Inequalities | Free SAT Math Course | The School of Mathematics

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<h1>SAT Math: Inequalities</h1>

Inequalities are an officially tested skill in the Algebra domain of the Digital SAT, and this free, complete lesson covers everything in the College Board's Inequalities question category: reading, writing, and graphing basic inequalities, the properties of inequalities, multi-step inequalities and real-world word problems, and/or compound inequalities, absolute value inequalities and absolute value equations, tolerance and margin-of-error applications, and linear inequalities in two variables, including boundary lines and matching graphs to their equations. Each idea is followed by a fully worked example, and then a set of original practice problems with instant feedback, so you can build fluency across the entire topic in one place. Everything here is free, and you can keep practicing afterward with the full SAT Math Question Bank linked below.

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<a class="cta-btn cta-primary" href="https://theschoolofmathematics.com/quiz/sat-inequalities-quiz-1">Practice Inequalities Free</a>

<a class="cta-btn cta-secondary" href="https://theschoolofmathematics.com/quiz/course/SAT-Math-Qbank">Explore the Full SAT Math Qbank</a>

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<h2>What's covered in this lesson</h2>

<ol>

<li><a href="#reading">Reading, Writing, and Graphing Inequalities</a></li>

<li><a href="#properties">Properties of Inequalities</a></li>

<li><a href="#multi-step">Multi-Step Inequalities</a></li>

<li><a href="#word-problems">Real-World Word Problems</a></li>

<li><a href="#compound">Compound Inequalities</a></li>

<li><a href="#absolute-value">Absolute Value Inequalities</a></li>

<li><a href="#applications">Tolerance and Margin-of-Error Applications</a></li>

<li><a href="#basics">What Absolute Value Means</a></li>

<li><a href="#solving">Solving Absolute Value Equations</a></li>

<li><a href="#no-solution">Recognizing No-Solution Equations</a></li>

<li><a href="#graph">The Absolute Value Function and Its Graph</a></li>

<li><a href="#solutions">Solutions to Two-Variable Inequalities</a></li>

<li><a href="#boundary">Boundary Lines and the Test-Point Method</a></li>

<li><a href="#matching">Matching Graphs to Inequalities</a></li>

<li><a href="#mistakes">Common Mistakes to Avoid</a></li>

<li><a href="#faq">Frequently Asked Questions</a></li>

</ol>

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<h2 id="reading">1. Reading, Writing, and Graphing Inequalities</h2>

<p>An <strong>inequality</strong> is a mathematical sentence that compares two expressions using a symbol other than equals: >, <, ≥, or ≤. On the Digital SAT, you'll often need to translate a verbal phrase into an inequality, or read a number line graph and identify the inequality it represents.</p>

<tr><th>Verbal phrase</th><th>Inequality</th></tr>

<tr><td>less than, fewer than, below</td><td><</td></tr>

<tr><td>greater than, more than, exceeds, above</td><td>></td></tr>

<tr><td>at most, no greater than, no more than, less than or equal to</td><td>≤</td></tr>

<tr><td>at least, no less than, no fewer than, greater than or equal to</td><td>≥</td></tr>

</table>

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<p class="figure-caption"><strong>x > 2</strong> – open circle, shaded right</p>

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<p class="figure-caption"><strong>x ≤ 2</strong> – closed circle, shaded left</p>

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<p><strong>Worked Example:</strong> Write each statement as an inequality.</p>

<p>a. The number x is no greater than −2.</p>

<p>b. The amount of calories n meets or exceeds 1,200.</p>

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<p class="prompt">1. Which inequality represents "the number y is no more than 12"?</p>

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<p><span class="label">Explanation:</span> "No more than" means the value can equal 12 or be smaller, so y ≤ 12.</p>

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<p class="prompt">2. Which inequality represents "the temperature t is at least 68 degrees"?</p>

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<p><span class="label">Explanation:</span> "At least" means the value can equal 68 or be larger, so t ≥ 68.</p>

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<p class="prompt">3. On a number line graph of x > −2, what type of dot is used at −2, and which direction is shaded?</p>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g1_pa-3',false)">A) Closed circle, shaded left</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g1_pa-3',false)">B) Closed circle, shaded right</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g1_pa-3',false)">C) Open circle, shaded left</button>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'g1_pa-3',true)">D) Open circle, shaded right</button>

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<p><span class="label">Explanation:</span> Since −2 itself is not included (the symbol is >, not ≥), use an open circle. Since x must be greater than −2, shade to the right.</p>

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<p class="prompt">4. Which inequality is graphed by a closed dot at 5 with shading to the left?</p>

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<p><span class="label">Explanation:</span> A closed dot means 5 is included, and shading left means values less than 5 are also included, so the inequality is x ≤ 5.</p>

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<h2 id="properties">2. Properties of Inequalities</h2>

<p>Inequalities follow rules similar to equations, with one critical exception. For all real numbers a, b, and c:</p>

<tr><th>Property</th><th>Rule</th></tr>

<tr><td>Addition and Subtraction</td><td>If a < b, then a + c < b + c and a − c < b − c</td></tr>

<tr><td>Multiplication and Division by a positive c</td><td>If a < b and c > 0, then ac < bc and <span class="frac"><span class="num">a</span><span class="den">c</span></span> < <span class="frac"><span class="num">b</span><span class="den">c</span></span></td></tr>

