Inequalities — Free Pre-Algebra Review Games.

The < symbol points left and means 'less than'.

3 > 2 is true, so x = 3 is a solution.

Subtract 5 from both sides: x > 3.

Add 2 to both sides: x <= 8.

4 is not >= 5, so it is not a solution.

Divide both sides by 3: x < 5.

Divide by -2 and flip the sign: x < -5.

Multiply both sides by 4: x >= -8.

'At most' means less than or equal to.

Subtract 7: x < 3 - 7 = -4.

-3x >= 9, divide by -3 and flip: x <= -3.

2x - 2 > 8, 2x > 10, x > 5.

Multiply by -3 and flip: x >= -12.

The open circle at -3 means greater than, the closed at 5 means at most.

-2x < 6, divide by -2 and flip: x > -3.

The symbol \(\geq\) combines a greater-than arrow with an equals bar, meaning the value can be greater than or exactly equal to the number. The symbol \(\leq\) is wrong because it points the opposite direction, indicating less than or equal to. Recognizing these symbols quickly is essential before you can write or solve any inequality.

Since \(x<4\) requires a strict inequality, \(x=4\) does not satisfy it because \(4\) is not strictly less than \(4\). The choice 'Yes, because \(4\) is less than \(4\)' is wrong because it contradicts the definition of a strict inequality, which excludes the boundary value. Always check whether an inequality is strict or includes equality before testing a boundary number.

An open circle on a number line shows that the boundary value is not included, which matches the strict inequality \(x>3\). The choice \(x\geq3\) is wrong because the 'or equal to' part means \(3\) is included, requiring a closed circle instead. Remembering that strict inequalities use open circles while inclusive inequalities use closed circles is key to correctly graphing solutions.

Adding \(4\) to both sides of \(x-4>1\) isolates \(x\), giving \(x>5\). The choice \(x>-3\) is wrong because it results from subtracting instead of adding \(4\), which does not correctly isolate the variable. Always perform the inverse operation on both sides of an inequality to solve for the variable, just like you would with an equation.

'No more than' means the value cannot exceed \(7\), but it can equal \(7\), so the correct symbol is \(\leq\). The symbol \(<\) is wrong because it excludes \(7\) entirely, which contradicts the phrase 'no more than.' Phrases like 'at most' and 'no more than' always translate to \(\leq\), a pattern worth memorizing for word problems.

'At least' means the number can be \(10\) or greater, which is represented by \(x\geq10\). The choice \(x>10\) is wrong because it excludes \(10\) itself, but 'at least' means \(10\) is a valid value. Translating key phrases like 'at least' and 'at most' correctly is essential for writing accurate inequalities from word problems.

Subtracting \(9\) from both sides of \(x+9\leq12\) isolates \(x\), giving \(x\leq3\). The choice \(x\leq21\) is wrong because it results from adding \(9\) instead of subtracting it, which is the incorrect inverse operation. Solving inequalities uses the same inverse operations as equations, so always double-check which operation undoes the one shown.

A closed circle indicates the value \(5\) is included, and shading to the right means all numbers greater than \(5\) are also solutions, matching \(x\geq5\). The choice \(x>5\) is wrong because it requires an open circle, not a closed one, since \(5\) itself is excluded. Matching the circle type and shading direction to the correct symbol is a core graphing skill for inequalities.

Adding \(6\) to both sides of \(x-6<-2\) isolates \(x\), giving \(x<4\). The choice \(x<-8\) is wrong because it comes from subtracting \(6\) instead of adding it, which is the wrong inverse operation. Always add the term that was subtracted to correctly isolate the variable in an inequality.

The number \(0\) is greater than \(-1\), so it does not satisfy \(x\leq-1\), making it not a solution. The choice \(-2\) is wrong as an answer here because \(-2\) is indeed less than \(-1\) and satisfies the inequality. Testing a candidate value by substituting it into the inequality is a reliable way to check whether it belongs to the solution set.

'Twice a number' translates to \(2n\), and 'more than' means strictly greater, giving \(2n>20\). The choice \(n+2>20\) is wrong because it represents adding \(2\) to the number rather than multiplying it by \(2\), misinterpreting 'twice.' Careful word-by-word translation of phrases into mathematical operations prevents common errors when writing inequalities.

