Compound inequalities joined by or, whose solutions need satisfy only one of the two parts. Includes graphing the union as two rays with a gap between them, solving each part independently with the methods of earlier lessons, reversing both symbols when dividing by a negative, and a velocity model in which direction is carried by a sign.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 6 — Solving and Graphing Linear Inequalities
Solving Compound Inequalities Involving “Or”
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. 6 Solving and Graphing Linear Inequalities — Lesson 6.5 Solving Compound Inequalities Involving “Or” §6.5, pp. 348-354 — the lesson these objectives are drawn from
Warm-up
Lesson 6.4 required both conditions. This lesson requires only one, and the graph changes shape as a result.
Discussion prompt
Graph x less than 1 and x greater than 2 on separate number lines. If a number needs to satisfy only one of them, which numbers qualify — and which do not?
Hint: Look at the numbers between one and two.
Answer:
\[ x < 1 \;\text{ or }\; x > 2 \]
Everything below one qualifies, and so does everything above two. The numbers from one to two inclusive satisfy neither condition, so the graph has a gap in the middle — two parts rather than one.
Concept
A number is a solution of a compound inequality with or if it is a solution of either inequality. The graph is everything the two parts cover between them, which usually has two pieces.
compound inequality with or — Two inequalities joined by or, solved by a number that satisfies at least one of them.
A number is rejected only when it fails both parts.
Figure (svg): A compound or inequality graphed as two separate rays
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.5 Solving Compound Inequalities Involving “Or” §6.5, pp. 348-349
Section
Section 1
Concept
With and, a number had to pass both tests. With or, passing either one suffices, so a number fails only if it fails both.
This is why an or inequality's solution set is at least as large as either part alone.
Figure (svg): A number tested against both parts of an or inequality
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.5 Solving Compound Inequalities Involving “Or” §6.5, pp. 349-349 — the Study Tip on what makes a number a solution with or
Picture it
Five, judged against both parts.
Figure (svg): A number tested against both parts of an or inequality
The left panel would have ended the matter under and. Under or it is simply irrelevant, because the right panel succeeded.
Worked example
Guided Practice 7 and 8. One condition is enough.
\[ \text{Is } 5 \text{ a solution of } \; x < 5 \;\text{ or }\; x > 4? \text{ And of } \; x \le 3 \;\text{ or }\; x > 0? \]
Test the first pair's left part
Why: Five is not less than five.
Test its right part
Why: Five is greater than four.
Conclude for the first
Why: One success is enough.
Test the second pair
Why: Five is not at most three; five is greater than zero.
Figure (svg): A number tested against both parts of an or inequality
\[ 5: \;\checkmark \text{ for both} \]
Verify: ask what would have happened with and
Why: Under and, both would have been rejected, since five fails the first part in each case. The same number and the same two conditions give opposite verdicts depending on the connecting word, which is the whole content of this lesson.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.5 Solving Compound Inequalities Involving “Or” §6.5, pp. 349-349
Sorting
For x less than 1 or x greater than 2, one part is enough.
Sort into buckets
Sort each number by whether it solves the compound or inequality.
The three rejected numbers are exactly those between one and two inclusive — the gap. Everything else on the line is a solution.
Worked example
A number must fail both parts to be excluded.
\[ \text{Which numbers are not solutions of } \; x < 1 \;\text{ or }\; x > 2? \]
Ask what fails the first part
Why: Everything at least one.
\[ x \ge 1 \]
Ask what fails the second part
Why: Everything at most two.
\[ x \le 2 \]
Find numbers failing both
Why: Between one and two, inclusive of both.
\[ 1 \le x \le 2 \]
State the exclusion
Why: Those are the only numbers left out.
Figure (svg): A compound or inequality graphed as two separate rays
\[ 1 \le x \le 2 \text{ are excluded} \]
Verify: notice what the excluded set is
Why: The numbers failing an or inequality form a compound and inequality — they must fail both parts. So the gap in an or graph is exactly the kind of bounded segment Lesson 6.4 produced, which is a neat connection between the two lessons.
Trap
\[ x < 5 \;\text{ or }\; x > 4 \quad \text{is } 5 \text{ a solution?} \]
Test the first part, find that 5 fails it, and answer no
Why: Under and, one failure settles the question, and the habit carries over.
With or the second part still has to be checked. Five is greater than four, so it passes and is a solution after all.
Under or, keep testing until one part passes, and only report no if both fail
Why: One success is enough, so a single failure decides nothing.
