Question

How do you solve $\left( {x - \dfrac{3}{{12}}} \right) + 2x - \dfrac{1}{{15}} = \left( {3x + \dfrac{1}{4}} \right)$?

Answer 1

Hint: To solve this question , we need to simplify it step by step . . Here we are going to perform some calculations to simplify the given equation by somewhere using equivalent equations and algebraic identities . Equivalent equations are said to be algebraic equations that may have the same solutions if we add or subtract the same number to both sides of an equation - Left hand side or Right hand side of the equal to sign . Or we can multiply or divide the same number to both sides of an equation - Left hand side or Right hand side of the equal to sign with the method of simplification .

Complete step by step solution:
We can observe that in the above question we are given the fractions in the equation . so, we can eliminate the fractions and we can easily work with the whole numbers . To make it simple we will remove the brackets .
$x - \dfrac{3}{{12}} + 2x - \dfrac{1}{{15}} = 3x + \dfrac{1}{4}$
Now as we can see there are different denominators in the given equation We need to make the like terms so that we can perform operations . We need to make our calculations easier and for that we will choose a number which is divisible by all the denominators , so that the certain number could be multiplied by each term and could help us in cancelling the denominator .
The number 60 is divisible by each denominator so by multiplying we get –
$
  60 \times x - 60 \times \dfrac{3}{{12}} + 60 \times 2x - 60 \times \dfrac{1}{{15}} = 60 \times 3x + 60 \times \dfrac{1}{4} \\
60x - 5 \times 3 + 120x - 4 = 180x + 15 \\
60x - 15 + 120x - 4 = 180x + 15 \\
180x - 19 = 180x + 15 \\
180x - 180x = 15 + 19 \\
0 = 34 \\
 $
There is no x−term left, but we have ended up with a false statement.
Therefore, there is no solution to this equation.

Note:
If we had ended up with 0=0, then it would have true for any value of x
In equivalent equations which have identical solutions we can perform multiplication or division by the same non-zero number both L.H.S. and R.H.S. of an equation .
In an equivalent equation which has an identical solution we can raise the same odd power to both L.H.S. and R.H.S. of an equation .
Cross check the answer and always keep the final answer simplified .
In order to determine the simplest form of the solution ,write the numerator and denominator in the form of their factors and try to cancel out factors from numerator and denominator, you’ll get your required result.