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<h4>Solving Equations and Creating Equivalent Equations</h4>
<p>One way to create equivalent equations is to solve the equation. In each step of solving, an equivalent equation is created.</p>
<p>Here is a general strategy for solving linear equations:</p>
<p><strong>Step 1 - </strong>Simplify each side of the equation as much as possible.</p>
<p>Use the Distributive Property to remove any parentheses.<br>
Combine like terms.</p>
<p><strong>Step 2 - </strong>Collect all the variable terms on one side of the equation.</p>
<p>Use the Addition or Subtraction Property of Equality.</p>
<p><strong>Step 3 - </strong> Collect all the constant terms on the other side of the equation.</p>
<p>Use the Addition or Subtraction Property of Equality.</p>
<p><strong>Step 4 - </strong>Make the coefficient of the variable term equal to 1.</p>
<p>Use the Multiplication or Division Property of Equality.<br>
State the solution to the equation.</p>
<p><strong>Step 5 - </strong>Check the solution.</p>
<p>Substitute the solution into the original equation to make sure the result is a true statement.</p>
<br>
<p class="os-raise-text-bold">Example 1</p>
<p><strong>Step 1 - </strong>Simplify each side of the equation as much as possible.</p>
<p>Use the Distributive Property to remove any parentheses.</p>
<p>\(\begin{array}{rcl}-6(x+3)&=&24 \\ -6x-18&=&24 \end{array}\)</p>
<p><strong>Step 2 - </strong>Collect all the variable terms on one side of the equation.</p>
<p>Nothing to do. All the variables of \(x\) are on the left side. </p>
<p><strong>Step 3 - </strong> Collect all the constant terms on the other side of the equation.</p>
<p>Use the Addition or Subtraction Property of Equality.</p>
<p>\(\begin{array}{rcl}-6x-18 {\style{color:red}+} {\style{color:red}1}{\style{color:red}8}&=&24 {\style{color:red}+} {\style{color:red}1}{\style{color:red}8} \\ -6x &=&42\end{array}\)</p>
<p><strong>Step 4 - </strong>Make the coefficient of the variable term equal to 1.</p>
<p>Use the Multiplication or Division Property of Equality.</p>
<p>\(\begin{array}{rcl}\frac{-6x}{6}&=&\frac{42}{-6} \\ x&=&7 \end{array}\)</p>
<p><strong>Step 5 - </strong>Check the solution.</p>
<p>Substitute the solution into the original equation to make sure the result is a true statement.</p>
<p>\(\begin{array}{rcl}-6(x+3)&=&24 \\ -6({\style{color:red}-}{\style{color:red}7}+3)&=&24 \\ -6(-4)&=&24 \\24&=&24 \checkmark \end{array}\)</p>
<br>
<p class="os-raise-text-bold">Example 2</p>
<p>Find the solution for \(3(4x - 1) - 2 = 8x + 3\).</p>
<p><strong>Step 1</strong> - Simplify each side of the equation as much as possible.</p>
<p>\(\begin{array}{rcl}3(4x - 1) - 2&=&8x + 3 \\ 12x - 3 - 2&=&8x + 3 \\ 12x - 5&=&8x + 3 \\\end{array}\)</p>
<p><strong>Step 2 -</strong> Collect all the variable terms on one side of the equation.</p>
<p>\(\begin{array}{rcl}12x - 5 - 8x&=&8x + 3 - 8x \\ 4x - 5 &=& 3\\\end{array}\)</p>
<p><strong>Step 3 -</strong> Collect all the constant terms on the other side of the equation.</p>