The two words demand opposite search strategies: under and you look for a failure, under or you look for a success.
Elimination
The inequality is joined by or.
Eliminate the wrong options
What makes a number fail a compound or inequality?
Survives elimination: A
Why: Only a number rejected by both conditions is excluded. Option D describes what usually happens when the two rays point away from each other, and it fails whenever the rays overlap — which is why the rule has to be stated in terms of the conditions rather than the picture.
Faded example
One success is enough.
Fill in the blanks
For x < 5 or x > 4 at x = 5: the first part is false and the second is true, so 5 is a solution.
Why: Five fails the first part and passes the second, and under or that is enough. Under and the same two results would have given the opposite verdict, which is why the connecting word must be read before any testing begins.
Socratic
And made it smaller.
Discussion prompt
Explain why joining two conditions with or produces a solution set at least as large as either part alone, while and produces one at most as large. Then say when the or set is exactly one of the two parts.
Hint: Ask whether a solution of one part can be lost.
Answer:
Every solution of either part is automatically a solution of the whole, since one success suffices — so nothing is lost and the set contains both parts entirely. With and, a solution of one part survives only if it also passes the other, so solutions can be lost and the set can only shrink.
The or set equals one of the parts when that part contains the other. For x greater than two or x greater than five, every number above five is already above two, so the answer is simply x greater than two — the larger ray swallows the smaller one. That is one of the three shapes an or graph can take.
Section
Section 2
Concept
To graph a compound or inequality, graph each part and keep everything either covers. When the two rays point away from each other, the result has two pieces separated by a gap.
Lesson 6.4's graphs had only one part; these usually have two.
Figure (svg): Three possible shapes for the graph of a compound or inequality
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.5 Solving Compound Inequalities Involving “Or” §6.5, pp. 348-348 — Example 1 and its Study Tip contrasting the two kinds of graph
Picture it
Apart, overlapping, or nested.
Figure (svg): Three possible shapes for the graph of a compound or inequality
The first shape is the usual one and the other two are worth recognising, because a question asking for two parts has a specific answer in mind and these do not provide it.
Worked example
This is Example 1 from the textbook.
\[ \text{Write and graph a compound inequality for all real numbers less than } 1 \text{ or greater than } 2. \]
Write the two conditions
Why: Less than one, or greater than two.
\[ x < 1\text{ or } x > 2 \]
Graph the first
Why: Open dot at one, shading left.
Graph the second
Why: Open dot at two, shading right.
Combine
Why: Keep both rays; the graph has two parts.
Figure (svg): A compound or inequality graphed as two separate rays
\[ x < 1 \;\text{ or }\; x > 2 \]
Verify: check a number in the gap
Why: One and a half is neither less than one nor greater than two, so it fails both parts and is correctly left unshaded. The gap is a real feature of the answer rather than a drawing artefact.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.5 Solving Compound Inequalities Involving “Or” §6.5, pp. 348-348
Discrimination
The connecting word and the directions both matter.
Sort into buckets
Sort each compound inequality by the number of parts in its graph.
Worked example
Guided Practice 4 to 6. Two are or and one is and.
\[ \text{Write and count the parts: at most } -3 \text{ or above } 0; \text{ below } 3 \text{ or above } 6; \text{ above } 2 \text{ and below } 7. \]
Take the first
Why: At most negative three or greater than zero.
Take the second
Why: Less than three or greater than six.
Take the third
Why: The word is and, so it is an overlap.
Note the pattern
Why: Or usually gives two parts; and gives one.
Figure (svg): The solution to Worked example how many parts shown as a ladder of expressions, one row per algebraic move
\[ x \le -3 \;\text{ or }\; x > 0; \quad x < 3 \;\text{ or }\; x > 6; \quad 2 < x < 7 \]
Verify: check the third is genuinely one part
Why: The and inequality has solutions between two and seven, a single bounded segment. Reading the connecting word before drawing anything is what settles how many parts to expect, and the word is the only difference between the three.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.5 Solving Compound Inequalities Involving “Or” §6.5, pp. 348-348
Error analysis
The student graphed x less than 1 or x greater than 2.
Annotate
On: \( \begin{aligned} &\text{open dot at } 1, \text{ open dot at } 2 \\ &\text{shade between them} \\ &\text{leave the outsides unshaded} \end{aligned} \)
Testing one number from outside the shaded region catches this immediately. Zero satisfies the first part, so any graph leaving zero unshaded is wrong.