Multiplying both sides of \(\frac{x}{2}<3\) by the positive number \(2\) isolates \(x\), giving \(x<6\). The choice \(x<1.5\) is wrong because it results from dividing by \(2\) instead of multiplying, which is the incorrect inverse operation. When solving inequalities with division, multiply both sides by the denominator to isolate the variable correctly.

The symbol \(\neq\) is formed by a slash through an equals sign, specifically meaning the two values are not the same. The choice \(\approx\) is wrong because it means 'approximately equal to,' a different concept entirely from strict inequality. Knowing all inequality and relation symbols helps you correctly interpret and write mathematical statements.

Adding \(3\) to both sides gives \(4x\leq12\), and dividing both sides by the positive number \(4\) gives \(x\leq3\). The choice \(x\leq12\) is wrong because it stops after adding \(3\) without completing the division step needed to isolate \(x\). Solving multi-step inequalities requires performing all necessary inverse operations in the correct order, just like solving equations.

Dividing both sides of \(-5x<20\) by \(-5\) requires flipping the inequality sign, resulting in \(x>-4\). The choice \(x<-4\) is wrong because it fails to flip the inequality symbol when dividing by a negative number. Always remember to reverse the inequality sign whenever you multiply or divide both sides by a negative value.

'The sum of a number and \(5\)' means \(n+5\), and 'less than or equal to \(14\)' translates directly to \(\leq14\), giving \(n+5\leq14\). The choice \(n-5\leq14\) is wrong because it uses subtraction instead of the addition implied by the word 'sum.' Recognizing operation keywords like 'sum,' 'difference,' and 'product' is critical for accurately translating word problems into inequalities.

Subtracting \(2x\) from both sides gives \(5>x-1\), and adding \(1\) to both sides gives \(6>x\), which is equivalent to \(x<6\). The choice \(x>6\) is wrong because it reverses the direction of the inequality without a valid reason, such as multiplying by a negative number. When variables appear on both sides, collect them on one side carefully and keep track of the inequality direction throughout.

Since \(x\leq-2\) includes \(-2\) itself and all smaller values, the correct graph uses a closed circle at \(-2\) with shading extending left. The choice with an open circle at \(-2\) shading left is wrong because the open circle would incorrectly exclude the value \(-2\) from the solution set. Matching the correct circle type to whether the inequality is strict or inclusive is essential for accurate graphing.

Multiplying both sides of \(\frac{x}{-3}\leq4\) by \(-3\) requires flipping the inequality sign, giving \(x\geq-12\). The choice \(x\leq-12\) is wrong because it fails to reverse the inequality direction despite multiplying by a negative number. Multiplying or dividing by a negative number always flips the inequality symbol, a rule that has no exceptions.

An open circle at \(3\) means \(3\) is excluded, and shading to the right means all values greater than \(3\) are included, matching \(x>3\). The choice \(x\geq3\) is wrong because it would require a closed circle to show that \(3\) is part of the solution set. Reading the circle type and shading direction together tells you exactly which inequality symbol to use.

Subtracting \(6\) from both sides gives \(-x>-4\), and dividing by \(-1\) flips the inequality to give \(x<4\). The choice \(x>4\) is wrong because it neglects to flip the inequality sign when dividing by the negative coefficient of \(x\). Whenever the variable has a negative coefficient, isolating it by dividing by a negative number requires reversing the inequality symbol.

'Five less than a number' translates to \(n-5\), and 'at least \(3\)' means \(\geq3\), giving \(n-5\geq3\). The choice \(5-n\geq3\) is wrong because it reverses the order of subtraction, computing '\(5\) minus the number' rather than 'the number minus \(5\).' Phrases like 'less than' require careful attention to word order since subtraction is not commutative.

Distributing gives \(3x+6\leq15\), subtracting \(6\) gives \(3x\leq9\), and dividing by \(3\) gives \(x\leq3\). The choice \(x\leq7\) is wrong because it likely comes from subtracting \(2\) instead of properly distributing the \(3\) across both terms in the parentheses first. Always distribute fully before combining like terms or isolating the variable in an inequality with parentheses.

