Solving Linear Equations | Free SAT Math Course

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<title>Solving Linear Equations - Free SAT Math Lesson & Practice | The School of Mathematics</title>

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<h1>Solving Linear Equations: Complete SAT Math Lesson</h1>

Solving linear equations is the single most tested skill in the Heart of Algebra domain of the Digital SAT, and it quietly underlies most of the Advanced Math domain too, since factoring, function evaluation, and system-solving all eventually reduce to isolating a variable. This free lesson walks through every equation type you will meet on test day: translating word problems into equations, applying the properties of equality, working through multi-step and distributive-property equations, solving equations with variables on both sides, recognizing when an equation has no solution or is an identity, and rearranging formulas to solve for a specific variable. Every rule below is followed by a fully worked example and then a set of original practice problems with instant feedback, so you can check your understanding topic by topic before moving on. All of the practice problems here are free, and once you've worked through the lesson you can keep building speed with the full SAT Math Question Bank linked below.

</p>

<a class="cta-btn cta-primary" href="https://theschoolofmathematics.com/quiz/sat-solving-equations-quiz-1">Practice Solving Equations 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="#writing-equations">Writing Equations from Word Sentences</a></li>

<li><a href="#properties-of-equality">Solving Equations with the Properties of Equality</a></li>

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

<li><a href="#both-sides">Equations with Variables on Both Sides</a></li>

<li><a href="#no-solution-identity">No-Solution and Identity Equations</a></li>

<li><a href="#literal-equations">Solving Formulas for a Specific Variable</a></li>

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

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

</ol>

</nav>

<h2 id="writing-equations">1. Writing Equations from Word Sentences</h2>

<p>An equation is a mathematical sentence built around an equal sign. On the SAT, many "solve for x" problems don't start as equations at all, they start as a sentence, and your first job is to define a variable and translate the sentence piece by piece. The table below shows how common phrases turn into symbols.</p>

<tr><th>Phrase</th><th>Symbol</th></tr>

<tr><td>increased by, more than, the sum of</td><td>+</td></tr>

<tr><td>decreased by, less than, the difference of</td><td>−</td></tr>

<tr><td>times, the product of, twice, triple</td><td>×</td></tr>

<tr><td>divided by, the quotient of, half of</td><td>÷</td></tr>

<tr><td>is, was, equals, is identical to, is the same as</td><td>=</td></tr>

</table>

<p><strong>Worked Example:</strong> Twice a number, increased by nine, is identical to twenty-three. Find the number.</p>

<p>Let <em>n</em> be the number.</p>

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<p class="prompt">1. Three times a number, decreased by seven, equals eleven. What is the number?</p>

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<p><span class="label">Explanation:</span> Let <em>n</em> be the number. "Three times a number, decreased by seven" translates to 3n − 7. Setting this equal to 11 gives 3n − 7 = 11, so 3n = 18, and n = 6.</p>

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<p class="prompt">2. The sum of five and twice a number is the same as the number increased by nine. What is the number?</p>

</div>

<p><span class="label">Explanation:</span> Let <em>n</em> be the number. The sentence becomes 5 + 2n = n + 9. Subtracting <em>n</em> from both sides gives 5 + n = 9, so n = 4.</p>

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<p class="prompt">3. The sum of three consecutive even integers is 96. What is the greatest of the three integers?</p>

</div>

<p><span class="label">Explanation:</span> Consecutive even integers differ by 2, so if <em>n</em> is the first, the three integers are n, n + 2, and n + 4. Their sum is n + (n + 2) + (n + 4) = 96, so 3n + 6 = 96, 3n = 90, and n = 30. The three integers are 30, 32, and 34, so the greatest is 34.</p>

</div>

<p class="prompt">4. Half of a number, plus three, is identical to the number decreased by five. What is the number?</p>

</div>

<p><span class="label">Explanation:</span> Let <em>n</em> be the number. The sentence becomes <span class="frac"><span class="num">1</span><span class="den">2</span></span>n + 3 = n − 5. Subtracting <span class="frac"><span class="num">1</span><span class="den">2</span></span>n from both sides gives 3 = <span class="frac"><span class="num">1</span><span class="den">2</span></span>n − 5, so 8 = <span class="frac"><span class="num">1</span><span class="den">2</span></span>n, and n = 16.</p>

</div>

<h2 id="properties-of-equality">2. Solving Equations with the Properties of Equality</h2>

<p>To solve an equation means to find every value of the variable that makes the equation a true statement. The core strategy is to isolate the variable, a coefficient of 1, on one side of the equal sign, by applying the same operation to both sides. These operations are called the properties of equality.</p>

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

<tr><td>Addition Property</td><td>If a = b, then a + c = b + c</td><td>x − 3 = 5 → x = 8</td></tr>

