slightly wrong, and the error grows rather than staying put.
\[ x = \tfrac{157}{38} \approx 4.13 \]
Keep the exact fraction until the final line, then round once
Why: One rounding introduces one error; several roundings compound.
A calculator holds far more digits than you need. Let it carry the exact value and round only what you write down as the answer.
Faded example
Give the exact value first and the rounded one second.
Fill in the blanks
38x = 157 \;\rightarrow\; x = \tfrac384.13} \approx ___
Why: Dividing by thirty-eight gives the exact fraction, and converting and rounding gives 4.13. Writing the exact value first means the rounding happens once, on a value that is still correct, rather than on a value already carrying an error.
Prediction
Rounding early is not merely untidy; it changes the answer.
Predict first
Solving 38x equals 157 by first rounding 157 over 38 to 4.1 and then reporting to two decimal places gives what?
Correct: 4.10, which is off by 0.03.
\[ \tfrac{157}{38} = 4.1315\ldots \approx 4.13 \quad \text{but} \quad 4.1 \approx 4.10 \]
Why: Rounding to one decimal place first discards information that the second rounding cannot recover, so the final answer is 4.10 rather than 4.13. The error is small here, but in a longer calculation each early rounding compounds — which is why the exact value is carried to the last line.
Socratic
Every equation in this lesson has an exact answer, even the messy ones.
Discussion prompt
Explain why solving a linear equation with whole-number coefficients always gives an exact answer that can be written as a fraction. Then say why a decimal answer is still often preferred.
Hint: Look at what the final step of solving always is.
Answer:
The last step is always a division of one number by another, and a quotient of two whole numbers is by definition a fraction. So the exact answer exists and can always be written down, however awkward its decimal expansion turns out to be.
A decimal is preferred when the answer has to be compared, measured or paid, because those activities work in decimals. Four point one three tells you immediately that the answer is a little over four; one hundred and fifty-seven over thirty-eight does not, without a division you have to perform yourself.
Section
Section 2
Concept
An equal sign claims exact equality. When a value has been rounded, the wavy approximately-equal symbol should be used instead, so that a reader knows the number is not exact.
\[ x = \tfrac{157}{38} \qquad x \approx 4.13 \]
Using an equal sign after rounding claims more precision than you have, which is a real error rather than a matter of style.
Figure (svg): The approximately-equal symbol contrasted with the equal sign
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.6 Solving Decimal Equations §3.6, pp. 163-163 — the note on using the approximately-equal symbol
Picture it
One asserts equality and the other asserts closeness.
Figure (svg): The approximately-equal symbol contrasted with the equal sign
Writing x equals 4.13 would be false, since 4.13 is not one hundred and fifty-seven over thirty-eight. The wavy symbol makes the statement true.
Worked example
One solution, several lines, and the symbol changes exactly once.
\[ \text{Write the solution of } 38x - 39 = 118 \text{ with the right symbol on every line.} \]
The first two lines are exact
Why: Adding thirty-nine and dividing by thirty-eight lose nothing, so equal signs are correct.
\[ 38 x = 157, x = \frac{157}{38} \]
The conversion to a decimal is where it changes
Why: The decimal expansion continues forever, so any written version is approximate.
\[ x \approx 4.131578... \]
The rounded answer is also approximate
Why: Rounding loses a little more.
\[ x \approx 4.13 \]
State the rule
Why: Every line before the rounding takes an equal sign; every line from the rounding onwards takes the wavy one.
Figure (svg): The solution to Worked example which symbol where shown as a ladder of expressions, one row per algebraic move
\[ 38x = 157, \; x = \tfrac{157}{38} \approx 4.13 \]
Verify: check that no line claims false equality
Why: The line x equals 157 over 38 is exactly true, and the line x approximately equals 4.13 is honestly stated. Had the last line used an equal sign it would have claimed that 4.13 and the fraction are the same number, which they are not.
Discrimination
Ask whether anything was discarded.
Sort into buckets
Sort each statement by which symbol it needs.
