Chapter 4.5, Problem 33PPS

To determine

To solve and check: the given equation

Answer to Problem 33PPS

The required solution is $y=\frac{5}{12}$

Explanation of Solution

Given: $\frac{2}{3}-6y=-1\frac{5}{6}$

Calculation:

To solve equation and check the solution

  $\frac{2}{3}-6y=-1\frac{5}{6}$

Convert mixed fraction into improper fraction

  $\frac{2}{3}-6y=-\frac{11}{6}$

Add both sides of equation with $\frac{11}{6}+6y$

  $\begin{array}{l}\frac{2}{3}-6y+\frac{11}{6}+6y=-\frac{11}{6}+\frac{11}{6}+6y\\ \frac{2}{3}+\frac{11}{6}=6y\end{array}$

LCD is 6. In left hand side of equation, make denominator 6

  $\begin{array}{l}\frac{2\cdot 2}{3\cdot 2}+\frac{11}{6}=6y\\ \frac{4}{6}+\frac{11}{6}=6y\\ \frac{15}{6}=6y\end{array}$

Divide both sides of equation with 6

  $\frac{15}{6}\cdot \frac{1}{6}=y$

Divide out the common terms

  $\begin{array}{l}\frac{\cancel{3}\cdot 5}{\cancel{3}\cdot 2\cdot 6}=y\\ y=\frac{5}{12}\end{array}$

Solution:- $y=\frac{5}{12}$

Check

Substitute $y=\frac{5}{12}$ in $\frac{2}{3}-6y=-1\frac{5}{6}$

  $\frac{2}{3}-6\left(\frac{5}{12}\right)=-\frac{11}{6}$

Make the denominator 12

  $\begin{array}{l}\frac{2\cdot 4}{3\cdot 4}-\frac{30}{12}=-\frac{11}{6}\\ \frac{8-30}{12}=-\frac{11}{6}\\ \frac{-22}{12}=-\frac{11}{6}\\ \frac{-11}{6}=-\frac{11}{6}\end{array}$

Hence verified. So solution is $y=\frac{5}{12}$

Conclusion:

Therefore, the required solution is $y=\frac{5}{12}$