<tr><td>Subtraction Property</td><td>If a = b, then a − c = b − c</td><td>x + 2 = 6 → x = 4</td></tr>

<tr><td>Multiplication Property</td><td>If a = b, then ca = cb</td><td><span class="frac"><span class="num">1</span><span class="den">2</span></span>x = 3 → x = 6</td></tr>

<tr><td>Division Property</td><td>If a = b and c ≠ 0, then <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><td>3x = 15 → x = 5</td></tr>

</table>

<p><strong>Worked Example:</strong> Solve −16 = 4y.</p>

<p class="step-math"><span class="frac"><span class="num">−16</span><span class="den">4</span></span> = <span class="frac"><span class="num">4y</span><span class="den">4</span></span>   (Divide each side by 4)</p>

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<p class="prompt">1. Solve for x: x + (−15) = −22</p>

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<p><span class="label">Explanation:</span> Add 15 to both sides: x + (−15) + 15 = −22 + 15, so x = −7.</p>

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<p class="prompt">2. Solve for y: −18 = 6y</p>

</div>

<p><span class="label">Explanation:</span> Divide both sides by 6: <span class="frac"><span class="num">−18</span><span class="den">6</span></span> = <span class="frac"><span class="num">6y</span><span class="den">6</span></span>, so y = −3.</p>

</div>

<p class="prompt">3. Solve for x: <span class="frac"><span class="num">2</span><span class="den">3</span></span>x = 10</p>

</div>

<p><span class="label">Explanation:</span> Multiply both sides by the reciprocal, <span class="frac"><span class="num">3</span><span class="den">2</span></span>: x = 10 × <span class="frac"><span class="num">3</span><span class="den">2</span></span> = 15.</p>

</div>

<p class="prompt">4. Solve for x: 5x − 3 = 22</p>

</div>

<p><span class="label">Explanation:</span> Add 3 to both sides: 5x = 25. Divide both sides by 5: x = 5.</p>

</div>

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

<p>Many SAT equations require more than one operation to isolate the variable. When grouping symbols like parentheses are involved, start by using the distributive property to remove them, then simplify each side before applying the properties of equality.</p>

<p><strong>Worked Example:</strong> Solve 6 − 2(3x − 1) = 26.</p>

<p class="step-math">6 − 6x + 2 = 26   (Distributive property)</p>

<p class="step-math">8 − 6x = 26   (Simplify)</p>

<p class="step-math">−6x = 18   (Subtract 8 from each side)</p>

</div>

<p class="prompt">1. Solve: <span class="frac"><span class="num">2</span><span class="den">3</span></span>(x − 6) − <span class="frac"><span class="num">1</span><span class="den">3</span></span>(x − 9) = 15</p>

</div>

<p><span class="label">Explanation:</span> Distribute: <span class="frac"><span class="num">2</span><span class="den">3</span></span>x − 4 − <span class="frac"><span class="num">1</span><span class="den">3</span></span>x + 3 = 15. Combine like terms: <span class="frac"><span class="num">1</span><span class="den">3</span></span>x − 1 = 15, so <span class="frac"><span class="num">1</span><span class="den">3</span></span>x = 16, and x = 48.</p>

</div>

<p class="prompt">2. Solve: 4(2x − 3) = 5x + 9</p>

</div>

<p><span class="label">Explanation:</span> Distribute: 8x − 12 = 5x + 9. Subtract 5x from each side: 3x − 12 = 9. Add 12: 3x = 21, so x = 7.</p>

</div>

<p class="prompt">3. Solve: 12 − 3(x − 2) = 2x + 3</p>

</div>

<p><span class="label">Explanation:</span> Distribute: 12 − 3x + 6 = 2x + 3, so 18 − 3x = 2x + 3. Add 3x to each side: 18 = 5x + 3. Subtract 3: 15 = 5x, so x = 3.</p>

</div>

<p class="prompt">4. Solve: <span class="frac"><span class="num">3</span><span class="den">4</span></span>(x + 8) = <span class="frac"><span class="num">1</span><span class="den">2</span></span>(x + 8) + 5</p>

</div>

<p><span class="label">Explanation:</span> Notice that (x + 8) appears on both sides. Subtracting <span class="frac"><span class="num">1</span><span class="den">2</span></span>(x + 8) from each side leaves <span class="frac"><span class="num">1</span><span class="den">4</span></span>(x + 8) = 5, so x + 8 = 20, and x = 12.</p>

</div>

<h2 id="both-sides">4. Equations with Variables on Both Sides</h2>

<p>Some equations have the variable on both sides of the equal sign. To solve these, first use the addition or subtraction property to rewrite an equivalent equation with all the variable terms on one side, then finish isolating the variable as usual. If the equation contains grouping symbols, distribute first.</p>