Worked example
Not every decimal answer is approximate.
\[ \text{Solve } \; 4x = 3 \; \text{ and decide which symbol the decimal answer needs.} \]
Divide both sides by 4
Why: The exact answer is three quarters.
\[ x = \frac{3}{4} \]
Convert to a decimal
Why: Three quarters is exactly 0.75, with nothing left over.
\[ 0.75 \]
Choose the symbol
Why: Nothing was rounded, so an equal sign is correct.
\[ x = 0.75 \]
Contrast with the earlier case
Why: One hundred and fifty-seven over thirty-eight does not terminate, so it needs the wavy symbol.
Figure (svg): The solution to Worked example a terminating decimal needs no wavy symbol shown as a ladder of expressions, one row per algebraic move
\[ 4x = 3 \;\Longrightarrow\; x = \tfrac{3}{4} = 0.75 \]
Verify: multiply the decimal back
Why: Four times 0.75 is exactly three, with no discrepancy at all. A terminating decimal reproduces the original exactly, which is precisely what distinguishes it from a rounded one.
Error analysis
The student solved and rounded three equations. Two of the symbol choices are wrong.
Annotate
On: \( x = \tfrac{157}{38} = 4.13 \qquad x = \tfrac{3}{4} = 0.75 \qquad x = \tfrac{1}{3} = 0.33 \)
The test is simple: did anything get discarded? If yes, the symbol changes; if no, it does not. Two of these three lines discarded something.
Elimination
The exact solution is 157 over 38.
Eliminate the wrong options
Which written answer is correct?
Survives elimination: A
Why: The wavy symbol records that the decimal is a rounding of the exact fraction. Option C is worth noticing as the opposite error: using an approximation symbol on an exact value understates your precision, which is less common but equally inaccurate.
Faded example
One line is exact and one is not.
Fill in the blanks
x = \tfrac@approx___ \qquad x ___ 4.13
Why: The fraction is exactly what the division produced, so an equal sign is correct there. The decimal discards digits, so the second line needs the approximately-equal symbol. The symbol changes exactly once in any solution, at the moment the rounding happens.
Socratic
Everyone knows 4.13 is rounded. Writing the symbol still matters.
Discussion prompt
Give one situation where using an equal sign after rounding would mislead a reader in a way that costs something real. Then say what the symbol tells a reader that the digits alone do not.
Hint: Think about someone using your answer in a further calculation.
Answer:
If someone takes your 4.13 and multiplies it by a thousand, they get 4130 when the true value is 4131.6 — an error of nearly two units. Had the symbol told them the value was approximate, they would have known to go back to the exact fraction before scaling it up.
The symbol tells a reader that further precision is available if they need it, and that the digits shown are not the whole story. Digits alone cannot convey that: 4.13 looks exactly as definite as 4.25, and only one of those is exact.
Section
Section 3
Concept
When a rounded answer is substituted back, the two sides will usually not be exactly equal. They should be close, and how close depends on how much was rounded away.
A gap far larger than the rounding could explain is a sign of a real error rather than of the rounding.
Figure (svg): A check with a rounded answer, where the two sides come out close but not identical
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.6 Solving Decimal Equations §3.6, pp. 163-163 — the CHECK following Example 1 and the note on approximate equality
Picture it
117.94 against 118, which the rounding explains.
Figure (svg): A check with a rounded answer, where the two sides come out close but not identical
The gap of 0.06 is about thirty-eight times the 0.0016 that was rounded away, which is exactly what multiplying by the coefficient of thirty-eight should do to it.
Worked example
The CHECK from Example 1, with the size of the gap explained.
\[ \text{Check } x \approx 4.13 \text{ in } \; 38x - 39 = 118. \]
Substitute the rounded value
Why: Thirty-eight times 4.13.
\[ 38(4.13) - 39 \]
Evaluate the left side
Why: 156.94 minus thirty-nine is 117.94.
\[ 117.94 \]
Compare with the right side
Why: 117.94 against 118, a gap of 0.06.
Judge whether the gap is explained
Why: The rounding discarded about 0.0016, and thirty-eight times that is about 0.06 — exactly the gap observed.
Figure (svg): A check with a rounded answer, where the two sides come out close but not identical
\[ 38(4.13) - 39 = 117.94 \approx 118 \]
Verify: check with more decimal places
Why: Using 4.1316 instead gives 157.0008 minus 39, which is 118.0008 — much closer. More decimal places give a closer check, which confirms that the gap came from the rounding rather than from a mistake in the solving.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.6 Solving Decimal Equations §3.6, pp. 163-163
Sorting
For an equation with a coefficient of about 38, judge each gap.
