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lesson 8 exit ticket - i can explain how a balanced hanger and an equat…
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Question

lesson 8 exit ticket

  • i can explain how a balanced hanger and an equation represent the same situation.
  • i can explain why some balanced hangers can be described by two different equations, one with parentheses and one without.
  • i can find an unknown weight on a hanger diagram and solve an equation that represents the diagram.
  • i can write an equation that describes the weights on a balanced hanger.

solve the equation 3(x + 4.5) = 36. you may consider using the diagram to help you.

Explanation:

Step1: Divide both sides by 3

To isolate the term \(x + 4.5\), we divide both sides of the equation \(3(x + 4.5)=36\) by 3.
\(\frac{3(x + 4.5)}{3}=\frac{36}{3}\)
Simplifying, we get \(x + 4.5 = 12\).

Step2: Subtract 4.5 from both sides

To solve for \(x\), we subtract 4.5 from both sides of the equation \(x + 4.5 = 12\).
\(x+4.5 - 4.5=12 - 4.5\)
Simplifying, we find that \(x = 7.5\).

Answer:

\(x = 7.5\)