<p><strong>Worked Example:</strong> Solve <span class="frac"><span class="num">7</span><span class="den">3</span></span>x − 8 = 6 + <span class="frac"><span class="num">1</span><span class="den">3</span></span>x.</p>

<p class="step-math"><span class="frac"><span class="num">7</span><span class="den">3</span></span>x − <span class="frac"><span class="num">1</span><span class="den">3</span></span>x − 8 = 6   (Subtract <span class="frac"><span class="num">1</span><span class="den">3</span></span>x from each side)</p>

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

</div>

<p class="prompt">1. Solve: 9 − 2(m − 3) = 4m + 3</p>

</div>

<p><span class="label">Explanation:</span> Distribute: 9 − 2m + 6 = 4m + 3, so 15 − 2m = 4m + 3. Add 2m to each side: 15 = 6m + 3. Subtract 3: 12 = 6m, so m = 2.</p>

</div>

<p class="prompt">2. Solve: 6x − 2(x + 1) = 3x + 8</p>

</div>

<p><span class="label">Explanation:</span> Distribute: 6x − 2x − 2 = 3x + 8, so 4x − 2 = 3x + 8. Subtract 3x from each side: x − 2 = 8, so x = 10.</p>

</div>

<p class="prompt">3. Solve: 3(2y + 5) = 7y − 5</p>

</div>

<p><span class="label">Explanation:</span> Distribute: 6y + 15 = 7y − 5. Subtract 6y from each side: 15 = y − 5, so y = 20.</p>

</div>

<p class="prompt">4. Solve: 8 − (3n + 1) = 2(n − 3) + 5</p>

</div>

<p><span class="label">Explanation:</span> Simplify each side: 7 − 3n = 2n − 1. Add 3n to each side: 7 = 5n − 1. Add 1: 8 = 5n, so n = <span class="frac"><span class="num">8</span><span class="den">5</span></span>.</p>

</div>

<h2 id="no-solution-identity">5. No-Solution and Identity Equations</h2>

<p>Not every equation has exactly one solution. It's possible for an equation to have no solution at all, meaning there is no value of the variable that makes it true. It's also possible for an equation to be true for every value of the variable, in which case it is called an identity. You can recognize both cases by fully simplifying each side: if the variable terms cancel out and you're left with a false numerical statement, there's no solution; if they cancel out and leave a true statement, the equation is an identity.</p>

<p><strong>Worked Example:</strong> Determine whether 2(1 − x) + 5x = 3(x + 1) has no solution, is an identity, or has one solution.</p>

<p class="step-math">2 − 2x + 5x = 3x + 3   (Distributive property)</p>

<p class="step-math">2 + 3x = 3x + 3   (Simplify)</p>

<p class="step-math">2 = 3   (Subtract 3x from each side)</p>

<p>Since 2 = 3 is false, the equation has <strong>no solution</strong>.</p>

</div>

<p class="prompt">1. Solve, if possible: 5x − 3(x − 2) = 2x + 10</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p5-1',true)">C) No solution</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-1',false)">D) Identity (all real numbers)</button>

</div>

<p><span class="label">Explanation:</span> Distribute: 5x − 3x + 6 = 2x + 10, so 2x + 6 = 2x + 10. Subtracting 2x from each side leaves 6 = 10, which is false, so the equation has no solution.</p>

</div>

<p class="prompt">2. Solve, if possible: 3(y − 4) = 3y − 12</p>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-2',false)">C) No solution</button>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p5-2',true)">D) Identity (all real numbers)</button>

</div>

<p><span class="label">Explanation:</span> Distribute the left side: 3y − 12 = 3y − 12. Both sides are identical, so this equation is true for every value of y, making it an identity.</p>

</div>

<p class="prompt">3. Solve, if possible: 6a − 4 = 2(3a − 2) + 1</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p5-3',true)">C) No solution</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-3',false)">D) Identity (all real numbers)</button>

</div>

<p><span class="label">Explanation:</span> Distribute the right side: 6a − 4 = 6a − 4 + 1, so 6a − 4 = 6a − 3. Subtracting 6a from each side leaves −4 = −3, which is false, so there is no solution.</p>

</div>

<p class="prompt">4. Solve, if possible: 4(2b + 3) − 5 = 8b + 7</p>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-4',false)">C) No solution</button>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p5-4',true)">D) Identity (all real numbers)</button>

</div>

<p><span class="label">Explanation:</span> Distribute: 8b + 12 − 5 = 8b + 7, so 8b + 7 = 8b + 7. Both sides are identical, so the equation is an identity.</p>

</div>

<h2 id="literal-equations">6. Solving Formulas for a Specific Variable</h2>

<p>A formula, or literal equation, is an equation that relates two or more variables, such as a geometry formula or a science formula. The SAT frequently asks you to solve a formula for one variable in terms of the others. The process is identical to solving a regular equation, treat every other letter as if it were a number, and isolate the variable you're solving for using the same properties of equality and the distributive property.</p>

