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Solve the equation below for xx in terms of aa.\newline4(ax+3)3ax=25+3a4(ax+3)-3ax = 25+3a

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Q. Solve the equation below for xx in terms of aa.\newline4(ax+3)3ax=25+3a4(ax+3)-3ax = 25+3a
  1. Expand Equation: Expand the left side of the equation.\newlineWe have the equation 4(ax+3)3ax=25+3a4(ax + 3) - 3ax = 25 + 3a. Let's distribute the 44 into the parentheses:\newline4ax+433ax=25+3a4 \cdot ax + 4 \cdot 3 - 3ax = 25 + 3a\newline4ax+123ax=25+3a4ax + 12 - 3ax = 25 + 3a
  2. Combine Like Terms: Combine like terms on the left side of the equation.\newlineNow we combine the terms with axax:\newline(4ax3ax)+12=25+3a(4ax - 3ax) + 12 = 25 + 3a\newline1ax+12=25+3a1ax + 12 = 25 + 3a
  3. Isolate x Term: Subtract 1212 from both sides of the equation to isolate the term with xx.\newlineax+1212=25+3a12ax + 12 - 12 = 25 + 3a - 12\newlineax=13+3aax = 13 + 3a
  4. Solve for x: Divide both sides by aa to solve for xx.axa=13+3aa\frac{ax}{a} = \frac{13 + 3a}{a}x=13a+3x = \frac{13}{a} + 3

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