Sort into buckets
Sort each check result by whether the rounding explains it.
The dividing line is roughly the coefficient times the rounding precision. Knowing that number in advance turns closeness from a judgement call into a calculation.
Worked example
Knowing how big a gap to expect is what makes the check useful.
\[ \text{A student solves } 38x - 39 = 118 \text{ and gets } x \approx 4.31. \text{ Check it.} \]
Substitute 4.31
Why: Thirty-eight times 4.31 is 163.78.
\[ 163.78 - 39 \]
Evaluate
Why: One hundred and twenty-four point seven eight.
\[ 124.78 \]
Compare
Why: 124.78 against 118, a gap of nearly seven.
Judge
Why: Rounding to two decimal places can shift the left side by at most about 0.19, so a gap of seven is far too large. The digits were transposed.
Figure (svg): The solution to Worked example a gap too large to be rounding shown as a ladder of expressions, one row per algebraic move
\[ 38(4.31) - 39 = 124.78 \;\text{, far from } 118 \]
Verify: compare with the correct answer's check
Why: The correct answer of 4.13 gave a gap of 0.06 and this one gives a gap of 6.78 — more than a hundred times larger. Knowing roughly how big a rounding gap should be is what turns closeness into a real test rather than a vague impression.
Trap
\[ 38(4.13) - 39 = 117.94 \neq 118 \]
Conclude that 4.13 is wrong because the two sides differ
Why: Every check so far has produced exact equality, so a gap looks like a failure.
The answer was rounded, so a small gap is expected. Rejecting it would mean rejecting every rounded answer ever produced.
\[ 38(4.13) - 39 = 117.94 \approx 118 \;\checkmark \]
Judge the size of the gap against how much was rounded away
Why: A gap of about the coefficient times the rounding is exactly what should happen.
Exact answers give exact checks; rounded answers give close ones. Knowing which kind of answer you have is what tells you which kind of check to expect.
Estimation
The expected gap can be predicted before the check is done.
Predict first
An answer is rounded to the nearest hundredth and then multiplied by a coefficient of 20. Roughly how far from exact should the check come out?
Correct: Within about 0.1.
\[ 0.005 \times 20 = 0.1 \]
Why: Rounding to the nearest hundredth changes the value by at most 0.005, and multiplying by twenty multiplies that error by twenty, giving at most about 0.1. Predicting this before checking is what lets you tell a rounding gap from a mistake.
Faded example
Substitute the rounded answer and compare.
Fill in the blanks
38(4.13) - 39 = 156.94 - 39 = 117.94 \approx 118
Why: The left side comes to 117.94, which is close to but not equal to 118. The gap of 0.06 is what rounding to two decimal places costs once it is multiplied by a coefficient of thirty-eight, so the check passes.
Socratic
The same rounding produces different gaps in different equations.
Discussion prompt
Explain why rounding the answer by the same amount produces a larger gap in an equation with a bigger coefficient. Then say what that means for how precisely you should round in such an equation.
Hint: Ask what the coefficient does to the error.
Answer:
The rounded answer differs from the exact one by a small amount, and the equation then multiplies that difference by the coefficient. A coefficient of thirty-eight magnifies the rounding error thirty-eight times, while a coefficient of two would only double it.
So an equation with a large coefficient needs more decimal places to reach the same accuracy in the check. If you need the two sides within 0.01 and the coefficient is a hundred, you need the answer to about four decimal places. Deciding the precision from the requirement rather than from habit is what this reasoning is for.
Section
Section 4
Concept
An equation whose coefficients are decimals is solved by exactly the four steps from Lesson 3.5. Alternatively, multiplying every term by a power of ten clears the decimals and turns it into a whole-number equation.
\[ 3.5x - 37.9 = 0.2x \;\rightarrow\; 35x - 379 = 2x \]
The power of ten is chosen from the largest number of decimal places present: one place needs ten, two places need a hundred.