<tr><td>Multiplication and Division by a negative c</td><td>If a < b and c < 0, then ac > bc and <span class="frac"><span class="num">a</span><span class="den">c</span></span> > <span class="frac"><span class="num">b</span><span class="den">c</span></span></td></tr>

</table>

<p>That last rule is the one to memorize carefully: <strong>whenever you multiply or divide both sides of an inequality by a negative number, you must flip the inequality symbol.</strong> Adding or subtracting never flips the symbol, only multiplying or dividing by a negative does.</p>

<p><strong>Worked Example:</strong> Solve −3x + 5 > 20.</p>

<p class="step-math">−3x > 15   (Subtract 5 from each side)</p>

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<p class="prompt">1. Solve: −2x + 7 ≤ 19</p>

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<p><span class="label">Explanation:</span> Subtract 7 from each side: −2x ≤ 12. Divide each side by −2, flipping the symbol: x ≥ −6.</p>

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<p class="prompt">2. Solve: 5 − x > 12</p>

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<p><span class="label">Explanation:</span> Subtract 5 from each side: −x > 7. Multiply each side by −1, flipping the symbol: x < −7.</p>

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</div>

<p><span class="label">Explanation:</span> Adding the same number to both sides never changes the direction of an inequality, so a + 5 < b + 5 is always true. Choice B reverses the inequality without a valid reason, C states the wrong direction, and D would mean a > b, which contradicts a < b.</p>

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<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g1_pb-4',false)">D) Cannot be determined</button>

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<p><span class="label">Explanation:</span> Multiplying both sides of a < b by a negative number c flips the inequality, giving ac > bc.</p>

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<h2 id="multi-step">3. Multi-Step Inequalities</h2>

<p>Solving a multi-step inequality follows the same process as solving a multi-step equation: distribute if needed, combine like terms, and isolate the variable using the properties above. The only new step to watch for is flipping the symbol if you multiply or divide by a negative number.</p>

<p><strong>Worked Example:</strong> Solve 5x − 6 ≥ 3x + 8.</p>

<p class="step-math">5x − 3x − 6 ≥ 8   (Subtract 3x from each side)</p>

<p class="step-math">2x − 6 ≥ 8   (Simplify)</p>

<p class="step-math">x ≥ 7   (Divide each side by 2; no flip needed since 2 is positive)</p>

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<p class="prompt">1. Solve: 7x + 4 < 3x + 24</p>

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<p><span class="label">Explanation:</span> Subtract 3x from each side: 4x + 4 < 24. Subtract 4: 4x < 20. Divide by 4: x < 5.</p>

</div>

<p class="prompt">2. Solve: 9 − 2x ≥ 4x − 15</p>

</div>

<p><span class="label">Explanation:</span> Add 2x to each side: 9 ≥ 6x − 15. Add 15: 24 ≥ 6x. Divide by 6: 4 ≥ x, which is the same as x ≤ 4.</p>

</div>

<p class="prompt">3. Solve: 3(2x − 1) < 5x + 9</p>

</div>

<p><span class="label">Explanation:</span> Distribute: 6x − 3 < 5x + 9. Subtract 5x from each side: x − 3 < 9. Add 3: x < 12.</p>

</div>

<p class="prompt">4. Solve: −4(x + 2) ≥ 2x − 4</p>

</div>

<p><span class="label">Explanation:</span> Distribute: −4x − 8 ≥ 2x − 4. Add 4x to each side: −8 ≥ 6x − 4. Add 4: −4 ≥ 6x. Divide by 6: −<span class="frac"><span class="num">2</span><span class="den">3</span></span> ≥ x, which is the same as x ≤ −<span class="frac"><span class="num">2</span><span class="den">3</span></span>.</p>

</div>

<h2 id="word-problems">4. Real-World Word Problems</h2>

<p>Many SAT inequality questions are wrapped inside a real-world scenario, often about budgets, fees, or limits. The key skill is translating the situation into an inequality using the same verbal phrases from the first section, then solving it just like any other inequality.</p>

<p><strong>Worked Example:</strong> A moving truck rental costs $35 for the day plus $0.50 per mile driven. If Deja has budgeted no more than $95 for the rental, what is the greatest number of miles she can drive?</p>

<p>Let m be the number of miles. The total cost is 35 + 0.5m, and it must be no more than 95:</p>

<p class="step-math">0.5m ≤ 60   (Subtract 35 from each side)</p>

<p class="step-math">m ≤ 120   (Divide each side by 0.5)</p>

<p>Deja can drive at most 120 miles.</p>

</div>

<p class="prompt">1. A parking garage charges $5 for the first hour and $3 for each additional hour. If Maria has at most $23 to spend, what is the greatest number of additional hours she can park?</p>

</div>

<p><span class="label">Explanation:</span> Let h be the number of additional hours. The total cost is 5 + 3h, and it must be at most 23: 5 + 3h ≤ 23, so 3h ≤ 18, and h ≤ 6. She can park at most 6 additional hours.</p>

</div>

<p class="prompt">2. A phone plan costs $30 per month plus $0.10 per text message beyond the included limit. Which inequality represents t, the number of extra text messages, that keeps the monthly bill under $50?</p>

</div>

<p><span class="label">Explanation:</span> The flat monthly fee, $30, plus $0.10 for each of the t extra texts, must total less than $50, giving 30 + 0.10t < 50.</p>

</div>

<p class="prompt">3. A gym membership costs a $40 sign-up fee plus $25 per month. Kevin has budgeted no more than $215 total. What is the greatest number of months he can attend under this budget?</p>

