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lesson 7 exit ticket - i can explain how a balanced hanger and an equat…

Question

lesson 7 exit ticket

  • i can explain how a balanced hanger and an equation represent the same situation.
  • 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. you may consider using a diagram to help you.

  1. $5x + \frac{1}{4} = \frac{61}{4}$

Explanation:

Step1: Subtract $\frac{1}{4}$ from both sides

To isolate the term with \(x\), we subtract \(\frac{1}{4}\) from both sides of the equation \(5x+\frac{1}{4}=\frac{61}{4}\). This gives us \(5x+\frac{1}{4}-\frac{1}{4}=\frac{61}{4}-\frac{1}{4}\). Simplifying both sides, the left side becomes \(5x\) (since \(\frac{1}{4}-\frac{1}{4}=0\)) and the right side becomes \(\frac{61 - 1}{4}=\frac{60}{4} = 15\). So now we have the equation \(5x=15\).

Step2: Divide both sides by 5

To solve for \(x\), we divide both sides of the equation \(5x = 15\) by 5. Using the division property of equality, we get \(\frac{5x}{5}=\frac{15}{5}\). Simplifying both sides, the left side becomes \(x\) (since \(\frac{5x}{5}=x\)) and the right side becomes \(3\) (since \(\frac{15}{5}=3\)).

Answer:

\(x = 3\)