Solving \(2x-1>7\) gives \(x>4\), and since \(5>4\), substituting \(x=5\) confirms \(2(5)-1=9>7\) is true. The choice \(4\) is wrong because substituting gives \(2(4)-1=7\), which is not strictly greater than \(7\) since the inequality is strict. Testing a candidate value by direct substitution is a reliable way to verify whether it satisfies the inequality.

Dividing all three parts of the compound inequality by the positive number \(2\) gives \(-3\leq x\leq5\). The choice \(-12\leq x\leq20\) is wrong because it results from multiplying by \(2\) instead of dividing, which is the incorrect inverse operation. In a compound inequality, whatever operation you perform must be applied to all three parts equally to keep the statement balanced.

Since the fee applies only when the weight exceeds \(50\) pounds and not at exactly \(50\), the correct inequality is the strict inequality \(w>50\). The choice \(w\geq50\) is wrong because it would incorrectly include a package weighing exactly \(50\) pounds, which the problem states does not incur a fee. Word problems often hinge on whether the boundary value itself is included, so read the wording carefully.

Multiplying both sides by \(5\) gives \(2x\geq20\), and dividing by \(2\) gives \(x\geq10\). The choice \(x\geq20\) is wrong because it stops after multiplying by \(5\) without completing the division by \(2\) needed to fully isolate \(x\). Multi-step inequalities require completing every inverse operation in sequence to reach the final simplified solution.

A closed circle at \(-4\) means \(-4\) is included, and shading to the left means values less than \(-4\) are also solutions, matching \(x\leq-4\). The choice \(x<-4\) is wrong because a strict inequality requires an open circle, which would exclude \(-4\) from the solution set. Always pair a closed circle with an inclusive symbol like \(\leq\) or \(\geq\) when graphing.

Subtracting \(3.5\) from both sides of \(x+3.5\geq7.2\) gives \(x\geq3.7\). The choice \(x\geq10.7\) is wrong because it results from adding \(3.5\) instead of subtracting it, which is the incorrect inverse operation. Decimal inequalities follow the same solving rules as whole-number inequalities, so isolate the variable using the correct inverse operation.

'A number multiplied by \(3\)' is \(3n\), 'increased by \(4\)' adds \(4\) to get \(3n+4\), and 'less than \(19\)' gives \(3n+4<19\). The choice \(3(n+4)<19\) is wrong because it multiplies \(3\) by the entire expression \(n+4\), misinterpreting the order of operations described in the sentence. Breaking a word problem into small phrases and translating each piece step by step prevents errors in the operation order.

Subtracting \(2\) from both sides of \(-7\leq x+2\) gives \(-9\leq x\), which is equivalent to \(x\geq-9\). The choice \(x\leq-9\) is wrong because it reverses the inequality direction without any valid mathematical reason, such as multiplying by a negative number. Rewriting an inequality with the variable on the left simply flips the whole statement, not the direction relative to the variable.

Since Ana can spend at most \$40 and each book costs \$8, the total cost \(8b\) must be less than or equal to \$40, giving \(8b\leq40\). The choice \(8b<40\) is wrong because 'at most' includes the possibility of spending exactly \$40, which requires the inclusive symbol \(\leq\) rather than a strict inequality. Translating budget constraints into inequalities requires matching phrases like 'at most' to the correct inclusive or exclusive symbol.

Distributing gives \(-2x+6<4x+6\), subtracting \(6\) gives \(-2x<4x\), and subtracting \(4x\) then dividing by \(-6\) (which flips the sign) gives \(x>0\). The choice \(x<0\) is wrong because it results from forgetting to flip the inequality symbol when dividing both sides by the negative number \(-6\). Multi-step inequalities with variables on both sides require careful tracking of sign flips whenever a negative coefficient is involved.

Subtracting \(2x\) from both sides leaves \(3>7\), which is a false statement regardless of the value of \(x\), so there is no solution. The choice 'All real numbers' is wrong because it applies when the resulting statement is always true, not when it is always false as it is here. When the variable cancels out and leaves a false numerical statement, the inequality has no solution; if it leaves a true statement, the solution is all real numbers.

Subtracting \(5x\) from both sides leaves \(-2\leq8\), which is always true no matter what value \(x\) takes, so every real number is a solution. The choice 'No solution' is wrong because it would only apply if the remaining statement were false, but \(-2\leq8\) is a true statement. Recognizing when the variable cancels out to leave a true or false statement tells you whether the answer is all real numbers or no solution.

