QuanticaAlgebra IOpen in the courseAn inequality with a letter in it is true at some values of that letter and false at others. Solving it means finding every value that makes it true. The moves are the ones chapter 4 used to solve equations, run under 8.1's rules, and one of those rules is the reason an inequality is not just an equation with a different symbol.
Problem
Every solution of sits at or above a single number, and that number is itself a solution. Enter that number.
Show a hint
- The boundary is the that makes exactly , and it counts as a solution because is true when the two sides are equal.
- Undo the subtraction. Adding to both sides is one of 8.1's moves that leaves the direction alone, so it clears the off the left.
Show the full solution
Adding to both sides gives . Adding the same amount to both sides leaves the direction alone. At the left side is exactly , which counts, and at it is , which does not.
Problem
The solutions of are exactly the numbers less than one cutoff value, and the cutoff itself is not a solution. Enter the cutoff value.
Show a hint
- Two moves, in the same order chapter 4 used on equations. Clear the first, then the coefficient .
- Subtracting from both sides leaves . Six is positive, so the direction of the inequality stays the same when you divide both sides by it.
Show the full solution
Subtract from both sides for , then divide by the positive for . At the left side is exactly , and is false, so is the cutoff and not a solution.
Problem
Enter the largest integer that satisfies .
Show a hint
- Describe all the solutions first, then pick the largest integer inside that description.
- Adding gives , and dividing by leaves a boundary that is not a whole number. Test the whole number just below it.
Show the full solution
Add to both sides for , then divide by the positive for . Since , the largest integer allowed is . Checking, is below while is not. The boundary is not a whole number, so no integer sits exactly on it and in place of would give the same answer.
Problem
A student solves . After subtracting from both sides, the student divides by and leaves the symbol pointing the same way, which makes the resulting statement about wrong. Enter the largest integer that fits the student's wrong statement.
Show a hint
- Follow the student's work rather than your own. Subtracting gives , and the student then divides by while keeping .
- The student's statement is , and . No integer equals , so take the largest integer below it, and on the negative side larger means closer to zero.
Show the full solution
Subtracting gives , and dividing by without reversing gives , so the student's largest integer is . Dividing by a negative reverses the symbol, so the true statement is and the smallest integer that works is . Testing in the original gives , above .
A negative coefficient is not forced on you. When the variable is on both sides, a variable term can be removed from either side, and one of those two removals leaves a positive coefficient behind. Removing a variable term is a subtraction, so it never reverses the direction. That choice comes up in each of the next two problems.
Problem
Enter the smallest integer that satisfies .
Show a hint
- Once has a value, is a number, so subtracting from both sides is 4.3's move and the direction holds.
- That leaves . Add , divide by , then take the smallest integer strictly above the boundary.
Show the full solution
Subtract from both sides for , add for , then divide by for , so the smallest integer is . At both sides equal , and does not allow equality.
Problem
In the variable terms can be gathered on either side. Use the choice that leaves a positive coefficient, so no reversal is needed. Enter the smallest value can take.
Show a hint
- Subtracting from both sides puts the variable on the right with a positive coefficient. Subtracting instead leaves on the left, so a division by a negative comes later.
- Take the first route. Subtracting and then from both sides gives .
Show the full solution
Subtract from both sides for , subtract for , and divide by the positive for , so the smallest value is . Gathering on the left instead gives , and dividing by reverses the symbol to . Both routes describe the same solutions, but the second has one more step where the direction can be lost.
Isolating comes last. Distribute to clear parentheses the 2.1 way, and combine like terms on each side the 2.4 way. Neither move changes the other side, so no question of direction comes up while a side is being cleaned. Clean both sides first, then gather the variable terms and isolate.
Problem
Enter the largest integer that satisfies .
Show a hint
- Simplify each side before gathering anything. Distribute both products, then combine like terms on each side separately.
- The two sides become and . Subtract rather than , since taking away the smaller coefficient leaves a positive coefficient on .
Show the full solution
Distributing and combining gives . Subtracting gives , adding gives , and dividing by gives . Since sits between and , the largest integer is . Distributing multiplies both terms inside the parentheses by , so is and not .
Problem
The inequality is true for every from one value upward, and that value itself makes it true. Clear the fractions first, then enter that value.
Show a hint
- The two denominators are and . Multiplying both sides by a number that both of them divide leaves no fraction anywhere.
- Multiplying by gives , and is positive, so the direction is unchanged.
Show the full solution
Multiply both sides by , a positive number, so the direction is unchanged, giving , then and . Checking the endpoint, , and is true, so itself is a solution.
Gathering the variable terms can remove every one of them, since the two sides can have the same variable part. What is left is a comparison between two numbers with no letter in it. A comparison like that reads the same at every value, true as it stands or false as it stands, so the original holds everywhere or nowhere.
Problem
The statement is a chain, and it says both and at once. Solve it by doing the same thing to all three parts at each step. Enter the smallest value of that makes the whole chain true.
Show a hint
- The middle part is , so the first move is subtracting , and that subtraction has to be done to and to as well.
- Subtracting from all three parts gives . Dividing all three by is dividing by a positive number, so both directions hold.
Show the full solution
Subtract from all three parts for , then divide all three by the positive for , so the smallest value is . Checking, , and is true. Any below makes less than , so the left comparison fails.
Problem
Enter how many of the values , , , , and satisfy .
Show a hint
- Testing each value one at a time is the long route. Simplify the left side and gather the variable terms first, then see what is left.
- The left side collects to , so both sides carry . Subtracting removes the letter entirely.
Show the full solution
Distributing gives on the left, so the inequality is . Subtracting from both sides leaves , which is true and contains no , so every real number is a solution and the count is . A true numerical statement at the end means the inequality holds everywhere, whatever the listed values happened to be.