<p><strong>Worked Example:</strong> Solve C = <span class="frac"><span class="num">5</span><span class="den">9</span></span>(F − 32) for F.</p>

<p class="step-math"><span class="frac"><span class="num">9</span><span class="den">5</span></span>C = F − 32   (Multiply each side by <span class="frac"><span class="num">9</span><span class="den">5</span></span>)</p>

</div>

<p class="prompt">1. Solve 2x − 3y = 12 for y.</p>

</div>

<p><span class="label">Explanation:</span> Subtract 2x from each side: −3y = 12 − 2x. Divide each side by −3: y = <span class="frac"><span class="num">12 − 2x</span><span class="den">−3</span></span>, which simplifies to y = <span class="frac"><span class="num">2x − 12</span><span class="den">3</span></span>.</p>

</div>

<p class="prompt">2. Solve V = <span class="frac"><span class="num">1</span><span class="den">3</span></span>πr<sup>2</sup>h for h.</p>

</div>

<p><span class="label">Explanation:</span> Multiply each side by 3: 3V = πr<sup>2</sup>h. Divide each side by πr<sup>2</sup>: h = <span class="frac"><span class="num">3V</span><span class="den">πr<sup>2</sup></span></span>.</p>

</div>

<p class="prompt">3. The area of a trapezoid is A = <span class="frac"><span class="num">1</span><span class="den">2</span></span>(b<sub>1</sub> + b<sub>2</sub>)h. Solve for b<sub>1</sub>.</p>

</div>

<p><span class="label">Explanation:</span> Multiply each side by 2: 2A = (b<sub>1</sub> + b<sub>2</sub>)h. Divide each side by h: <span class="frac"><span class="num">2A</span><span class="den">h</span></span> = b<sub>1</sub> + b<sub>2</sub>. Subtract b<sub>2</sub> from each side: b<sub>1</sub> = <span class="frac"><span class="num">2A</span><span class="den">h</span></span> − b<sub>2</sub>.</p>

</div>

<p class="prompt">4. The surface area of a cylinder is S = 2πr<sup>2</sup> + 2πrh. Solve for h.</p>

</div>

<p><span class="label">Explanation:</span> Subtract 2πr<sup>2</sup> from each side: S − 2πr<sup>2</sup> = 2πrh. Divide each side by 2πr: h = <span class="frac"><span class="num">S − 2πr<sup>2</sup></span><span class="den">2πr</span></span>.</p>

</div>

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

<li><strong>Forgetting to distribute the negative sign.</strong> In an expression like 8 − (3n + 1), the subtraction applies to both terms inside the parentheses, giving 8 − 3n − 1, not 8 − 3n + 1.</li>

<li><strong>Only performing an operation on one side of the equation.</strong> Whatever you add, subtract, multiply, or divide, it has to happen to the entire left side and the entire right side, not just the term nearest the variable.</li>

<li><strong>Stopping too early on no-solution or identity questions.</strong> If the variable terms cancel out, don't assume you made an error, finish simplifying the numbers on each side. A false statement means no solution; a true statement means an identity.</li>

<li><strong>Losing track of a negative sign when dividing.</strong> When dividing by a negative coefficient, every term on the other side of the equation needs to flip sign, not just the first one.</li>

<li><strong>Mixing up which variable to solve for in a literal equation.</strong> Before doing any algebra, circle the target variable so you always know which letter you're isolating, especially when a formula has three or more letters.</li>

</ul>

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

<h3>Do I need to memorize the properties of equality by name for the SAT?</h3>

<p>No. The SAT never asks you to name a property, it only tests whether you can apply them correctly. What matters is the habit: whatever you do to one side of the equation, do the exact same thing to the other side.</p>

</div>

<h3>How can I tell quickly whether an equation has no solution or is an identity?</h3>

<p>Fully distribute and combine like terms on each side first. If the variable disappears entirely, look at what's left. A false number statement, like 6 = 10, means no solution. A true number statement, like 7 = 7, means the equation is an identity and true for every value of the variable.</p>

</div>

<h3>What's the difference between solving an equation and solving a formula for a variable?</h3>

<p>The algebra steps are identical. The only difference is that in a formula, you're isolating one letter while treating every other letter as a constant, so your final answer is an expression rather than a single number.</p>

</div>

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

<p>The <a href="https://theschoolofmathematics.com/quiz/course/SAT-Math-Qbank">SAT Math Question Bank</a> has dedicated Solving Equations quizzes with additional original problems, along with topic quizzes covering every other Heart of 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-solving-equations-quiz-1">Practice Solving Equations 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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