Question

How do you solve \[\dfrac{3}{{2x - 1}} - \dfrac{4}{{3x - 1}} = 1\]

Answer 1

Hint: We solve for the value of unknown variable x by taking LCM on the left side of the equation and then cross multiplying the values from both sides of the equation. Shift the required terms to opposite sides and divide with a suitable number to calculate the value of x.
* LCM is the least common multiple of two or more numbers. We write each number in the form of its prime factors and write the LCM of the numbers as multiplication of prime factors with their highest powers.

Complete step by step answer:
We are given the equation \[\dfrac{3}{{2x - 1}} - \dfrac{4}{{3x - 1}} = 1\] … (1)
Since the equation has only one variable i.e. x we will calculate the value of the unknown variable ‘x’.
We will take LCM on the left side of the equation.
The fractions on left hand side have denominators \[2x - 1\] and \[3x - 1\], so we will calculate LCM of \[2x - 1\] and \[3x - 1\]
Take LCM and write equation (1) as
\[ \Rightarrow \dfrac{{3(3x - 1) - 4(2x - 1)}}{{(2x - 1)(3x - 1)}} = 1\]
Multiply the values in the numerator of left hand side of the equation
\[ \Rightarrow \dfrac{{9x - 3 - 8x + 4}}{{(2x - 1)(3x - 1)}} = 1\]
Since terms in numerator are having variables associated with them as x, we can add the terms of x together and add the constant values together.
\[ \Rightarrow \dfrac{{(9x - 8x) + (4 - 3)}}{{(2x - 1)(3x - 1)}} = 1\]
\[ \Rightarrow \dfrac{{x + 1}}{{(2x - 1)(3x - 1)}} = 1\]
Multiply the brackets in the denominator
\[ \Rightarrow \dfrac{{x + 1}}{{2x(3x - 1) - 1(3x - 1)}} = 1\]
\[ \Rightarrow \dfrac{{x + 1}}{{6{x^2} - 2x - 3x + 1}} = 1\]
\[ \Rightarrow \dfrac{{x + 1}}{{6{x^2} - 5x + 1}} = 1\]
Cross multiply the values from left side of the equation to right side of the equation
\[ \Rightarrow x + 1 = 6{x^2} - 5x + 1\]
Shift all values to right hand side of the equation
\[ \Rightarrow 0 = 6{x^2} - 5x - x + 1 - 1\]
\[ \Rightarrow 6{x^2} - 6x = 0\]
Take common 6x from both terms
\[ \Rightarrow 6x(x - 1) = 0\]
Now we know the product of two terms is 0 if one of them is 0 or both is 0.
Either \[6x = 0\]or \[(x - 1) = 0\]
Shift constant values to right hand side of the equation
Either \[x = 0\]or \[x = 1\].

The solution of the equation \[\dfrac{3}{{2x - 1}} - \dfrac{4}{{3x - 1}} = 1\] is \[x = 0\] or \[x = 1\].

Note: Many students make mistake of writing the LCM as \[(2x - 1)(3x - 1)\] but forget to multiply the terms in the numerator with the respective factors of the multiple ( factor multiplied by denominator gives LCM ). Keep in mind we can transform the fractions into fractions with the same denominator by multiplying required value to both numerator and denominator.