Question

How do you solve the equation $4\dfrac{2}{3}b+\dfrac{4}{5}=2\dfrac{1}{7}b-\dfrac{1}{5}b-\dfrac{1}{3}$ ?

Answer 1

Hint: Now we are given with a linear equation in b. to solve the equation we will first convert the mixed fractions into fractions and then convert the fractions into integers. Now we will simplify the equation by bringing all the variables on one side and the constants on the other side. Hence find the value of b.

Complete step-by-step solution:
Now consider the given equation $4\dfrac{2}{3}b+\dfrac{4}{5}=2\dfrac{1}{7}b-\dfrac{1}{5}b-\dfrac{1}{3}$
Now we can see that the given equation is a linear equation in b.
To solve the equation we will first convert the mixed fraction into simple fractions.
Now we know that $a\dfrac{p}{q}=\dfrac{aq+p}{q}$ Hence using this in the given equation we get,
$\begin{align}
  & \Rightarrow \dfrac{4\times 3+2}{3}b+\dfrac{4}{5}=\dfrac{2\times 7+1}{7}b-\dfrac{1}{5}b-\dfrac{1}{3} \\
 & \Rightarrow \dfrac{14}{3}b+\dfrac{4}{5}=\dfrac{15}{7}-\dfrac{1}{5}b-\dfrac{1}{3} \\
\end{align}$
Now we will simplify the equation further by converting these fractions into integers. Hence to do so we will multiply the equation with LCM of all the denominators.
Now we want to find the LCM of 3, 5, 7.
Since all are prime, the LCM is 3 × 7 × 5 = 105.
Hence multiplying the equation with 105 and simplifying we get,
$\left( 35\times 14 \right)b+\left( 21\times 4 \right)=\left( 15 \right)\left( 15 \right)-\left( 21 \right)b-\left( 35 \right)$
Now on simplifying we get,
$490b+84=225-21b-35$
Now we will bring all the variables on one side of the equation and the constant on other side of the equation hence we get,
$\begin{align}
  & \Rightarrow 490b+21b=225-35-84 \\
 & \Rightarrow 511b=106 \\
 & \Rightarrow b=\dfrac{106}{511} \\
\end{align}$
Hence the solution of the given equation is $b=\dfrac{106}{511}$ .

Note: Now note that after converting the mixed fraction into fraction we can solve the equation using fractions too. To avoid addition and subtraction in fractions we have converted the fractions into integers before simplification. Also note that while doing addition or subtraction in fractions the denominator must be the same.