Check a finished description with two substitutions. Put the boundary value into the original and see whether it holds, which settles whether the boundary itself counts. Then test one value from the described side. A missed reversal shows up on that second test, since with the direction backwards every value in the description fails the original.
Problem
Enter how many integers with satisfy . The inequality on its own has infinitely many integer solutions, so only the ones inside the stated range are counted.
Show a hint
- Both sides have an term, so gather them on one side before anything else. Collect them where the coefficient comes out positive and no step will divide by a negative.
- Adding and then to both sides gives . That sets only the top of the allowed range, since the bottom is the stated . The symbol is strict, so check whether the cutoff value itself counts.
Show the full solution
Adding and to both sides gives , so . Inside the allowed integers run from to , and . Checking, gives , true, and gives , false. Adding rather than subtracting keeps the coefficient positive and removes the reversal step.
Problem
Enter how many integers satisfy . Both ends of the solved form are integers, so test each one in the original inequality before counting.
Show a hint
- Clear the fraction first by multiplying all three parts by . That multiplier is positive, so neither direction reverses.
- Subtracting from all three parts leaves . Dividing all three by reverses both directions, so the end becomes the lower end.
Show the full solution
Multiplying all three parts by gives , subtracting gives , and dividing all three by reverses both directions to , which reads . The integers are through , so the count is . Test both ends in the original. At the middle part is , and is true, so is in. At the middle part is , and is false, so is out.
Every answer here was described in words and symbols, a boundary together with the direction that goes with it. Lesson 8.4, Graphing Inequalities, is where those descriptions are drawn instead. On a number line the boundary is one marked point and the solutions are everything on one side of it, and with two letters the picture is a region of the plane.
Practice these ideas
Practice
A number satisfies . Enter the largest value can be.
Show the solution
Subtracting from both sides leaves the direction unchanged, giving . Since means less than or equal to, is allowed, and checking, .
Practice
Enter the smallest integer that satisfies .
Show the solution
Subtracting gives , and multiplying both sides by the positive gives , so the smallest integer is . At the left side equals exactly, which is not greater than .
Practice
Enter the smallest integer that satisfies .
Show the solution
Dividing both sides by reverses the direction to , so the smallest integer is . Checking, while is not below .
Practice
Solve . The solutions are exactly the values of at or above a single number, and that number is itself a solution. Enter that number.
Show the solution
Adding to both sides gives , and dividing both sides by leaves the direction alone, so . At the left side is exactly , and includes that boundary.
Practice
Enter the largest integer that satisfies .
Show the solution
Subtracting gives , and dividing by gives , so the largest integer is . Checking, while is not below .
Practice
A number satisfies . Enter the largest value can be.
Show the solution
Subtracting gives , and dividing by reverses the direction to . At the left side is , which is allowed since the comparison is , while gives . Skipping the reversal would give , which is false at , where the left side is .
Practice
Enter the smallest integer that satisfies .
Show the solution
Subtracting gives , adding gives , and dividing by gives , so the smallest integer is . At both sides equal , and the bound is strict, so is ruled out.
Practice
Enter how many integers with satisfy .
Show the solution
Subtracting gives , so . Inside the integers allowed run from to , and . Subtracting the two endpoints without adding gives and misses one of the two ends.
Practice
Enter the largest value of that satisfies , as a fraction in lowest terms.
Show the solution
Subtracting gives , subtracting gives , and dividing by the positive number keeps the direction, giving . Both sides equal at , and since the comparison is rather than , that boundary value is itself a solution.
Practice
The inequality holds for every at or below one value, and that value satisfies it too. Enter that value.
Show the solution
Distributing and combining gives . Subtracting from both sides gives , and subtracting gives , so . At both sides equal , and a statement is true when the two sides are equal, so itself is a solution.
Practice
Solve . Enter the largest number that is not a solution.
Show the solution
Multiplying both sides by the positive number keeps the direction, giving , so and . Every number above is a solution and is not, since at the left side is , which is not greater than .
Practice
Enter the smallest integer that satisfies .
Show the solution
Subtracting from all three parts gives , and dividing all three by the positive gives , so the smallest integer is . The left end is strict, so is excluded even though sits right at it.
Practice
Enter the largest integer that satisfies .
Show the solution
Subtracting from all three parts gives , and dividing all three by reverses both symbols to , which reads . The largest integer is . Checking, satisfies the chain while falls below .
Practice
Test the five values , , , and in . Enter how many of them satisfy the inequality.
Show the solution
Distributing gives , so the inequality reads . Subtracting from both sides leaves , which is false, so no value of works and the count is . Once the variable cancels and a false number statement is left, the inequality has no solutions no matter which values are tested.
Practice
A student solves . After distributing and gathering the variable terms on the left, the student divides by the negative coefficient and keeps the symbol pointing the same way. Enter the largest integer that actually satisfies the original inequality.
Show the solution
Distributing gives , subtracting and from both sides gives , and dividing by reverses the direction to , so the largest integer is . Checking, gives while gives , which is false. Without the reversal the description comes out as , the same boundary pointing the wrong way.
Practice
Enter the largest integer that satisfies .
Show the solution
Multiplying by the positive gives , which is . Subtracting and adding gives , so and the largest integer is . Distributing over gives . Writing there instead changes the boundary and the answer with it.
Practice
Enter how many integers satisfy .
Show the solution
Multiplying all three parts by gives , subtracting gives , and dividing all three parts by reverses both directions to , which reads . The integers run from to , so the count is . Both boundaries are fractions here, so no integer sits exactly on either end.