</div>

<p><span class="label">Explanation:</span> Let m be the number of months. The total cost is 40 + 25m, and it must be no more than 215: 40 + 25m ≤ 215, so 25m ≤ 175, and m ≤ 7. He can attend for at most 7 months.</p>

</div>

<h2 id="compound">5. Compound Inequalities</h2>

<p>Two or more inequalities joined by the word <strong>and</strong> or <strong>or</strong> form a <strong>compound inequality</strong>. A compound inequality with "and" is true only if both individual inequalities are true, its graph is the <strong>intersection</strong> of the two solution sets. A compound inequality with "or" is true if at least one of the inequalities is true, its graph is the <strong>union</strong> of the two solution sets.</p>

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<p class="figure-caption"><strong>−6 < x < −3</strong> ("and") – the segment between the two points</p>

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<p class="figure-caption"><strong>x ≤ 4 or x ≥ 5</strong> ("or") – the two outer rays</p>

</div>

<p><strong>Worked Example:</strong> Solve 9 − 3x > 18 and 5 + 2x > −7.</p>

<p class="step-math">x < −3   and   x > −6   (dividing by −3 flips the first symbol)</p>

<p>Combined, the solution is −6 < x < −3.</p>

</div>

<p class="prompt">1. Solve: x + 4 < 9 and x − 2 > −6</p>

</div>

<p><span class="label">Explanation:</span> x + 4 < 9 gives x < 5. x − 2 > −6 gives x > −4. Combined with "and," the solution is −4 < x < 5.</p>

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<p class="prompt">2. A number line graph shows open circles at −3 and 4, with the segment between them shaded. Which compound inequality does this represent?</p>

</div>

<p><span class="label">Explanation:</span> Open circles mean −3 and 4 are excluded, and shading between them means only values strictly between the two are included: −3 < x < 4.</p>

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<p class="prompt">3. Solve: 2x − 1 ≤ 7 or 3x + 5 > 20</p>

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<p><span class="label">Explanation:</span> 2x − 1 ≤ 7 gives x ≤ 4. 3x + 5 > 20 gives x > 5. Combined with "or," any x satisfying either condition works: x ≤ 4 or x > 5.</p>

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<p class="prompt">4. A number x satisfies −5 < x ≤ 8. Which of the following is NOT a possible value of x?</p>

</div>

<p><span class="label">Explanation:</span> The strict inequality −5 < x means −5 itself is excluded. Since the other symbol is ≤, 8 is included. Only −5 is not a possible value.</p>

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<p class="prompt">5. The solution set of a compound inequality joined by "and" is called the ___ of the two individual solution sets.</p>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g1_pe-5',false)">A) union</button>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'g1_pe-5',true)">B) intersection</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g1_pe-5',false)">C) complement</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g1_pe-5',false)">D) product</button>

</div>

<p><span class="label">Explanation:</span> "And" requires both conditions to be true at once, which describes the intersection, the overlap, of the two solution sets. "Or" describes the union instead.</p>

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<h2 id="absolute-value">6. Absolute Value Inequalities</h2>

<p>Since |x| represents the distance from x to zero on a number line, absolute value inequalities translate directly into compound inequalities. For any positive number c:</p>

<tr><th>Inequality</th><th>Equivalent compound inequality</th></tr>

</table>

<p>In the table above, < can be replaced with ≤, and > can be replaced with ≥. The intuition: |x| < c means x is within c units of zero, so x is trapped between −c and c. And |x| > c means x is farther than c units from zero, so x is either far to the left or far to the right.</p>

<p><strong>Worked Example:</strong> Solve |2x − 3| ≤ 9.</p>

<p class="step-math">−6 ≤ 2x ≤ 12   (Add 3 to each expression)</p>

<p class="step-math">−3 ≤ x ≤ 6   (Divide each expression by 2)</p>

</div>

<p class="prompt">1. Solve |x| < 7.</p>

</div>

<p><span class="label">Explanation:</span> |x| < 7 means x is within 7 units of zero, so −7 < x < 7.</p>

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<p class="prompt">2. Solve |x| > 5.</p>

</div>

<p><span class="label">Explanation:</span> |x| > 5 means x is farther than 5 units from zero, so x < −5 or x > 5.</p>

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<p class="prompt">3. Solve |3x + 6| ≤ 15.</p>

</div>

<p><span class="label">Explanation:</span> Rewrite as −15 ≤ 3x + 6 ≤ 15. Subtract 6 from each expression: −21 ≤ 3x ≤ 9. Divide each expression by 3: −7 ≤ x ≤ 3.</p>

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<p class="prompt">4. Solve |4x − 5| > 11.</p>

</div>

<p><span class="label">Explanation:</span> Rewrite as 4x − 5 < −11 or 4x − 5 > 11. First case: 4x < −6, so x < −<span class="frac"><span class="num">3</span><span class="den">2</span></span>. Second case: 4x > 16, so x > 4.</p>

</div>

<p class="prompt">5. For which value of c does |x + 4| < c have no solution?</p>

</div>

<p><span class="label">Explanation:</span> Since |x + 4| is never negative, the inequality |x + 4| < c has no solution whenever c is zero or negative. Only c = −2 fits that condition.</p>

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<h2 id="applications">7. Tolerance and Margin-of-Error Applications</h2>

