When there are fractions in equations, the standard procedure is to remove the fractions by multiplying both sides of the equation by an appropriate whole number. The next step is to remove the brackets. \begin{examplebox} \begin{questions} \questionpoints[2] Solve: \begin{parts} \part \(2 x+\frac{1}{2}=\frac{2}{3}\) \begin{solutionordottedlines}[1in] \[ 2 x+\frac{1}{2}=\frac{2}{3} \] \[ 6\left(2 x+\frac{1}{2}\right)=6 \times \frac{2}{3} \quad \begin{gathered} \text { (Multiply both sides by } 6, \text { the lowest common multiple of the } \\ \text { denominators. }) \end{gathered} \] \[ 12 x+3=4 \] \[ \begin{aligned} 12 x & =1 \\ x & =\frac{1}{12} \end{aligned} \] \end{solutionordottedlines} \part \(3 x-\frac{1}{4}=\frac{4}{5}\) \begin{solutionordottedlines}[1in] \(\quad 3 x-\frac{1}{4}=\frac{4}{5}\) \[ \begin{aligned} 20\left(3 x-\frac{1}{4}\right) & =20 \times \frac{4}{5} \\ 60 x-5 & =16 \\ 60 x & =21 \quad \text { (Multiply both sides by 20.) } \\ x & =\frac{21}{60} \\ & =\frac{7}{20} \end{aligned} \] \end{solutionordottedlines} \end{parts} \questionpoints[2] Solve: \begin{parts} \part \(\frac{2 x}{3}+\frac{1}{5}=4\) \begin{solutionordottedlines}[1in] \end{solutionordottedlines} \part \(10-\frac{a+3}{4}=6\) \begin{solutionordottedlines}[1in] \end{solutionordottedlines} \end{parts} \questionpoints[1] Solve \(2.1 x+3.5=9.4\) \begin{solutionordottedlines}[1in] \[ \begin{aligned} 2.1 x+3.5 & =9.4 \\ 21 x+35 & =94 \quad \text { (Multiply both sides by 10.) } \\ 21 x & =59 \\ x & =\frac{59}{21} \end{aligned} \] \end{solutionordottedlines} \questionpoints[1] Solve \(\frac{2 x}{3}-3=x+\frac{3}{4}\) \begin{solutionordottedlines}[1in] \[ \begin{aligned} \frac{2 x}{3}-3 & =x+\frac{3}{4} \\ 8 x-36 & =12 x+9 \quad \text { (Multiply both sides by 12.) } \\ -4 x & =45 \\ x & =-\frac{45}{4} \end{aligned} \] \end{solutionordottedlines} \questionpoints[1] Solve \(\frac{a+5}{4}=\frac{a+3}{3}\) \begin{solutionordottedlines}[1in] \[ \begin{aligned} \frac{a+5}{4} & =\frac{a+3}{3} \\ 3(a+5) & =4(a+3) \quad \text { (Multiply both sides by 12.) } \\ 3 a+15 & =4 a+12 \\ 15 & =a+12 \\ 3 & =a \\ a & =3 \end{aligned} \] \end{solutionordottedlines} \end{questions} \end{examplebox} \fbox{\parbox{\linewidth}{ \large{\textbf{Crash Course:}} \begin{itemize} \item Remove all fractions by multiplying both sides of the equation by the lowest common multiple of the denominators. \item Remove all brackets. \item Collect like terms and solve the equation. \end{itemize} } } \begin{exercisebox} \subsection{Exercises:} \begin{questions} \questionpoints[9] Solve for the unknown pronumeral: \begin{parts}\begin{multicols}{2} \part \(2 x-\frac{1}{2}=\frac{1}{4}\) \begin{solutionordottedlines}[2cm] \end{solutionordottedlines} \part \(\frac{7}{2}-3 y=\frac{2}{3}\) \begin{solutionordottedlines}[2cm] \end{solutionordottedlines} \part \(-2 x+\frac{1}{3}=\frac{1}{5}\) \begin{solutionordottedlines}[2cm] \end{solutionordottedlines} \part \(\frac{a}{3}+5=3\) \begin{solutionordottedlines}[2cm] \end{solutionordottedlines} \part \(\frac{3 b}{7}+6=2\) \begin{solutionordottedlines}[2cm] \end{solutionordottedlines} \part \(\frac{2}{3}(m-3)=1\) \begin{solutionordottedlines}[2cm] \end{solutionordottedlines} \part \(\frac{2 y+1}{3}+4=7\) \begin{solutionordottedlines}[2cm] \end{solutionordottedlines} \part \(\frac{3 y-1}{2}+2=9\) \begin{solutionordottedlines}[2cm] \end{solutionordottedlines} \part \(2+\frac{2 x-1}{5}=5\) \begin{solutionordottedlines}[2cm] \end{solutionordottedlines} \part \(e+1.8=2.9\) \begin{solutionordottedlines}[2cm] \end{solutionordottedlines} \part \(1.2 u=15.6\) \begin{solutionordottedlines}[2cm] \end{solutionordottedlines} \part \(1.2 x+4=10\) \begin{solutionordottedlines}[2cm] \end{solutionordottedlines} \part \(\frac{3 x}{4}-1=\frac{x}{2}+3\) \begin{solutionordottedlines}[2cm] \end{solutionordottedlines} \part \(\frac{x}{3}-4=6-\frac{2 x}{5}\) \begin{solutionordottedlines}[2cm] \end{solutionordottedlines} \end{multicols}\end{parts} \begin{parts}\begin{multicols}{2} \part \(1.6 x+10=0.9 x+12\) \begin{solutionordottedlines}[2cm] \end{solutionordottedlines} \part \(4.8-1.3 x=23+1.3 x\) \begin{solutionordottedlines}[2cm] \end{solutionordottedlines} \part \(\frac{a+3}{2}=\frac{a+1}{5}\) \begin{solutionordottedlines}[2cm] \end{solutionordottedlines} \part \(\frac{2 a+1}{3}=\frac{3 a+1}{4}\) \begin{solutionordottedlines}[2cm] \end{solutionordottedlines} \part \(\frac{3 a+2}{4}+2=a\) \begin{solutionordottedlines}[2cm] \end{solutionordottedlines} \part \(\frac{3 a-2}{4}=\frac{a-5}{2}\) \begin{solutionordottedlines}[2cm] \end{solutionordottedlines} \part \(\frac{4 a-1}{3}+a=2\) \begin{solutionordottedlines}[2cm] \end{solutionordottedlines} \part \(5 a+9=24\) \begin{solutionordottedlines}[2cm] \end{solutionordottedlines} \part \(\frac{c}{3}-2=4\) \begin{solutionordottedlines}[2cm] \end{solutionordottedlines} \part \(\frac{4 e}{3}+\frac{1}{2}=2\) \begin{solutionordottedlines}[2cm] \end{solutionordottedlines} \part \(2 g+5=7 g-6\) \begin{solutionordottedlines}[2cm] \end{solutionordottedlines} \part \(2(i-1)=5(i+6)\) \begin{solutionordottedlines}[2cm] \end{solutionordottedlines} \part \(2(k+1)-3(k-2)=7\) \begin{solutionordottedlines}[2cm] \end{solutionordottedlines} \part \(\frac{m+1}{3}=\frac{m-2}{5}\) \begin{solutionordottedlines}[2cm] \end{solutionordottedlines} \part \(\frac{2 q-2}{5}+1=4 q\) \begin{solutionordottedlines}[2cm] \end{solutionordottedlines} \end{multicols}\end{parts} \end{questions} \end{exercisebox}