Adding \(1\) to all three parts gives \(-2\leq2x<6\), and dividing all parts by the positive number \(2\) gives \(-1\leq x<3\). The choice \(-2\leq x<6\) is wrong because it stops after adding \(1\) without completing the division step needed to fully isolate \(x\). Solving compound inequalities requires applying the same operation to all three parts in every step until the variable is isolated.

Setting up \(25+0.20m\leq65\), subtracting \(25\) gives \(0.20m\leq40\), and dividing by \(0.20\) gives \(m\leq200\), so the maximum is \(200\) miles. The choice \(100\) is wrong because it likely results from an arithmetic error, such as dividing \(40\) by \(0.40\) instead of \(0.20\). Real-world budget problems require translating the situation into an inequality first, then solving carefully to find the boundary value.

Multiplying both sides by \(4\) gives \(3-2x\geq4\), subtracting \(3\) gives \(-2x\geq1\), and dividing by \(-2\) flips the inequality to give \(x\leq-\frac{1}{2}\). The choice \(x\geq-\frac{1}{2}\) is wrong because it fails to reverse the inequality direction when dividing both sides by the negative coefficient \(-2\). Even in fraction-based inequalities, dividing by a negative number still requires flipping the inequality symbol.

An absolute value inequality \(|x|<5\) means \(x\) is within \(5\) units of zero on both sides, which translates to the compound inequality \(-5<x<5\). The choice '\(x<-5\) or \(x>5\)' is wrong because that describes \(|x|>5\), where \(x\) is far from zero, not close to it. Absolute value less-than inequalities always become compound 'and' statements, while greater-than inequalities become 'or' statements.

Distributing gives \(4x-8\geq3x+3-5\), which simplifies to \(4x-8\geq3x-2\), and subtracting \(3x\) then adding \(8\) to both sides gives \(x\geq6\). The choice \(x\leq6\) is wrong because it reverses the inequality direction, which is not justified since no multiplication or division by a negative number occurred in this problem. Combining like terms fully before isolating the variable prevents sign errors in multi-step inequalities.

Two separate shaded rays pointing outward from open circles represent a disjoint 'or' compound inequality, giving \(x<-2\) or \(x>3\). The choice \(-2<x<3\) is wrong because it describes a single shaded region between two values, which is an 'and' statement, not two separate outward rays. Graphs with two disjoint shaded rays always correspond to 'or' compound inequalities, while a single shaded segment corresponds to an 'and' statement.

Multiplying all three parts by \(3\) gives \(-12<x+2\leq6\), and subtracting \(2\) from all three parts gives \(-14<x\leq4\). The choice \(-10<x\leq8\) is wrong because it results from adding \(2\) instead of subtracting it in the final step, which is the incorrect inverse operation. Solving compound inequalities with fractions requires clearing the denominator first, then isolating the variable across all three parts consistently.

Consecutive integers are \(n\) and \(n+1\), and their sum is \(n+(n+1)=2n+1\), which must be at least \(15\), giving the inequality \(2n+1\geq15\). The choice \(n+1\geq15\) is wrong because it only accounts for the second integer alone, ignoring the smaller integer \(n\) in the sum. Setting up consecutive integer problems requires expressing all unknowns in terms of a single variable before writing the inequality.

Adding \(7\) to both sides gives \(2x\geq3x\), and subtracting \(3x\) then dividing by \(-1\) flips the inequality to give \(x\leq0\). The choice \(x\geq0\) is wrong because it fails to reverse the inequality symbol when dividing both sides by \(-1\) to isolate the variable. Whenever isolating a negative \(x\) term requires dividing by \(-1\), remember that the inequality direction must be reversed.

Setting up \(12t+600>3000\), subtracting \(600\) gives \(12t>2400\), and dividing by \(12\) gives \(t>200\), so the smallest whole number satisfying this strict inequality is \(201\). The choice \(200\) is wrong because the inequality is strict, meaning \(t=200\) gives exactly \$3,000, which does not satisfy 'more than \$3,000.' When a real-world problem uses a strict inequality with whole-number solutions, the answer must be the next whole number beyond the calculated boundary.