Faded example
The excluded numbers fail both parts.
Fill in the blanks
For x < 1 or x > 2, a number is excluded when it is at least 1 and at most 2.
Why: Failing the first part means being at least one, and failing the second means being at most two, so the excluded numbers form the segment from one to two inclusive. The gap in an or graph is itself an and condition.
Prediction
The inequality is x less than 5 or x greater than 2.
Predict first
What is the graph?
Correct: The whole number line.
\[ x \le 2 \;\Longrightarrow\; x < 5 \;\checkmark \qquad x \ge 5 \;\Longrightarrow\; x > 2 \;\checkmark \]
Why: Every number is either below five or above two, and many are both. Since one success suffices, nothing is left out and the solution set is all real numbers. The second option describes what happens when the two rays point away from each other, which is the usual case but not this one — reading the directions is what distinguishes them.
Socratic
Two or inequalities can produce opposite-looking graphs.
Discussion prompt
Explain how the directions of the two rays decide whether the graph has a gap, covers everything, or reduces to one ray. Then say how to tell which case you are in without drawing.
Hint: Compare the two boundaries and the two directions.
Answer:
If the rays point away from each other, with the left-shading ray's boundary below the right-shading one's, they leave a gap. If they point towards each other, they overlap and cover everything. If they point the same way, one contains the other and the union is just the larger.
Without drawing, compare the boundaries after noting the directions. For x less than a or x greater than b, a gap appears when a is at most b, and everything is covered when a exceeds b. That single comparison predicts the shape, and it is worth doing before drawing rather than after.
Section
Section 3
Concept
To solve a compound or inequality, solve each part on its own using the methods of Lessons 6.1 to 6.3, then join the two answers with the word or.
The two parts never interact, so no combined form exists.
Figure (svg): Both parts of an or inequality solved side by side
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.5 Solving Compound Inequalities Involving “Or” §6.5, pp. 349-349 — Examples 2 and 3, solving each part with the earlier methods
Picture it
Nothing crosses between the columns.
Figure (svg): Both parts of an or inequality solved side by side
Unlike an and inequality, there is no single-line method here. The two parts are separate problems from start to finish, and only their answers are joined.
Worked example
This is Example 2 from the textbook.
\[ \text{Solve } \; x - 4 \le 3 \;\text{ or }\; 2x > 18 \; \text{ and graph the solution.} \]
Solve the left part
Why: Add four to each side.
\[ x \le 7 \]
Solve the right part
Why: Divide each side by two.
\[ x > 9 \]
Join with or
Why: Either answer is enough.
\[ x \le 7\text{ or } x > 9 \]
Graph both
Why: Solid dot at seven shading left, open dot at nine shading right.
Figure (svg): The solution to Worked example a one-step compound or inequality shown as a ladder of expressions, one row per algebraic move
\[ x \le 7 \;\text{ or }\; x > 9 \]
Verify: test a number in the gap
Why: Eight gives four on the left, which is not at most three, and sixteen on the right, which is not greater than eighteen. So eight fails both parts and is correctly excluded — the gap from just above seven to nine inclusive.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.5 Solving Compound Inequalities Involving “Or” §6.5, pp. 349-349
Translation
Two independent solves, joined at the end.
Match the pairs
Why: The first three leave gaps and the fourth covers everything, because its two rays overlap. Solving each part is the same work in all four; only the shape of the combined answer differs.
Worked example
This is Example 3 from the textbook.
\[ \text{Solve } \; 3x + 1 < 4 \;\text{ or }\; 2x - 5 > 7. \]
Solve the left part
Why: Subtract one, then divide by three.
\[ x < 1 \]
Solve the right part
Why: Add five, then divide by two.
\[ x > 6 \]
Join with or
Why: Less than one or greater than six.
\[ x < 1\text{ or } x > 6 \]
Graph both
Why: Two open dots and two rays pointing outwards.
Figure (svg): Both parts of an or inequality solved side by side
\[ x < 1 \;\text{ or }\; x > 6 \]
Verify: test one number in each part and one in the gap
Why: Zero gives one on the left, which is less than four, so zero is a solution. Seven gives nine on the right, which is greater than seven, so seven is too. Three gives ten and one, failing both, so it is correctly excluded.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.5 Solving Compound Inequalities Involving “Or” §6.5, pp. 349-349
Trap
\[ x < 1 \;\text{ or }\; x > 6 \]
Compress it to 1 > x > 6
Why: Compound inequalities were written on one line in Lesson 6.4, so the habit carries over.