Figure (svg): An equation with decimals multiplied through by 10 to clear them
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.6 Solving Decimal Equations §3.6, pp. 164-164 — Example 2, Solve an Equation that Contains Decimals
Picture it
Every term multiplied by ten, and the decimals vanish.
Figure (svg): An equation with decimals multiplied through by 10 to clear them
The move is optional. It costs one line and buys whole-number arithmetic for the rest of the solution, which is usually worth it.
Worked example
This is Example 2 from the textbook. Round to the nearest tenth.
\[ \text{Solve } \; 3.5x - 37.9 = 0.2x \; \text{ to the nearest tenth.} \]
Collect the variable terms
Why: Three point five is greater than nought point two, so subtract 0.2x from both sides.
\[ 3.3 x - 37.9 = 0 \]
Isolate the variable term
Why: Add 37.9 to each side.
\[ 3.3 x = 37.9 \]
Divide
Why: 37.9 over 3.3, which a calculator gives as 11.4848 and continuing.
\[ 11.4848... \]
Round to the nearest tenth
Why: The second decimal place is eight, which rounds the first place up.
\[ x \approx 11.5 \]
Figure (svg): The solution to Worked example solve with decimals directly shown as a ladder of expressions, one row per algebraic move
\[ 3.5x - 37.9 = 0.2x \;\Longrightarrow\; x \approx 11.5 \]
Verify: check the rounded answer on both sides
Why: The left side is 3.5 times 11.5 minus 37.9, which is 40.25 minus 37.9, or 2.35. The right side is 0.2 times 11.5, which is 2.3. The two are close, and the gap of 0.05 is what rounding to one decimal place costs here.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.6 Solving Decimal Equations §3.6, pp. 164-164
Matching
The power of ten comes from the largest number of decimal places.
Match the pairs
Why: The multiplier is ten raised to the largest number of decimal places anywhere in the equation. One place needs ten, two places need a hundred, three need a thousand. Using a smaller power leaves some decimals behind; using a larger one is harmless but makes the numbers unnecessarily big.
Worked example
The same equation solved the other way, to compare the two routes.
\[ \text{Solve } \; 3.5x - 37.9 = 0.2x \; \text{ by clearing the decimals.} \]
Count the decimal places
Why: Every number has one decimal place, so one factor of ten clears them all.
\[ \text{multiply by } 10 \]
Multiply every term by 10
Why: Thirty-five x minus 379 equals 2x.
\[ 35 x - 379 = 2 x \]
Collect and isolate
Why: Subtract 2x to get 33x minus 379 equals zero, then add 379.
\[ 33 x = 379 \]
Divide and round
Why: 379 over 33 is 11.4848 and continuing, the same value as before.
\[ x \approx 11.5 \]
Figure (svg): An equation with decimals multiplied through by 10 to clear them
\[ 35x - 379 = 2x \;\Longrightarrow\; x = \tfrac{379}{33} \approx 11.5 \]
Verify: compare the two routes' exact answers
Why: The first route gave 37.9 over 3.3 and the second gave 379 over 33, which are the same fraction with both parts multiplied by ten. Clearing the decimals did not change the answer, only the appearance of the arithmetic along the way.
Trap
\[ 3.5x - 37.9 = 0.2x \]
Multiply the decimal coefficients by 10 and leave the constant alone
Why: The coefficients are what look untidy, so they attract the correction.
\[ 35x - 37.9 = 2x \quad \text{(wrong)} \]
Multiplying both sides means multiplying every term on both sides. Leaving the constant untouched changes what the equation says.
\[ 10(3.5x) - 10(37.9) = 10(0.2x) \;\Longrightarrow\; 35x - 379 = 2x \]
Multiply every term on both sides, exactly as distributing requires
Why: The move is multiplying each whole side by ten, and a side is multiplied term by term.
A quick check: the answer should be unchanged by clearing. If it is not, some term was missed.
Faded example
Multiply every term by the right power of ten.
Fill in the blanks
3.5x - 37.9 = 0.2x \;\rightarrow\; 35x - 379 = 2x
Why: Multiplying every term by ten gives thirty-five x minus 379 equals two x. Every term must be multiplied, including the constant — leaving it as 37.9 would change what the equation claims.
Elimination
The equation is 0.4x plus 1.2 equals 2.8.