<p>\(\begin{array}{rcl}4x - 5 + 5 &=& 3 + 5 \\ 4x &=& 8\end{array}\)</p>
<p><strong>Step 4 -</strong> Make the coefficient of the variable term equal to 1.</p>
<p>\(\begin{array}{rcl}\frac{4x}{4} &=& \frac84 \\ x &=& 2\end{array}\)</p>
<p><strong>Step 5 -</strong> Check the solution.</p>
<p>\(\begin{array}{rcl}3(4x - 1) - 2&=&8x + 3 ? \\ 3(4(2) - 1) - 2&=&8(2) + 3 ? \\ 3(8 - 1) - 2&=&16 + 3 ? \\ 3(7) - 2&=&19 ? \\ 21 - 2&=&19 ? \\ 19&=&19 \checkmark\end{array}\)</p>
<br>
<p class="os-raise-text-bold">Example 3</p>
<p>Solve \(12x + 3(x + 7) = 10x - 24\).</p>
<p><strong>Step 1 -</strong> Simplify each side of the equation as much as possible.</p>
<p>\(\begin{array}{rcl}12x + 3(x + 7) = 10x - 24 \\ 12x + 3x + 21 = 10x - 24 \\ 15x + 21 = 10x - 24 \end{array}\)</p>
<p><strong>Step 2 -</strong> Collect all the variable terms on one side of the equation.</p>
<p>\(\begin{array}{rcl}15x + 21 -10x&=&10x-24-10x \\ 5x + 21&=&- 24\end{array}\)</p>
<p><strong>Step 3 -</strong> Collect all the constant terms on the other side of the equation.</p>
<p>\(\begin{array}{rcl}5x+21-21&=&-24-21 \\ 5x&=&-45\end{array}\)</p>
<p><strong>Step 4 -</strong> Make the coefficient of the variable term equal to 1.</p>
<p>\(\begin{array}{rcl}\frac{5x}{5}&=&-\frac{45}{5} \\ x &=& -9\end{array}\)</p>
<p><strong>Step 5 -</strong> Check the solution.</p>
<p>\(\begin{array}{rcl}12x + 3(x + 7)&=&10x - 24 ? \\ 12(-9) + 3(-9 + 7)&=&10(-9) - 24 ? \\ -108 + 3(-2)&=&-90 - 24 ? \\ -108 + -6&=&-86 ? \\ -114&=&-114 \checkmark\end{array}\)</p>
<br>
<h4>Video: Acceptable Moves</h4>
<p>Watch the following video to learn more about acceptable moves. </p>
<div class="os-raise-d-flex-nowrap os-raise-justify-content-center">
<div class="os-raise-video-container"><video controls="true" crossorigin="anonymous">
<source src="https://k12.openstax.org/contents/raise/resources/84fd718f1e7fa4c23f38cb1f2a45b7a21744d55f">
<track default="true" kind="captions" label="On" src="https://k12.openstax.org/contents/raise/resources/86ba592db56095fbc646f9f369cf34545df8eb23 " srclang="en_us">https://k12.openstax.org/contents/raise/resources/84fd718f1e7fa4c23f38cb1f2a45b7a21744d55f
</video></div>
</div>
<br>
<br>
<h4>Try It: Solving Equations and Creating Equivalent Equations</h4>
<div class="os-raise-ib-cta" data-button-text="Solution" data-fire-event="Reveal1" data-schema-version="1.0">
<div class="os-raise-ib-cta-content">
<ol class="os-raise-noindent" >
<li>Write two equivalent equations to \( 2(x+4)= 12 \).</li>
</ol>
</div>
<div class="os-raise-ib-cta-prompt">
<p>Write down your answer, then select the <strong>solution</strong> button to compare your work.</p>
</div>
</div>
<div class="os-raise-ib-content" data-schema-version="1.0" data-wait-for-event="Reveal1">
<p>Compare your answer:</p>
<p>Your answer may vary, but here is a sample.</p>
<ul>
<li>\(2x + 8 = 24 \)</li>
<li>\(x + 4 = 6 \)</li>
<li>\(2x = 16\)<br>
</li>
</ul>
</div>
<div class="os-raise-ib-cta" data-button-text="Solution" data-fire-event="Reveal2" data-schema-version="1.0">
<div class="os-raise-ib-cta-content">
<ol class="os-raise-noindent" start="2">