<p>One of the most common real-world uses of absolute value inequalities is describing how far a measured value is allowed to differ from a target value, often called a <strong>tolerance</strong> or <strong>margin of error</strong>. If a quantity x must stay within a tolerance of t units from a target value v, that's written |x − v| ≤ t.</p>

<p><strong>Worked Example:</strong> A recipe calls for 200 grams of flour, and a scale reading r is considered acceptable if it's within 8 grams of the target. Write an inequality for the acceptable readings, and find the resulting range.</p>

</div>

<p class="prompt">1. A machine fills bags of flour to a target weight of 50 grams. A bag is acceptable if its weight w differs from the target by no more than 0.5 grams. Which inequality represents the acceptable weights?</p>

</div>

<p><span class="label">Explanation:</span> The weight w must be within 0.5 grams of the target, 50, and "no more than" includes equality, so |w − 50| ≤ 0.5.</p>

</div>

<p class="prompt">2. Using the inequality from the previous problem, what is the range of acceptable weights?</p>

</div>

<p><span class="label">Explanation:</span> |w − 50| ≤ 0.5 becomes −0.5 ≤ w − 50 ≤ 0.5. Add 50 to each expression: 49.5 ≤ w ≤ 50.5.</p>

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<p class="prompt">3. A thermostat keeps a room's temperature T within 3 degrees of 70°F. What is the resulting range of acceptable temperatures?</p>

</div>

<p><span class="label">Explanation:</span> The condition is |T − 70| ≤ 3, which becomes −3 ≤ T − 70 ≤ 3. Add 70 to each expression: 67 ≤ T ≤ 73.</p>

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<p class="prompt">4. The inequality |x − 6| ≤ 4 is equivalent to which compound inequality?</p>

</div>

<p><span class="label">Explanation:</span> Rewrite as −4 ≤ x − 6 ≤ 4. Add 6 to each expression: 2 ≤ x ≤ 10.</p>

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<h2 id="basics">8. What Absolute Value Means</h2>

<p>The <strong>absolute value</strong> of a number is its distance from zero on a number line. Distance is never negative, so absolute value is always zero or positive, no matter what sign the original number had. The absolute value of a number n is written |n|.</p>

<text x="105" y="10" font-size="11" fill="#4285f4">distance 8</text>

<text x="168" y="10" font-size="11" fill="#ea4335">distance 8</text>

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<p class="figure-caption">|−8| = 8 and |8| = 8, since both are 8 units from zero.</p>

</div>

<p><strong>Worked Example:</strong> The distance between −3 and the origin is 3, so |−3| = 3. The distance between 3 and the origin is also 3, so |3| = 3. Therefore, if |x| = 3, then x = 3 or x = −3.</p>

</div>

</div>

<p><span class="label">Explanation:</span> −9 is 9 units from zero on the number line, so |−9| = 9.</p>

</div>

<p class="prompt">2. What is the distance between −12 and 0 on the number line?</p>

</div>

<p><span class="label">Explanation:</span> Distance is always nonnegative, so the distance between −12 and 0 is 12, which matches |−12| = 12.</p>

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<p class="prompt">3. If |x| = 15, what are the possible values of x?</p>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g2_pa-3',false)">D) There is no solution</button>

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<p><span class="label">Explanation:</span> Both 15 and −15 are exactly 15 units from zero, so both values make the equation true.</p>

</div>

<p class="prompt">4. Which value of x makes |x − 4| equal to 0?</p>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g2_pa-4',false)">D) x can be any number</button>

</div>

<p><span class="label">Explanation:</span> The only number with an absolute value of 0 is 0 itself, so x − 4 = 0, which means x = 4. This is the only solution, since a quantity is 0 units from zero only when it equals exactly zero.</p>

</div>

<h2 id="solving">9. Solving Absolute Value Equations</h2>

<p>To solve an absolute value equation like |expression| = a positive number, remember that whatever is inside the bars could have been positive or negative before the absolute value was applied. So you split the equation into two cases: one where the expression equals the positive value, and one where it equals the negative value, then solve each case separately.</p>

<p><strong>Worked Example:</strong> Solve |4x − 3| = 13.</p>

</div>

<p class="prompt">1. Solve |2x + 1| = 9.</p>

</div>

<p><span class="label">Explanation:</span> Split into two cases: 2x + 1 = 9 gives 2x = 8, so x = 4. And 2x + 1 = −9 gives 2x = −10, so x = −5.</p>

</div>

<p class="prompt">2. Solve |5x − 2| = 18.</p>

</div>

<p><span class="label">Explanation:</span> Split into two cases: 5x − 2 = 18 gives 5x = 20, so x = 4. And 5x − 2 = −18 gives 5x = −16, so x = −<span class="frac"><span class="num">16</span><span class="den">5</span></span>.</p>

</div>

<p class="prompt">3. Solve |x + 7| = 0.</p>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g2_pb-3',false)">D) There is no solution</button>

</div>

<p><span class="label">Explanation:</span> Since |0| = 0 is the only case, there's only one equation to solve, not two: x + 7 = 0, so x = −7. When an absolute value expression equals 0, there's exactly one solution instead of two.</p>

</div>

<p class="prompt">4. Solve 2|3x − 1| = 16.</p>

</div>

<p><span class="label">Explanation:</span> First divide both sides by 2: |3x − 1| = 8. Then split into two cases: 3x − 1 = 8 gives 3x = 9, so x = 3. And 3x − 1 = −8 gives 3x = −7, so x = −<span class="frac"><span class="num">7</span><span class="den">3</span></span>.</p>