Eliminate the wrong options
Which version has the decimals correctly cleared?
Survives elimination: A
Why: Every term is multiplied by ten, giving whole numbers throughout and the same solution of four. Option D is worth noticing: it is a legal transformation that reaches the same answer, so the error there is one of efficiency rather than correctness.
Socratic
Both routes reach the same answer, so the choice is about effort.
Discussion prompt
Give one situation where clearing the decimals is clearly worth the extra line, and one where it is not. Use the number of decimal places in your reasoning.
Hint: Think about how large the cleared numbers become.
Answer:
It is worth doing when the decimals have one or two places and the resulting whole numbers stay small — 3.5x minus 37.9 becomes 35x minus 379, which is easy arithmetic. Every step after that is whole-number work with no place-value slips available.
It is not worth doing when the decimals have many places, since the multiplier becomes large and the numbers unwieldy: an equation with 0.0625 in it would need multiplying by ten thousand. In that case solving with the decimals directly, and letting a calculator carry them, is much less work.
Section
Section 5
Concept
Using a rounded solution in a real situation can produce a small discrepancy called rounding error. It is usually harmless and occasionally matters, and knowing which is part of answering the question.
Figure (svg): Three shares of a pizza bill, each rounded up, totalling one cent more than the bill
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.6 Solving Decimal Equations §3.6, pp. 164-164 — the Rounding Error paragraph and Example 3
Picture it
Each share rounded to the nearest cent, and the total overshoots.
Figure (svg): Three shares of a pizza bill, each rounded up, totalling one cent more than the bill
Three lots of 4.30 is 12.90 against a bill of 12.89. The extra penny came from rounding three times, and somebody has to decide who pays it.
Worked example
This is Example 3 from the textbook. The pizza costs 12.89 dollars.
\[ \text{Solve } \; 3x = 12.89 \; \text{ for each person's share, to the nearest cent.} \]
Divide both sides by 3
Why: The exact answer is 4.29666 continuing, a repeating decimal.
\[ x = 4.29666... \]
Round to the nearest cent
Why: The third decimal place is six, which rounds the second place up.
\[ x \approx 4.30 \]
State the share
Why: Each person pays four dollars thirty.
\[ 4.30\text{ dollars} \]
Notice the rounding error
Why: Three times 4.30 is 12.90, which is one cent more than the bill.
Figure (svg): Three shares of a pizza bill, each rounded up, totalling one cent more than the bill
\[ 3x = 12.89 \;\Longrightarrow\; x \approx 4.30 \text{ each} \]
Verify: compute the discrepancy exactly
Why: The exact share is 4.29666 and the rounded one is 4.30, a difference of about a third of a cent. Three of those make a whole cent, which is exactly the overshoot observed. The error is fully explained by the rounding.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.6 Solving Decimal Equations §3.6, pp. 164-164
Sorting
The quantity decides, not the calculator.
Sort into buckets
Sort each quantity by how it should be rounded.
Three different treatments, none of them decided by the arithmetic. The calculator will always offer more digits than the situation can use.
Worked example
The question does not always state the precision, and then the quantity decides.
\[ \text{A } 7 \text{ metre plank is cut into } 3 \text{ equal pieces. How long is each?} \]
Solve the equation
Why: Three L equals seven, so L is seven thirds, or 2.3333 continuing.
\[ L = \frac{7}{3} \]
Ask what precision the situation supports
Why: A tape measure reads to about a millimetre, so three decimal places in metres is the most that is meaningful.
Round accordingly
Why: Two point three three three metres, or 2333 millimetres.
\[ 2.333 m \]
Check the total
Why: Three times 2.333 is 6.999 metres, one millimetre short of seven.
\[ 1 \text{mm}\text{ short} \]
Figure (svg): The solution to Worked example choosing the precision shown as a ladder of expressions, one row per algebraic move
\[ L = \tfrac{7}{3} \approx 2.333 \text{ metres} \]
Verify: compare the discrepancy with the measurement precision
Why: The one millimetre shortfall is exactly at the limit of what a tape measure can distinguish, so reporting more decimal places would claim precision the instrument cannot deliver. The rounding error and the measurement error are the same size, which is the right place to stop.
