So far, we’ve only seen equations with a single variable. There are equations that have variables in more than one place. For example, \( 3x + 4 = x\). How do we solve these? The first video will explain some of the tools we will use, then the second video will show how to solve these kinds of equations.

Video Source (03:22 mins) | Transcript

Video Source (09:43 mins) | Transcript

Tools taught in the first video include the following:

• Combine like terms (add things that have the same variable).
• Distribute when needed (multiply each of the things inside the parentheses).
• Add the additive inverse of terms to both sides.
• Multiply by the multiplicative inverse to both sides.

When faced with a problem, start by combining any like terms on the same side of the equation. Then combine like terms from both sides of the equation. After that, use the things we learned in last week’s lesson of adding or multiplying by the inverse as needed. Remember, we can add, subtract, multiply, or divide all we want, as long as we do it to both sides of the equation.

Additional Resources

• Khan Academy: Combining Like Terms (03:05 mins, Transcript)
• Khan Academy: Introduction to Equations with Variables on Both Sides (08:52 mins, Transcript)
• Khan Academy: Linear Equations 3 (06:44 mins, Transcript)

Practice Problems

1. \(2 - 7{\text{g}} = -9{\text{g}}\) (
   Solution

   x

   Solution:
   \(-1\)
   )
2. \(12 + 3{\text{W}} = -4 + {\text{W}}\) (
   Solution

   x

   Solution: \(-8\)
   Details:
   We do the order of operations backwards to solve for W.
   
   Step 1: Combine like terms using addition or subtraction. Right now we have terms containing W on both sides of the equation:
   
   \(12 + 3{\text{W}} = -4 + {\text{W}}\)
   
   First, we’ll subtract W from both sides to gather all terms containing W to the left side of the equation:
   
   
   
   Which gives us:
   
   \(12 + 2{\text{W}} = -4 {\color{Red} + 0}\)
   
   Which is equal to:
   
   \(12 + 2{\text{W}} = -4\)
   
   Next, we’ll subtract 12 from both sides of the equation:
   
   
   
   Which gives us:
   
   \(2{\text{W}} = {\color{Red} - 16}\)
   
   Step 2: Undo any multiplication using the multiplicative inverse or division to isolate W.
   
   \({\color{Cyan} \left [ \dfrac{1}{2} \right ]}2{\text{W}}=-16{\color{Cyan} \left [ \dfrac{1}{2} \right ]}\)
   
   On the left side of the equation: \(\left (\dfrac{1}{2} \right )\left ( 2 \right )\) gives us 1W.
   
   On the right side of the equation: \(-16\dfrac{1}{2}=\dfrac{-16}{1}\times\dfrac{1}{2}=\dfrac{-16\times1}{1\times2}=\dfrac{-16}{2}=-8\)
   
   So our final answer is:
   
   \({\text{W}} = {\color{Red} - 8}\)
   )
3. \({\text{m}} {-} 3 = 2{\text{m}} - 3\) (
   Video Solution

   x

   Solution: 0
   Details:
   

   
   (Video Source | Transcript)
   )
4. \(3 - 6{\text{P}} = -6 - 7{\text{P}}\) (
   Solution

   x

   Solution: \(-9\)
   Details:
   We do the order of operations backwards to solve for W.
   
   Step 1: Combine like terms using addition or subtraction.
   
   Right now we have terms containing P on both sides of the equation:
   
   \(3 - 6{\text{P}} = -6 - 7{\text{P}}\)
   
   First, we’ll add 7P to both sides to gather all terms containing P to the left side of the equation:
   
   
   
   Which gives us:
   
   
   
   Next, we’ll subtract 3 from both sides of the equation:
   
   
   
   Which gives us:
   
   \({\text{P}} = {\color{Red} -9}\)
   )
5. \(6{\text{x}} {-} 1 = -5 + 7{\text{x}}\) (
   Solution

   x

   Solution:
   4
   )
6. \(7 - 5{\text{C}} = -9 - 9{\text{C}}\) (
   Video Solution

   x

   Solution: \(-4\)
   Details:
   

   
   (Video Source | Transcript)
   )

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