</div>

<h2 id="no-solution">10. Recognizing No-Solution Equations</h2>

<p>Since absolute value can never produce a negative result, an equation like |expression| = a negative number has no solution at all, no matter what the expression is. You can recognize these instantly without doing any algebra: just check the sign of the number on the other side of the equal sign.</p>

<p><strong>Worked Example:</strong> Solve |x − 6| = −4.</p>

<p>Since −4 is negative, and an absolute value can never equal a negative number, this equation has <strong>no solution</strong>. There's no need to attempt splitting it into cases.</p>

</div>

<p class="prompt">1. Solve |2x + 5| = −3.</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'g2_pc-1',true)">C) No solution</button>

</div>

<p><span class="label">Explanation:</span> The right side, −3, is negative. Since absolute value is never negative, this equation has no solution.</p>

</div>

<p class="prompt">2. Which of the following equations has no solution?</p>

</div>

<p><span class="label">Explanation:</span> Only |x + 1| = −6 sets an absolute value expression equal to a negative number, so it's the only equation with no solution.</p>

</div>

<p class="prompt">3. Which value of k guarantees that |x + 3| = k has no solution?</p>

</div>

<p><span class="label">Explanation:</span> The equation has no solution whenever k is negative. Only k = −2 is negative; k = 0 gives exactly one solution, and k = 5 or k = 10 each give two solutions.</p>

</div>

<p class="prompt">4. How many solutions does |x| = −7 have?</p>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g2_pc-4',false)">D) Infinitely many</button>

</div>

<p><span class="label">Explanation:</span> No value of x has an absolute value of −7, since absolute value is always zero or positive. This equation has 0 solutions.</p>

</div>

<h2 id="graph">11. The Absolute Value Function and Its Graph</h2>

<p>The <strong>absolute value function</strong> is written f(x) = |x|. Its graph is a V-shape: it comes down from the upper left, touches its lowest point at the origin, and rises back up to the upper right. A more general absolute value function, y = a|x − h| + k, has its vertex, the sharp point of the V, at (h, k). If a > 0, the graph opens upward and the vertex is the minimum point; if a < 0, the graph opens downward and the vertex is the maximum point.</p>

<text x="115" y="185" font-size="11" fill="#202124">Vertex (0, 0)</text>

</svg>

<p class="figure-caption">The graph of f(x) = |x|, a V-shape with vertex at the origin.</p>

</div>

<p><strong>Worked Example:</strong> What is the vertex of f(x) = 2|x − 3| + 5, and does it open upward or downward?</p>

<p>Comparing to y = a|x − h| + k, h = 3 and k = 5, so the vertex is (3, 5). Since a = 2 is positive, the graph opens upward and the vertex is the minimum point.</p>

</div>

<p class="prompt">1. What is the vertex of the graph of f(x) = |x + 4| − 2?</p>

</div>

<p><span class="label">Explanation:</span> Rewrite as |x − (−4)| + (−2) to match y = a|x − h| + k. This gives h = −4 and k = −2, so the vertex is (−4, −2).</p>

</div>

<p class="prompt">2. What is the vertex of g(x) = −3|x − 1| + 6, and does the graph open upward or downward?</p>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g2_pd-2',false)">A) (1, 6), opens upward</button>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'g2_pd-2',true)">B) (1, 6), opens downward</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g2_pd-2',false)">C) (−1, 6), opens downward</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g2_pd-2',false)">D) (1, −6), opens upward</button>

</div>

<p><span class="label">Explanation:</span> The vertex is (1, 6), read directly from h = 1 and k = 6. Since a = −3 is negative, the graph opens downward.</p>

</div>

<p class="prompt">3. What is the value of f(x) = |x − 5| when x = 2?</p>

</div>

<p><span class="label">Explanation:</span> Substitute x = 2: f(2) = |2 − 5| = |−3| = 3.</p>

</div>

<p class="prompt">4. For the graph of y = a|x| + k, if a > 0, the vertex is the graph's...</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'g2_pd-4',true)">A) minimum point</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g2_pd-4',false)">B) maximum point</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g2_pd-4',false)">C) x-intercept only</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g2_pd-4',false)">D) y-intercept only</button>

</div>

<p><span class="label">Explanation:</span> When a > 0, the graph opens upward like a V, so the vertex is the lowest point on the graph, the minimum.</p>

</div>

<h2 id="solutions">12. Solutions to Two-Variable Inequalities</h2>

<p>A <strong>linear inequality in two variables</strong> can be written in one of the forms ax + by < c, ax + by ≤ c, ax + by > c, or ax + by ≥ c. An ordered pair (a, b) is a <strong>solution</strong> to the inequality if the inequality is true when a and b are substituted for x and y. Unlike a linear equation, which has infinitely many solutions forming a line, a linear inequality has infinitely many solutions filling an entire half of the coordinate plane.</p>

<p><strong>Worked Example:</strong> Is (3, −1) a solution to 2x − y ≤ 5?</p>

<p>Is 7 ≤ 5? No, so (3, −1) is <strong>not</strong> a solution.</p>

</div>

<p class="prompt">1. Is (2, 4) a solution to y < 3x − 1?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'g3_pa-1',true)">A) Yes, since 4 < 5</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g3_pa-1',false)">B) No, since 4 < 5 is false</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g3_pa-1',false)">C) Yes, since 2 < 4</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g3_pa-1',false)">D) No, since 4 > 5</button>

