Practice Solving Equations with Variables on Both Sides

1. Solve for \(h\)
   

   \(17+4h+2=1-5h\)

2. Solve for \(c\)
   

   \(12c−4=14c−10\)

3. Solve for \(r\)
   

   \(16−2r=-3r+6r+1\)

4. Solve for \(d\)
   \(2d+ 4=10+5d\)

Source: Khan Academy, https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:solve-equations-inequalities/x2f8bb11595b61c86:linear-equations-variables-both-sides/e/linear_equations_3
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1. We need to manipulate the equation to get \(h\) by itself.

   \(17+4h+2=1-5h\)
   \(19+4h=1-5h\)	Combine like terms
   \(19+4h+5h=1-5h+5h\)	Add \(5h\) to each side
   \(9h+19=1\)	Combine like terms
   \(9h+19-19=1-19\)	Subtract 19 from each side
   \(9h=-18\)	Combine like terms
   \(\frac{9h}{9}  = \frac{-18}{9}\)	Divide each side by 9
   \(h  = -2\)	Simplify

   
   The answer:

   \(h=-2\)

2. We need to manipulate the equation to get \(c\) by itself.

   \(12c−4=14c−10\)
   \(12c−4-14c=14c−10-14c\)	Subtract \(14c\) from each side
   \(-2c−4=−10\)	Combine like terms
   \(-2c−4+4=−10+4\)	Add 4 to each side
   \(-2c=−6\)	Combine like terms
   \(\frac{-2c}{-2} = \frac{-6}{-2} \)	Divide each side by -2
   \(c = 3 \)	Simplify

   
   

   The answer:

   \(c=3\)

3. We need to manipulate the equation to get \(r\) by itself.

   \(16−2r=-3r+6r+1\)
   \(16−2r=3r+1\)	Combine like terms
   \(16−2r-3r=3r+1-3r\)	Subtract \(3r\) from each side
   \(16−5r=1\)	Combine like terms
   \(16−5r-16=1-16\)	Subtract 16 from each side
   \(−5r=-15\)	Combine like terms
   \(\frac{-5r}{-5} = \frac{-15}{-5}\)	Divide each side by -5
   \(r=3\)	Simplify

   
   The answer:

   \(r=3\)

4. We need to manipulate the equation to get \(d\) by itself.

   \(2d+ 4=10+5d\)
   \(2d+ 4-5d=10+5d-5d\)	Subtract \(5d\) from each side
   \(-3d+4=10\)	Combine like terms
   \(-3d+4-4=10-4\)	Subtract 4 from each side
   \(-3d=6\)	Combine like terms
   \(\frac{-3d}{-3} = \frac{6}{-3}\)	Divide each side by -3
   \(d=-2\)	Simplify

   
   The answer:
   
   \(d=-2\)