<li>Solve \(2(x+4)=12\).</li>
</ol>
</div>
<div class="os-raise-ib-cta-prompt">
<p>Write down your answer, then select the <strong>solution</strong> button to compare your work.</p>
</div>
</div>
<div class="os-raise-ib-content" data-schema-version="1.0" data-wait-for-event="Reveal2">
<p>Compare your answer:</p>
<p>
<strong>Step 1 -</strong> Simplify each side of the equation as much as possible. Use the Distributive Property to remove any parentheses.</p>
<p> \(\begin{array}{rcl}2(x + 4) &=& 12 \\
2x + 8 &= &12\end{array}\)</p>
<p> <strong>Step 2 -</strong> Collect all the variable terms on one side of the equation. Nothing to do. All the variables of x are on the left side.</p>
<p> <strong>Step 3 - </strong>Collect all the constant terms on the other side of the equation.</p>
<p>Use the Addition or Subtraction Property of Equality.</p>
<p> \(\begin{array}{rcl}2x + 8{\style{color:red}-}{\style{color:red}8} &=& 12{\style{color:red}-}{\style{color:red}8 }\\
2x &=& 4\end{array}\)</p>
<p> <strong>Step 4 -</strong> Make the coefficient of the variable term equal to 1. Use the Multiplication or Division Property of Equality.</p>
<p> \(\begin{array}{rcl}\frac{2x}{2} &=& \frac42 \\
x &=& 2\end{array}\)</p>
<p> <strong>Step 5 -</strong> Check the solution.</p>
<p>Substitute the solution into the original equation to make sure the result is a true statement.</p>
<p>\(\begin{array}{rcl}2(x + 4) &=& 12 \\
2({\style{color:red}2} + 4) &=& 12 \\
2(6) &=& 12 \\
12 &=& 12 \checkmark\end{array}\)</p>
</div>
<div class="os-raise-ib-cta" data-button-text="Solution" data-fire-event="Reveal3" data-schema-version="1.0">
<div class="os-raise-ib-cta-content">
<ol class="os-raise-noindent" start="3">
<li>
Solve \(-12+8(x−5)=-4+3(5x-2) \)</li>
</ol>
</div>
<div class="os-raise-ib-cta-prompt">
<p>Write down your answer, then select the <strong>solution</strong> button to compare your work.</p>
</div>
</div>
<div class="os-raise-ib-content" data-schema-version="1.0" data-wait-for-event="Reveal3">
<p>Compare your answer:</p>
<p><strong>Step 1 -</strong> Simplify each side of the equation as much as possible.</p>
<p>\(\begin{array}{rcl}-12 + 8(x - 5) &=& -4 + 3(5x - 2) \\ -12 + 8x - 40 &=& -4 + 15x - 6 \\ 8x - 52 &=& 15x - 10\end{array}\)</p>
<p><strong>Step 2 -</strong> Collect all the variable terms on one side of the equation.</p>
<p>\(\begin{array}{rcl}8x - 52 - 15x &=& 15x - 10 - 15x \\ -7x -52 &=& - 10\end{array}\)</p>
<p><strong>Step 3 -</strong> Collect all the constant terms on the other side of the equation.</p>
<p>\(\begin{array}{rcl}-7x - 52 + 52 &=& -10 + 52 \\ -7x &=& 42\end{array}\)</p>
<p><strong>Step 4 - </strong>Make the coefficient of the variable term equal to 1.</p>
<p>\(\begin{array}{rcl}\frac{-7x}{-7} &=& \frac {42}{-7} \\ x &=& -6\end{array}\)</p>
<p><strong>Step 5 -</strong> Check the solution.</p>
<p>\(\begin{array}{rcl}-12 + 8(x - 5) &=& -4 + 3(5x - 2) ? \\ -12 + 8(-6 - 5) &=& -4 + 3(5(-6) - 2) ? \\ -12 + 8(-11) &=& -4 + 3(-30 - 2) ? \\ -12 + -88 &=& -4 + 3(-32) ? \\ -100 &=& -4 + -96 \\ -100 &=& -100 \checkmark\end{array}\)</p>
</div>