Question

Solve the given equation: $\dfrac{{x - 4}}{{x - 5}} + \dfrac{{x - 6}}{{x - 7}} = \dfrac{{10}}{3}$ . Find $x$.

Answer 1

Hint: In this question, we will first add both the fractions by taking the L.C.M of the denominators and then by cross multiplication we get the quadratic equation. We will factorise the quadratic equation by middle term factorisation. We know that for the factorisation of quadratic polynomials of the form ${x^2} + bx + c$ we have to find numbers $p$ and $q$ such that $p + q = b$ and $pq = c$. This is called factorisation by splitting the middle term.

Complete step by step answer:
Here we have $\dfrac{{x - 4}}{{x - 5}} + \dfrac{{x - 6}}{{x - 7}} = \dfrac{{10}}{3}$ .
We will add them by taking the L.C.M :
$\dfrac{{(x - 4)(x - 7) + (x - 6)(x - 5)}}{{(x - 5)(x - 7)}} = \dfrac{{10}}{3}$
By breaking the brackets and multiplying them it gives:
$ \Rightarrow \dfrac{{x(x - 7) - 4(x - 7) + x(x - 5) - 6(x - 5)}}{{x(x - 7) - 5(x - 7)}} = \dfrac{{10}}{3}$
\[ \Rightarrow \dfrac{{{x^2} - 7x - 4x + 28 + {x^2} - 5x - 6x + 30}}{{{x^2} - 7x - 5x + 35}} = \dfrac{{10}}{3}\]
We can group the similar terms together and then we add them:
\[ \Rightarrow \dfrac{{{x^2} + {x^2} - 7x - 4x - 5x - 6x + 28 + 30}}{{{x^2} - 7x - 5x + 35}} = \dfrac{{10}}{3}\]
\[ \Rightarrow \dfrac{{2{x^2} - 22x + 58}}{{{x^2} - 12x + 35}} = \dfrac{{10}}{3}\]

We will take the common factor out from the numerator in the left hand side of the equation i.e.
\[ \Rightarrow \dfrac{{2({x^2} - 11x + 29)}}{{{x^2} - 12x + 35}} = \dfrac{{2 \times 5}}{3}\]
By cancelling out $2$ from the numerators in both side of the equation ,
\[ \Rightarrow \dfrac{{{x^2} - 11x + 29}}{{{x^2} - 12x + 35}} = \dfrac{5}{3}\]
We cross multiply them:
$3({x^2} - 11x + 29) = 5({x^2} - 12x + 35)$
On multiplying it can be written as
$3{x^2} - 33x + 87 = 5{x^2} - 60x + 175$
We can transfer the similar terms together to the left hand side of the equation and add them;
$5{x^2} - 3{x^2} - 60x + 33x + 175 - 87 \\
\Rightarrow 2{x^2} - 27x + 88 = 0$

Now we have a quadratic equation, so we will use middle term factorisation to solve them:
Here we have equation: $2{x^2} - 27x + 88 = 0$ , we have to split the middle term i.e. $( - 27x)$in such numbers that that the product of the numbers will be equal to $176{x^2}$. We can write
$ - 27x = - 6x - 11x$
As,
$( - 16x)( - 11x) = - 176{x^2}$.
Therefore by putting them in the equation,
$2{x^2} - 16x - 11x + 88 = 0$,
Now take out the common factor and simplify it;
$2x(x - 8) - 11(x - 8) \Rightarrow (2x - 11)(x - 8) = 0$.
We will equate the factors now with zero, and it gives;
$2x - 11 = 0$
So,
$2x = 11 \\
\Rightarrow x = \dfrac{{11}}{2}$
Similarly by equating the another factor we can write:
$x - 8 = 0 \\
\therefore x = 8$.

Hence the required values of $x$ are: $x = \dfrac{{11}}{2}$ or, $x = 8$.

Note: As we know that factorising is the reverse of expanding brackets, it is an important way of solving equations. The first step of factoring an expression is to take out any common factors which the terms have. So if we were asked to factor the expression,
${x^2} + x$ , since $x$ goes into both terms, we would write $x(x + 1)$. We should keep in mind while solving this kind of middle term factorisation that we use correct identities to factorise the given algebraic expressions and keep checking the negative and positive sign otherwise it will give wrong answer