That chain claims one is greater than six, which is false, and it describes an overlap rather than a union. An or inequality has to stay as two statements.
\[ x < 1 \;\text{ or }\; x > 6 \quad \text{written as two statements} \]
Keep the word or between the two parts
Why: The combined form means and, so using it would change the meaning entirely.
The combined notation is reserved for and, because it traps the variable between two bounds — which is precisely what or does not do.
Faded example
Each part on its own.
Fill in the blanks
3x + 1 < 4 \;\Longrightarrow\; x < 1 \qquad 2x - 5 > 7 \;\Longrightarrow\; x > 6
Why: Each part is a two-step inequality solved exactly as in Lesson 6.3, and neither affects the other. The answers are joined by or, giving a graph with two parts and a gap from one to six inclusive.
Elimination
And inequalities were written on one line.
Eliminate the wrong options
What goes wrong if you write x < 1 or x > 6 as a chain?
Survives elimination: A
Why: The chain notation traps the variable between two bounds, which is the and relationship, and it also asserts that the left bound is below the right one. Both claims are wrong for a union with a gap, which is why or inequalities stay as two statements.
Socratic
In an and inequality, one line could be operated on as a whole.
Discussion prompt
Explain why the two parts of an or inequality have to be solved independently, while the two parts of an and inequality could sometimes be handled together. Then say what that means for how many reversals a solve can contain.
Hint: Ask what the combined form was doing.
Answer:
The combined form worked because both parts contained the same expression, so operating on that expression served both at once. An or inequality is a statement about two separate conditions with no shared middle expression to work on, so there is nothing to operate on jointly and each part must be solved on its own terms.
It means the two parts can require different numbers of reversals. One part might divide by a positive and the other by a negative, so one symbol turns and the other does not. The reversal decision belongs to each part separately, which is a genuine difference from the and case where one operation on the whole line reversed both symbols together.
Section
Section 4
Concept
When a step divides or multiplies each part by a negative number, both symbols reverse. Each reversal is decided by that part's own operation.
The textbook's Study Tip warns to reverse both when both are divided by a negative.
Figure (svg): A compound or inequality in which both parts reverse
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.5 Solving Compound Inequalities Involving “Or” §6.5, pp. 350-350 — Example 5 and its Study Tip on reversing both inequalities
Picture it
Two divisions, two reversals.
Figure (svg): A compound or inequality in which both parts reverse
Both symbols turned here because both parts were divided by negative thirty-two. Had one part had a positive coefficient, only one of the two symbols would have moved.
Worked example
This is Example 5 from the textbook.
\[ \text{For } v = -32t + 64, \text{ find } t \text{ with } v > 32 \;\text{ or }\; v < -32. \]
Write both parts
Why: Substituting the velocity expression into each.
\[ -32 t + 64 > 32\text{ or } < -32 \]
Subtract 64 from each part
Why: No reversal from a subtraction.
\[ -32 t > -32\text{ or } -32 t < -96 \]
Divide the first by -32 and reverse
Why: Negative divisor.
\[ t < 1 \]
Divide the second by -32 and reverse
Why: Negative divisor again.
\[ t > 3 \]
Figure (svg): A compound or inequality in which both parts reverse
\[ t < 1 \;\text{ or }\; t > 3 \]
Verify: test one value in each part
Why: At t equal to zero the velocity is sixty-four, which is greater than thirty-two, so zero satisfies the first part. At t equal to four it is negative sixty-four, which is less than negative thirty-two, satisfying the second. And at t equal to two the velocity is zero, failing both, which is why the gap exists.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.5 Solving Compound Inequalities Involving “Or” §6.5, pp. 350-350
Faded example
Each division by a negative turns its own symbol.
Fill in the blanks
-32t > -32 \;\Longrightarrow\; t < 1 \qquad -32t < -96 \;\Longrightarrow\; t > 3
Why: Both parts are divided by negative thirty-two, so both symbols turn. The result is t less than one or t greater than three, with a gap covering the time when the ball is near the top of its flight.