</div>

<p><span class="label">Explanation:</span> Substitute: 3(2) − 1 = 5. Is 4 < 5? Yes, so (2, 4) is a solution.</p>

</div>

<p class="prompt">2. Is (−1, 2) a solution to 2x + y ≥ 0?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'g3_pa-2',true)">A) Yes, since 0 ≥ 0</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g3_pa-2',false)">B) No, since 0 is not greater than 0</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g3_pa-2',false)">C) Yes, since −1 < 2</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g3_pa-2',false)">D) No, since the sum is negative</button>

</div>

<p><span class="label">Explanation:</span> Substitute: 2(−1) + 2 = −2 + 2 = 0. Is 0 ≥ 0? Yes, this is true (the symbol includes equality), so (−1, 2) is a solution.</p>

</div>

<p class="prompt">3. Which of the following points is NOT a solution to y ≤ −2x + 5?</p>

</div>

<p><span class="label">Explanation:</span> For (1, 4): −2(1) + 5 = 3. Is 4 ≤ 3? No, so this point fails. Checking the others: (0,0) gives 0 ≤ 5, true; (2,0) gives 0 ≤ 1, true; (−1,6) gives 6 ≤ 7, true.</p>

</div>

<p class="prompt">4. For the inequality 3x − 2y > 6, which point is a solution?</p>

</div>

<p><span class="label">Explanation:</span> For (4, 0): 3(4) − 2(0) = 12. Is 12 > 6? Yes. Checking the others: (0,0) gives 0 > 6, false; (1,1) gives 1 > 6, false; (0,3) gives −6 > 6, false.</p>

</div>

<h2 id="boundary">13. Boundary Lines and the Test-Point Method</h2>

<p>The equation you get by replacing the inequality symbol with an equal sign is called the <strong>boundary line</strong>. It divides the coordinate plane into two half-planes. If the inequality symbol is strict (< or >), the boundary line is <strong>dashed</strong>, since points on the line itself aren't included. If the symbol includes equality (≤ or ≥), the boundary line is <strong>solid</strong>.</p>

<p>To figure out which half-plane is the solution set, use the <strong>test-point method</strong>: pick any point not on the boundary line, usually the origin (0, 0) since it's easy to compute with, and substitute it into the original inequality. If the result is true, shade the half-plane containing that point. If false, shade the other half-plane. If the boundary line passes through the origin, pick a different test point instead.</p>

</svg>

<p class="figure-caption"><strong>Dashed line, shaded above</strong> – a strict inequality like y > mx + b</p>

</div>

</svg>

<p class="figure-caption"><strong>Solid line, shaded below</strong> – a non-strict inequality like y ≤ mx + b</p>

</div>

<p><strong>Worked Example:</strong> Determine whether the boundary line for 2y + x ≤ 4 is solid or dashed, and which side of the line contains the shaded solution set.</p>

<p>Since the symbol is ≤, the boundary line is solid.</p>

<p>Testing the origin: 2(0) + 0 = 0. Is 0 ≤ 4? Yes, so the shaded region is the side of the line containing the origin.</p>

</div>

<p class="prompt">1. Is the boundary line for y > 3x − 2 solid or dashed?</p>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g3_pb-1',false)">A) Solid</button>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'g3_pb-1',true)">B) Dashed</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g3_pb-1',false)">C) Depends on the value of x</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g3_pb-1',false)">D) Cannot be determined</button>

</div>

<p><span class="label">Explanation:</span> The symbol > is strict, meaning points on the line itself are not solutions, so the boundary line is dashed.</p>

</div>

<p class="prompt">2. For the inequality 4x − y ≥ 8, should the side containing the origin be shaded?</p>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g3_pb-2',false)">A) Yes, since 0 ≥ 8 is true</button>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'g3_pb-2',true)">B) No, since 0 ≥ 8 is false</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g3_pb-2',false)">C) Yes, the origin is always a solution</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g3_pb-2',false)">D) Cannot be determined without graphing</button>

</div>

<p><span class="label">Explanation:</span> Testing the origin: 4(0) − 0 = 0. Is 0 ≥ 8? No, so the origin is not a solution, and the side of the line containing the origin should not be shaded.</p>

</div>

<p class="prompt">3. Is (0, 0) a valid test point for 3x − 2y > 2, and what does testing it tell you?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'g3_pb-3',true)">A) Valid; since 0 > 2 is false, shade the side not containing the origin</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g3_pb-3',false)">B) Valid; since 0 > 2 is true, shade the side containing the origin</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g3_pb-3',false)">C) Invalid, since (0, 0) lies on the boundary line</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g3_pb-3',false)">D) Invalid, since the inequality has no constant term</button>

</div>

<p><span class="label">Explanation:</span> The boundary line 3x − 2y = 2 does not pass through (0, 0), since 3(0) − 2(0) = 0 ≠ 2, so (0, 0) is a valid test point. Substituting gives 0 > 2, which is false, so the shaded region is the side of the line that does not contain the origin.</p>

</div>

<p class="prompt">4. Why can't (0, 0) be used as a test point for the inequality y > 2x?</p>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g3_pb-4',false)">A) Because (0, 0) makes the inequality true</button>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'g3_pb-4',true)">B) Because (0, 0) lies on the boundary line itself</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g3_pb-4',false)">C) Because (0, 0) is not a real ordered pair</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g3_pb-4',false)">D) Because the inequality is strict</button>

