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Solve for pp.\newline2(p+4)+3=152(p + 4) + 3 = 15\newlinep=p = _____

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Q. Solve for pp.\newline2(p+4)+3=152(p + 4) + 3 = 15\newlinep=p = _____
  1. Distribute and Simplify: First, distribute the 22 into the parentheses.\newline2(p+4)+3=152(p + 4) + 3 = 15\newline2p+2×4+3=152p + 2\times4 + 3 = 15\newline2p+8+3=152p + 8 + 3 = 15
  2. Combine Constant Terms: Next, combine the constant terms on the left side.\newline2p+8+3=152p + 8 + 3 = 15\newline2p+11=152p + 11 = 15
  3. Isolate Variable Term: Now, subtract 1111 from both sides to isolate the term with the variable pp. \newline2p+1111=15112p + 11 - 11 = 15 - 11\newline2p=42p = 4
  4. Solve for p: Finally, divide both sides by 22 to solve for pp.2p2=42\frac{2p}{2} = \frac{4}{2}p=2p = 2

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