Worked example
The reversal is decided part by part.
\[ \text{Solve } \; -2x > 6 \;\text{ or }\; 3x > 12. \]
Solve the first part
Why: Divide by negative two and reverse.
\[ x < -3 \]
Solve the second part
Why: Divide by three; no reversal.
\[ x > 4 \]
Join with or
Why: Two rays pointing outwards.
\[ x < -3\text{ or } x > 4 \]
Note the asymmetry
Why: One symbol turned and the other did not.
Figure (svg): The solution to Worked example only one part reverses shown as a ladder of expressions, one row per algebraic move
\[ x < -3 \;\text{ or }\; x > 4 \]
Verify: test a number in each part
Why: Negative four gives eight, which is greater than six, so it satisfies the first part. Five gives fifteen, which exceeds twelve, satisfying the second. Zero gives zero and zero, failing both, so the gap is genuine.
Trap
\[ -32t > -32 \;\text{ or }\; -32t < -96 \]
Divide both by -32 and reverse the first symbol only
Why: The reversal is remembered once and applied to the line rather than to each part.
\[ t < 1 \;\text{ or }\; t < 3 \quad \text{(wrong)} \]
The second part's symbol should have turned too. As written the answer says t is less than three, which wrongly includes t equal to two — where the velocity is zero and neither condition holds.
\[ t < 1 \;\text{ or }\; t > 3 \]
Decide each part's reversal from its own division
Why: Two divisions by a negative means two reversals.
Testing a number in the gap catches it: at t equal to two the velocity is zero, which is neither above thirty-two nor below negative thirty-two.
Sorting
Count the divisions by a negative.
Sort into buckets
Sort each compound or inequality by how many of its symbols reverse when solved.
The count is whatever the two parts happen to require, from zero to two. Deciding it for the line as a whole rather than part by part is what produces the standard error.
Elimination
Solve -32t + 64 > 32 or -32t + 64 < -32.
Eliminate the wrong options
Which is the solution?
Survives elimination: A
Why: Both parts divide by negative thirty-two, so both symbols turn. Testing t equal to two settles it: the velocity there is zero, which is neither above thirty-two nor below negative thirty-two, and only option A excludes it.
Socratic
In an and inequality one operation reversed both symbols together.
Discussion prompt
Explain why a compound or inequality can have one symbol reverse and the other not, while an and inequality solved in combined form always reverses both together. Then say what this means for checking your work.
Hint: Ask what operation is being performed on what.
Answer:
In the combined form of an and inequality there is one operation applied to all three expressions at once, so it affects both comparisons simultaneously and both symbols turn together. An or inequality is two separate problems, each with its own operations, so their reversals are independent — one part may divide by a negative while the other divides by a positive.
For checking, it means the two parts have to be verified separately: test a number that satisfies only the first part, and another that satisfies only the second. A single test number can confirm at most one part, so a complete check of an or inequality needs at least two — plus one in the gap if there is one.
Section
Section 5
Concept
Velocity uses positive numbers for upward motion and negative for downward. A question about speed in either direction becomes a compound inequality with or, since two separate ranges qualify.
The ball's velocity is zero at the top of its flight, which lies in the gap.
Figure (svg): A table of a baseball's velocity as it rises and falls
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.5 Solving Compound Inequalities Involving “Or” §6.5, pp. 350-350 — Examples 4 and 5, on the velocity of a baseball
Picture it
The sign changes at the top.
Figure (svg): A table of a baseball's velocity as it rises and falls
The velocity falls steadily from sixty-four to negative sixty-four, passing through zero at two seconds. That the model is linear is why every technique in this chapter applies to it.
Worked example
This is Example 4 from the textbook.
\[ \text{For } v = -32t + 64, \text{ tabulate } v \text{ at } t = 0, 1, 2, 3, 4 \text{ and describe the motion.} \]
Substitute the whole-number times
Why: Sixty-four, thirty-two, zero, negative thirty-two, negative sixty-four.
Describe the first half
Why: The ball rises and slows down.
Describe the middle
Why: At two seconds the velocity is zero, the highest point.
\[ v = 0\text{ at } t = 2 \]
Describe the second half
Why: The velocity goes negative and grows in size as the ball falls.
Figure (svg): A table of a baseball's velocity as it rises and falls
\[ v = 64, 32, 0, -32, -64 \]
Verify: check the constant step
Why: Each value is thirty-two less than the one before, matching the coefficient of negative thirty-two in the model. A constant step confirms the whole table at once, exactly as in Lesson 4.2.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.5 Solving Compound Inequalities Involving “Or” §6.5, pp. 350-350
Faded example
Substitute the model, then solve each part.