</div>

<p><span class="label">Explanation:</span> The boundary line y = 2x passes directly through the origin, since 2(0) = 0. A test point can't lie on the boundary line itself, so a different point, like (1, 0), should be used instead.</p>

</div>

<p class="prompt">5. What is the boundary line equation for 2y + x ≤ 4, written in slope-intercept form?</p>

</div>

<p><span class="label">Explanation:</span> Solve for y: 2y = 4 − x, so y = −<span class="frac"><span class="num">1</span><span class="den">2</span></span>x + 2.</p>

</div>

<h2 id="matching">14. Matching Graphs to Inequalities</h2>

<p>On the Digital SAT, this skill is tested by describing or showing a graph and asking which inequality matches it, or the reverse: giving an inequality and asking which graph matches. Reading a graph means noting three things: the slope and y-intercept of the boundary line, whether the line is solid or dashed, and which side is shaded.</p>

<p><strong>Worked Example:</strong> A graph shows a dashed line passing through (0, −3) and (1, −1), with the region above the line shaded. Which inequality does this graph represent?</p>

<p>Find the slope: m = <span class="frac"><span class="num">−1 − (−3)</span><span class="den">1 − 0</span></span> = 2. The y-intercept is −3, so the boundary line is y = 2x − 3.</p>

<p>Since the line is dashed, the symbol is strict. Since the shading is above the line, the symbol is >.</p>

</div>

<p class="prompt">1. A graph shows a dashed line passing through (0, −3) and (1, −1), with the region above the line shaded. Which inequality does this graph represent?</p>

</div>

<p><span class="label">Explanation:</span> This is the worked example above: slope 2, y-intercept −3, dashed line (strict), shaded above (>), giving y > 2x − 3.</p>

</div>

<p class="prompt">2. A graph shows a solid line passing through (4, 0) and (0, 4), with the region containing the origin shaded. Which inequality does this graph represent?</p>

</div>

<p><span class="label">Explanation:</span> The line through (4, 0) and (0, 4) has slope −1 and equation x + y = 4. The line is solid, so the symbol includes equality. Testing the origin: 0 + 0 = 0, and since the origin is shaded, 0 must satisfy the inequality, so the symbol is ≤, giving x + y ≤ 4.</p>

</div>

<p class="prompt">3. Which description matches the graph of y ≤ −3x + 6?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'g3_pc-3',true)">A) Solid line, y-intercept 6, slope −3, shaded below the line</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g3_pc-3',false)">B) Dashed line, y-intercept 6, slope −3, shaded below the line</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g3_pc-3',false)">C) Solid line, y-intercept 6, slope −3, shaded above the line</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g3_pc-3',false)">D) Solid line, y-intercept −6, slope 3, shaded below the line</button>

</div>

<p><span class="label">Explanation:</span> The symbol ≤ includes equality, so the line is solid. The line has slope −3 and y-intercept 6. Since y is less than or equal to the line's expression, the shading is below the line.</p>

</div>

<p class="prompt">4. The point (2, 5) is tested in the inequality y > 4x − 1. Is (2, 5) a solution?</p>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g3_pc-4',false)">A) Yes, because 5 > 7</button>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'g3_pc-4',true)">B) No, because 5 is not greater than 7</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g3_pc-4',false)">C) Yes, because 2 < 5</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'g3_pc-4',false)">D) No, because 2 is too small</button>

</div>

<p><span class="label">Explanation:</span> Substitute: 4(2) − 1 = 7. Is 5 > 7? No, so (2, 5) is not a solution.</p>

</div>

<p class="prompt">5. A graph shows a dashed vertical line at x = 3, with shading to the right of the line. Which inequality does this graph represent?</p>

</div>

<p><span class="label">Explanation:</span> A dashed line means the symbol is strict, and shading to the right means x-values greater than 3 are included, so the inequality is x > 3.</p>

</div>

<h2 id="mistakes">Common Mistakes to Avoid</h2>

<li><strong>Forgetting to flip the inequality symbol.</strong> This only happens when you multiply or divide both sides by a negative number, never when you add or subtract. It's the single most common inequality error on the SAT.</li>

<li><strong>Misreading "at most" and "at least."</strong> "At most" means less than or equal to (≤), the value has a ceiling. "At least" means greater than or equal to (≥), the value has a floor.</li>

<li><strong>Using an open circle when the symbol includes "or equal to."</strong> A solid, closed circle means that value is included in the solution (≤ or ≥). An open circle means it isn't (< or >).</li>

<li><strong>Mixing up when to use "and" versus "or."</strong> |expression| < c always becomes an "and" style compound inequality (a bounded range). |expression| > c always becomes an "or" style compound inequality (two separate rays).</li>

<li><strong>Forgetting to flip both inequality symbols when isolating x in a three-part inequality.</strong> If you multiply or divide all three parts of −c < ax + b < c by a negative number, both symbols flip, not just one.</li>

<li><strong>Assuming every absolute value inequality has a solution.</strong> |expression| < a negative number has no solution, since absolute value can never be negative.</li>

<li><strong>Forgetting the negative case entirely.</strong> It's easy to solve only expression = positive value and stop. Always write out both cases before solving.</li>

<li><strong>Not checking the sign on the other side first.</strong> Before splitting into cases, glance at the number the absolute value is set equal to. If it's negative, you can immediately write "no solution" and skip the algebra.</li>

