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Rahul solved the equation 2(x – ) – 2 left-parenthesis x minus StartFraction 1 Over 8 EndFraction right-parenthesis minus StartFraction 3 Over 5 EndFraction x equals StartFraction 55 Over 4 EndFraction x = 2 left-parenthesis x minus StartFraction 1 Over 8 EndFraction right-parenthesis minus StartFraction 3 Over 5 EndFraction x equals StartFraction 55 Over 4 EndFraction . In which step did he use the addition property of equality?

A table titled Rahul's Solution with 2 columns and 5 rows. The first column, Steps, has the entries 1, 2, 3, 4. The second column, Resulting equations, has the entries, 2 x minus StartFraction 1 Over 4 EndFraction minus StartFraction 3 Over 5 EndFraction x equals StartFraction 55 Over 4 EndFraction, StartFraction 7 Over 5 EndFraction x minus StartFraction 1 Over 4 EndFraction equals StartFraction 55 Over 4 EndFraction, StartFraction 7 Over 5 EndFraction x equals StartFraction 56 Over 4 EndFraction, x equals 10.

Step 1
Step 2
Step 3
Step 4

1 answer

To determine in which step Rahul used the addition property of equality, let's review each step of the solution:

  1. Step 1: The equation is simplified to \(2x - \frac{1}{4} - \frac{3}{5}x = \frac{55}{4}\).
  2. Step 2: Rahul appears to combine like terms, leading to \(\frac{7}{5}x - \frac{1}{4} = \frac{55}{4}\).
  3. Step 3: Another operation is performed to isolate the \(x\) term, resulting in \(\frac{7}{5}x = \frac{56}{4}\).
  4. Step 4: Finally, \(x\) is solved to get \(x = 10\).

The addition property of equality states that if you add the same number to both sides of an equation, the two sides remain equal.

  • In Step 2, Rahul does not appear to use the addition property; instead, he combines terms.
  • In Step 3, when he adds \(\frac{1}{4}\) to both sides to isolate \(\frac{7}{5}x\), this is where he applies the addition property of equality.

Thus, the addition property of equality is used in Step 3.