Fill in the blanks
-32t + 64 > 32 \;\Longrightarrow\; -32t > -32 \;\Longrightarrow\; t < 1
Why: Subtracting sixty-four gives negative 32t greater than negative thirty-two, and dividing by negative thirty-two reverses the symbol to give t less than one. At t equal to zero the velocity is sixty-four, which confirms it.
Worked example
Guided Practice 11. The same ball, different thresholds.
\[ \text{Find } t \text{ with } v > 0 \;\text{ or }\; v < -32. \]
Write the first part
Why: Negative 32t plus 64 greater than zero.
\[ -32 t > -64 \]
Solve it
Why: Divide by negative thirty-two and reverse.
\[ t < 2 \]
Write and solve the second part
Why: Negative 32t less than negative ninety-six, so reverse.
\[ t > 3 \]
Join and interpret
Why: The ball is rising, or falling faster than thirty-two feet per second.
\[ t < 2\text{ or } t > 3 \]
Figure (svg): The solution to Worked example a different pair of conditions shown as a ladder of expressions, one row per algebraic move
\[ t < 2 \;\text{ or }\; t > 3 \]
Verify: interpret the gap
Why: Between two and three seconds the ball is falling but has not yet reached a speed of thirty-two feet per second downwards. That is a real physical period, which is why the gap in the graph corresponds to something rather than being an artefact.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.5 Solving Compound Inequalities Involving “Or” §6.5, pp. 350-350
Trap
Find the times when the ball is moving fast.
Write v > 32, since fast means a large velocity
Why: A large number is what fast suggests, and negative numbers look small.
A velocity of negative sixty-four is fast — it is the ball falling at sixty-four feet per second. Only asking about the positive side describes the rise and ignores the fall entirely.
\[ v > 32 \;\text{ or }\; v < -32 \]
Ask about both directions separately and join them with or
Why: The sign carries direction, so fast in either direction is two conditions.
Lesson 6.7 will write this pair as a single statement using absolute value, which is exactly the notation for size regardless of direction.
Elimination
The answer to Example 5 was t less than 1 or t greater than 3.
Eliminate the wrong options
What is happening between 1 and 3 seconds?
Survives elimination: A
Why: The two conditions asked for speeds above thirty-two in either direction, so the gap is where the speed is at most thirty-two. That includes rising slowly, being momentarily still, and starting to fall — three different motions with one thing in common.
Hypothesis
Predict before you check.
Predict first
At what time does the ball's velocity change sign, and what is happening there?
Correct: At t = 2, where the ball reaches its highest point.
\[ -32t + 64 = 0 \;\Longrightarrow\; t = 2 \]
Why: Setting the velocity to zero gives negative 32t plus 64 equal to zero, so t is two. That is where the ball stops rising and begins to fall, so the sign of the velocity changes from positive to negative. Finding it is the x-intercept computation from Lesson 4.4 applied to a physical model.
Socratic
The question asks about one quantity.
Discussion prompt
Explain why asking for a velocity above 32 or below negative 32 requires or rather than and. Then say what an and version of the question would mean.
Hint: Ask whether a single velocity could satisfy both.
Answer:
No velocity is simultaneously above thirty-two and below negative thirty-two, so an and version would have no solutions at all. The question is about two separate ranges — fast upwards and fast downwards — and a moment qualifies by being in either one, which is exactly what or means.
An and version would be asking for a velocity in the overlap of the two ranges, which is empty. Any question about a quantity being far from some central value in either direction is an or question, and Lesson 6.7 will show that absolute value gives it a compact notation.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| And (6.4) | Or (6.5) | |
|---|---|---|
| A solution must | pass both parts | pass at least one part |
| The graph is | the overlap | everything either covers |
| Usual number of parts | one | two |
| Combined one-line form | available | not available |
Every row flips between the columns, and all four differences follow from the single change in the connecting word.
Pattern
Whether the parts are one-step or multi-step, the same five moves cover it.
Step five needs at least two test numbers, since a single number can confirm only one part of a union.
OpenStax Intermediate Algebra 2e, §2.6 Solve Compound Inequalities §2.6
Check
One part is enough.
Check your understanding
Is 3 a solution of x < 1 or x > 2?
Answer: A
Why: Three fails the first part and passes the second, and with or one success suffices. A number is excluded only when both parts reject it.
Check
Two independent solves.
Check your understanding
Solve x + 4 < 8 or x - 3 > 5.