<li><strong>Distributing a negative sign incorrectly when writing the second case.</strong> The second case is expression = −(the number), not −expression = the number. Keep track of which side the negative sign belongs on.</li>

<li><strong>Assuming every absolute value equation has two solutions.</strong> An equation like |x + 7| = 0 has only one solution, and an equation with a negative value on the other side has zero solutions. Only a positive value on the other side guarantees two solutions.</li>

<li><strong>Reading the vertex coordinates with the wrong sign for h.</strong> In y = a|x − h| + k, a plus sign inside the bars, like |x + 4|, means h = −4, not 4. Rewrite addition as subtracting a negative before identifying h.</li>

<li><strong>Using the origin as a test point when it lies on the boundary line.</strong> If the line passes through (0, 0), which happens whenever there's no constant term, substituting the origin gives 0 = 0 and tells you nothing about which side to shade. Pick a different point, like (1, 0) or (0, 1), instead.</li>

<li><strong>Mixing up which side "above" and "below" refer to.</strong> "Above the line" means larger y-values for a given x, which is the side of the graph containing points with y greater than the line's expression. It's easy to misjudge this visually when the line has a steep or negative slope.</li>

<li><strong>Forgetting that a solid line means the boundary itself is part of the solution.</strong> Points exactly on a solid boundary line satisfy the inequality; points on a dashed boundary line do not.</li>

<li><strong>Reversing the inequality symbol when converting standard form to slope-intercept form.</strong> Just like with equations, dividing an inequality by a negative number to isolate y requires flipping the inequality symbol.</li>

<li><strong>Assuming a point that works in the boundary equation also works in the inequality.</strong> A point can satisfy the boundary equation exactly (be on the line) without satisfying a strict inequality, since a strict inequality excludes the line itself.</li>

</ul>

<h2 id="faq">Frequently Asked Questions</h2>

<h3>Is solving inequalities actually tested on the Digital SAT?</h3>

<p>Yes. "Linear inequalities in 1 or 2 variables" is one of the officially listed question types in the Algebra domain of the Digital SAT Math section, alongside linear equations, linear functions, and systems of equations. Compound and absolute value inequalities aren't singled out as a separate named category and appear less often, usually through a tolerance word problem.</p>

</div>

<h3>Why does multiplying by a negative number flip the inequality?</h3>

<p>Think of a number line: 2 < 5 is true. Multiply both sides by −1, and you get −2 and −5. But −2 is actually greater than −5, since it's farther to the right. So the inequality has to flip, −2 > −5, to stay true.</p>

</div>

<h3>How can I tell quickly whether an absolute value inequality needs "and" or "or"?</h3>

<p>|x| < c describes values within a fixed distance of zero, a bounded range that needs "and" (or, equivalently, a three-part inequality). |x| > c describes values farther than a fixed distance from zero, two separate unbounded pieces that need "or."</p>

</div>

<h3>Why does an absolute value equation usually have two solutions?</h3>

<p>Because two different numbers, one positive and one negative, can have the same absolute value. For example, both 5 and −5 have an absolute value of 5, so any equation that simplifies to |something| = 5 needs to account for both possibilities.</p>

</div>

<h3>Can I always tell how many solutions an absolute value equation has without solving it?</h3>

<p>Yes. Look at what the absolute value expression is set equal to. If it's positive, expect two solutions. If it's exactly 0, expect one solution. If it's negative, expect no solution. Checking this first can save you from doing unnecessary algebra.</p>

</div>

<h3>How is the absolute value function different from a linear function?</h3>

<p>A linear function graphs as a single straight line, while the absolute value function is really two linear pieces joined at the vertex, one with a positive slope and one with a negative slope. That's why its graph is a V-shape instead of a single straight line.</p>

</div>

<h3>Do I need to memorize how to sketch these graphs for the SAT?</h3>

<p>No. Since the Digital SAT never asks you to draw a graph, the more valuable skill is reading a graph that's already shown to you, or reasoning about an inequality algebraically using test points, exactly what this lesson focuses on.</p>

</div>

<h3>Is there a shortcut for figuring out which side to shade without picking a test point?</h3>

<p>If the inequality is already solved for y, like y > mx + b, the symbol tells you directly: > or ≥ means shade above the line, and < or ≤ means shade below the line. The test-point method is most useful as a double-check, or when the inequality is still in standard form.</p>

</div>

<h3>What's the difference between a solution to an equation and a solution to an inequality?</h3>

<p>A linear equation's solutions form a single line, every point exactly on that line. A linear inequality's solutions fill an entire half of the plane, either every point on one side of the line, or every point on that side plus the line itself.</p>

</div>

<h3>Where can I practice more problems like these?</h3>

<p>The <a href="https://theschoolofmathematics.com/quiz/sat-inequalities-quiz-1">Inequalities quizzes</a> in the SAT Math Question Bank include additional original problems on this topic, along with quizzes covering every other Algebra, Advanced Math, Problem-Solving and Data Analysis, and Geometry and Trigonometry skill on the Digital SAT.</p>

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<a class="cta-btn cta-primary" href="https://theschoolofmathematics.com/quiz/sat-inequalities-quiz-1">Practice Inequalities Free</a>

<a class="cta-btn cta-secondary" href="https://theschoolofmathematics.com/quiz/course/SAT-Math-Qbank">Explore the Full SAT Math Qbank</a>

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