Trap
\[ 3x = 12.89 \;\rightarrow\; x = 4.296666\ldots \]
Report the answer to six decimal places, since the calculator supplied them
Why: More digits look more accurate, and the calculator produced them without being asked.
Nobody can pay 4.296666 dollars. The extra digits describe a quantity of money that does not exist, so they carry no information at all.
\[ x \approx 4.30 \text{ dollars} \]
Round to the precision the quantity actually has
Why: Money exists in cents, so two decimal places is the finest meaningful answer.
Extra digits are not free. They suggest a precision that the situation cannot support, which misleads a reader in the opposite direction from rounding too soon.
Estimation
Predicting the discrepancy is part of reporting the answer honestly.
Predict first
Four people share a 20.30 dollar bill and each pays a rounded share. Roughly how far from 20.30 will the total collected be?
Correct: Within a cent or two.
\[ \tfrac{20.30}{4} = 5.075 \approx 5.08, \quad 4(5.08) = 20.32 \]
Why: Each share is rounded by at most half a cent, and four such roundings can total at most about two cents. The exact share is 5.075, which rounds to 5.08, and four of those is 20.32 — two cents over. The number of roundings bounds the total error, which is why sharing between many people accumulates more discrepancy than sharing between few.
Elimination
Three people each pay 4.30 towards a 12.89 bill, collecting one cent too much.
Eliminate the wrong options
Which is the most sensible practical answer?
Survives elimination: A
Why: Splitting the discrepancy so that one person pays a penny less makes the total exactly 12.89. Rounding error in real situations is usually resolved by adjusting one of the parts rather than by changing the precision, and noticing that the adjustment is needed is the part the algebra cannot do for you.
Socratic
One rounding is harmless. Many roundings need watching.
Discussion prompt
Explain why sharing a bill between twenty people accumulates more rounding error than sharing it between three. Then describe a situation where accumulated rounding error would matter enough to change how you calculate.
Hint: Count how many roundings happen in each case.
Answer:
Each share is rounded once, so twenty shares means twenty roundings, each up to half a cent. The total discrepancy can therefore reach about ten cents rather than the one or two cents that three roundings can produce. More parts means more opportunities for the error to build.
It matters in any repeated financial calculation — interest applied daily for a year, or a payroll rounding thousands of amounts. There the standard practice is to carry full precision through every step and round only the final figure, or to track the accumulated discrepancy and correct it periodically. That is the same principle as this lesson's rule to round at the end, applied at a scale where it stops being merely tidy.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Exact answer | Rounded answer | |
|---|---|---|
| Usual form | a fraction | a decimal |
| Symbol to use | an equal sign | the wavy symbol |
| What the check gives | exact equality | closeness explained by the rounding |
Both kinds of answer are correct. Which you report depends on whether the question wants precision or usability, and the symbol has to match the choice.
Pattern
Whether the equation contains decimals or produces one, the same five moves cover it.
Step two says keep every value exact. A calculator holds far more digits than you need — let it carry them, and round only what you write down.
OpenStax Elementary Algebra 2e, §2.5 Solve Equations with Fractions or Decimals §2.5
Check
Solve exactly, then round.
Check your understanding
Solve 7x minus 5 equals 20 to the nearest hundredth.
Answer: A
Why: Adding five gives 7x equals 25, so x is twenty-five sevenths, which is 3.5714 and continuing. Rounding to two decimal places gives 3.57, and the approximately-equal symbol records the rounding.
Check
Clearing decimals. Multiply every term.
Check your understanding
Multiplying every term of 0.6x plus 2.4 equals 5.4 by 10 gives which equation?
Answer: A
Why: Every term on both sides is multiplied by ten, giving whole numbers throughout. Solving it gives 6x equals 30, so x is five — the same answer the decimal version produces.
Check
Rounding error. Check whether the parts add up.
Check your understanding
Four people share a 10.30 dollar bill equally, each paying a share rounded to the nearest cent. What happens to the total?
Answer: A
Why: The exact share is 2.575, which rounds up to 2.58, and four of those is 10.32. Each rounding added half a cent, and four half-cents make two cents, which is exactly the overshoot.
Real world
A shop applies a 17.5 percent tax to a pre-tax price. An item's final price is 47 dollars, and you want to know the pre-tax price.