Answer: A
Why: Subtracting four from the first part gives x less than four, and adding three to the second gives x greater than eight. The two answers are joined by or, leaving a gap from four to eight inclusive.
Check
Each part decides its own reversal.
Check your understanding
Solving -2x > 6 or 3x > 12, how many symbols reverse?
Answer: A
Why: The first part divides by negative two and reverses to x less than negative three; the second divides by three and keeps its symbol, giving x greater than four. The two parts are independent problems.
Real world
A machine part is rejected in quality control if its diameter is more than 0.02 mm away from the target of 12.00 mm in either direction.
Discussion prompt
Write the rejection condition as a compound inequality, graph it, and say what the acceptable range is. Then say why this is an or question and what an and version would mean.
Hint: Too big or too small are two separate conditions.
Answer:
\[ d > 12.02 \;\text{ or }\; d < 11.98 \]
The acceptable range is the gap: from 11.98 to 12.02 millimetres inclusive, which is a compound and inequality. As in Lesson 6.4, the numbers failing an or condition always form an and condition, and here that gap is exactly what quality control is trying to achieve.
It has to be or because a single diameter cannot be both too large and too small, so an and version would reject nothing at all. Any tolerance question — a measurement being too far from a target in either direction — has this shape, and Lesson 6.7 will write it as a single absolute-value statement.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
What is the graph of x < 5 or x > 2?
Correct: All real numbers, since every number satisfies at least one part.
\[ x \le 2 \Rightarrow x < 5 \;\checkmark \qquad x \ge 5 \Rightarrow x > 2 \;\checkmark \]
\[ \text{compare } x < 2 \;\text{ or }\; x > 5: \; \text{a genuine gap} \]
Why: Any number at most two is less than five, and any number at least five is greater than two, so every number passes at least one condition and nothing is left out. The first option assumes the usual shape without checking the directions: these two rays point towards each other and overlap, rather than pointing away and leaving a gap. Comparing the two boundaries against the two directions before drawing is what distinguishes the cases.
Explain it
They have just learned and inequalities and are treating or the same way.
Discussion prompt
In no more than four sentences, explain how or differs from and and what that does to the graph. Then tell them the check that catches a mistaken shading.
Hint: Both against either.
Answer:
A usable answer: with and a number has to pass both tests, so you keep only the overlap and get one bounded piece. With or it only has to pass one, so you keep everything either test accepts — which usually gives two pieces with a gap where a number fails both.
To check, pick a number from the part you did not shade and test it in both conditions. If it passes either one it should have been shaded, so your graph is wrong — and that single test catches the commonest error, which is shading the gap instead of the two rays.
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: Testing is fixed by remembering that under or you look for a success, not a failure. Shading is fixed by testing a number from the unshaded region in both conditions. Per-part reversals are fixed by solving each part in its own column with its own decision. Spotting full coverage is fixed by comparing the two boundaries against the two directions before drawing anything. 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 write two conditions joined by or whose rays point away from each other, graph each on its own number line, and beneath them draw a third line showing everything either covers, labelling the gap and writing the and inequality that describes it. In the middle, solve one multi-step compound or inequality by working the two parts in separate columns, writing beside each column whether that part's symbol reversed and why. Underneath, write three test numbers — one from each shaded part and one from the gap — with the verdicts of both conditions for each. In the lower half, write an or inequality whose two rays overlap, graph it, and write one sentence saying why its answer is all real numbers. Finally, in the margin, write the two-column summary of how and differs from or in what a solution must do and in the shape of the graph.
Your gap-number should fail both conditions and each of your other two test numbers should pass exactly one. If a number from the gap passes either condition, the shading of that part is wrong.
Recap
Five things, and the first is the one that flips every rule from the last lesson.
| If the question says | Your first move is |
|---|---|
| Less than a or greater than b | Write two statements joined by or |
| Is this number a solution | Look for one part it passes |
| Graph the compound inequality | Shade everything either part covers |
| Both parts have negative coefficients | Reverse both symbols, separately |
| Too far in either direction | Two conditions joined by or |
Lesson 6.6 introduces the notation that writes such pairs compactly. Absolute value measures distance regardless of direction, and an absolute-value equation turns out to be two ordinary equations joined by or.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 6 Solving and Graphing Linear Inequalities — Lesson 6.5 Solving Compound Inequalities Involving “Or” §6.5, pp. 348-354 — everything on these slides traces back here
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