Question

How do you solve $7\dfrac{x}{8}-3\dfrac{x}{5}=\dfrac{11}{2}$ ? \[\]

Answer 1

Hint: We convert the mixed fractions in the left hand side to improper fraction using the working rule $w\dfrac{a}{b}=\dfrac{bw+a}{b}$. We then add the fractional expressions at the left hand side by finding the least common multiple of denominators and converting them into equivalent fractions.\[\]

Complete step by step answer:
We are asked to solve for $x$in the following equation
\[7\dfrac{x}{8}-3\dfrac{x}{5}=\dfrac{11}{2}\]
We see that there are two mixed fractions in the left hand side. We convert them into improper fractions using the working rule $w\dfrac{a}{b}=\dfrac{bw+a}{b}$ to have;
\[\begin{align}
  & \Rightarrow \dfrac{7\times 8+x}{8}-\dfrac{3\times 5+x}{5}=\dfrac{11}{2} \\
 & \Rightarrow \dfrac{56+x}{8}-\dfrac{15+x}{5}=\dfrac{11}{2} \\
\end{align}\]
We see that we have two fractional expressions with denominators 8 and 5 in the left hand side. We know that we can only add or subtract two fractions only when they have the same denominators. So we need to find the equivalent fractions of the above fractions such that they have the same denominator. The denominator will be the least common multiple of 8 and 5 which is $\operatorname{LCM}\left( 8,5 \right)=40$. So we multiply 5 in both numerator and denominator of first fraction and similarly we multiply 8 in both numerator and denominator of second fraction to have;
\[\begin{align}
  & \Rightarrow \dfrac{56+x}{8}\times \dfrac{5}{5}-\dfrac{15+x}{5}\times \dfrac{8}{8}=\dfrac{11}{2} \\
 & \Rightarrow \dfrac{280+5x}{40}-\dfrac{120+8x}{40}=\dfrac{11}{2} \\
\end{align}\]
We subtract the numerators while keeping the denominators’ same to have ;
\[\begin{align}
  & \Rightarrow \dfrac{280+5x-\left( 120+8x \right)}{40}=\dfrac{11}{2} \\
 & \Rightarrow \dfrac{280+5x-120-8x}{40}=\dfrac{11}{2} \\
 & \Rightarrow \dfrac{-3x+160}{40}=\dfrac{11}{2} \\
\end{align}\]
We cross multiply to have;
\[\begin{align}
  & \Rightarrow 2\left( -3x+160 \right)=11\times 40 \\
 & \Rightarrow -6x+320=440 \\
\end{align}\]
We subtract both sides of the above equation by 320 to have ;
\[\begin{align}
  & \Rightarrow -6x+320-320=440-320 \\
 & \Rightarrow -6x=120 \\
\end{align}\]
We divide the above equation by 6 to have;
\[\begin{align}
  & \Rightarrow \dfrac{-6x}{6}=\dfrac{120}{6} \\
 & \Rightarrow -x=20 \\
\end{align}\]
We multiply $-1$ both sides of equation to have;
\[\begin{align}
  & \Rightarrow -x\times \left( -1 \right)=20\times \left( -1 \right) \\
 & \Rightarrow x=-20 \\
\end{align}\]

So the solution of the given equation is $x=-20$. \[\]

Note: We note that equivalent fractions of $\dfrac{a}{b}$ are given by $\dfrac{a\times k}{b\times k}$ where $k$ is any integer of our choice except zero. We can alternatively use elimination of denominators to solve the equation. We can multiply the lcm of denominators 8 and 5 which is 40 both sides of the given equation to eliminate the denominators. We should remember that the equation remains balanced even if we add, subtract, multiply or divide (except zero) the same number on both sides.