Discussion prompt
Write and solve an equation for the pre-tax price, giving both the exact fraction and an answer rounded to the nearest cent. Then check your rounded answer and say why the check does not come out exactly 47, and by how much you should expect it to miss.
Hint: The final price is the pre-tax price multiplied by 1.175.
Answer:
\[ 1.175p = 47 \;\Longrightarrow\; p = \tfrac{47}{1.175} = 40 \text{ exactly} \]
This one happens to come out exactly forty, so no rounding is needed at all and the check gives precise equality. That is worth noticing: not every decimal equation produces a messy answer.
\[ \text{with a final price of } 50: \quad p = \tfrac{50}{1.175} \approx 42.55 \]
At a final price of fifty the answer is 42.5531 continuing, which rounds to 42.55. Checking gives 1.175 times 42.55, which is 49.996 — four thousandths short of fifty, because the rounding discarded about 0.003 and the multiplier of 1.175 barely magnified it. A coefficient near one produces a very small gap, which is the opposite of the coefficient-38 case in this lesson.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
When you substitute a rounded answer back into the original equation and the two sides differ slightly, what does that mean?
Correct: The gap is expected, and its size should be judged against the rounding.
\[ 38(4.13) - 39 = 117.94 \approx 118 \quad \text{gap } 0.06 \;\checkmark \]
\[ 38(4.31) - 39 = 124.78 \quad \text{gap } 6.78 \;\text{— far too large} \]
Why: A rounded answer is not the exact solution, so substituting it cannot give exact equality. What matters is whether the gap is about the size the rounding would produce — roughly the coefficient times the amount discarded. A gap of 0.06 with a coefficient of thirty-eight is exactly right; a gap of seven would signal a real error. Judging the size rather than demanding zero is what makes checking a rounded answer meaningful.
Explain it
They can solve equations and have always had whole-number answers.
Discussion prompt
In no more than four sentences, explain why you should not round until the very end, and what symbol to use once you have. Then tell them what to expect when they check a rounded answer.
Hint: The expectation is different from every check they have done before.
Answer:
A usable answer: rounding throws a little of the number away, and any step after that works with a value already slightly wrong. So keep the exact value — usually a fraction — all the way to the last line, and round only what you write down as the answer. Once you have rounded, use the wavy equals sign to say so.
When they check a rounded answer, the two sides will come out close but not identical. That is expected and is not a failure — what matters is whether the gap is small. A gap of a few hundredths is fine; a gap of several units means something actually went wrong.
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: Rounding at the end is fixed by writing the exact fraction on its own line before converting. The symbol is fixed by asking whether anything was discarded. Clearing is fixed by writing the multiplier against every term before simplifying any of them. Judging a gap is fixed by predicting its size first — roughly the coefficient times the amount rounded away. 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 solve one equation whose answer does not terminate, writing every line and marking the exact line where the symbol changes from equals to approximately equals. Underneath, substitute your rounded answer back into the original, compute both sides, and write the size of the gap beside your prediction of what it should have been. In the middle, take a decimal equation and solve it twice — once directly and once by clearing the decimals — and confirm the two answers agree. Near the bottom, share an awkward total between three or four people, round each share, and write down the discrepancy and how you would resolve it. Finally, in the margin, list three kinds of quantity and the precision each one calls for.
Your predicted gap and your measured gap should be close. If the measured one is much larger, the error is in the solving rather than in the rounding, and the check has done its job.
Recap
Five things, and the first one is a habit rather than a technique.
| If the question says | Your first move is |
|---|---|
| Round to the nearest hundredth | Solve exactly first, then round once |
| Solve 3.5x - 37.9 = 0.2x | Optionally multiply every term by 10 |
| Check your solution | Expect closeness, and predict the gap |
| Each person's share | Round to the nearest cent, then check the total |
| x = 157/38 | Write the wavy symbol before the decimal |
Lesson 3.7 turns to formulas, where the goal is not a number at all but a rearranged formula — solving for one variable in terms of the others, using exactly the moves from these six lessons.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 3 Solving Linear Equations — Lesson 3.6 Solving Decimal Equations §3.6, pp. 163-168 — everything on these slides traces back here
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