Question

Find the value of x in the function-
$\dfrac{\mathrm x-3}{\mathrm x-4}+\dfrac{\mathrm x-5}{\mathrm x-6}=\dfrac{10}3,\;\mathrm x\neq4,\;6$

Answer 1

Hint:This is a direct question of a linear equation in one variable where we have to find the value of the variable. General algebraic formulas will be used. Also, quadratic formula can be used which is given by-
$=\dfrac{-\mathrm b\pm\sqrt{\mathrm b^2-4\mathrm{ac}}}{2\mathrm a}$

Complete step-by-step answer:
It is given that
$\dfrac{\mathrm x-3}{\mathrm x-4}+\dfrac{\mathrm x-5}{\mathrm x-6}=\dfrac{10}3$
Taking the LCM, we can write that-
\dfrac{\left(\mathrm x-3\right)\left(\mathrm x-6\right)+\left(\mathrm x-4\right)\left(\mathrm x-5\right)}{\left(\mathrm x-4\right)\left(\mathrm x-6\right)}=\dfrac{10}3\\\\\\\
On cross multiplying and opening the brackets we get,
$3\left( {{{\text{x}}^2} - 9{\text{x}} + 18 + {{\text{x}}^2} - 9{\text{x}} + 20} \right)\; = \;10\left( {{{\text{x}}^2} - 10{\text{x}} + 24} \right)$
$6{{\text{x}}^2} - 54{\text{x}} + 114 = 10{{\text{x}}^2} - 100{\text{x}} + 240$
$4{{\text{x}}^2} - 46{\text{x}} + 126 = 0$

$Dividing\;the\;equation\;by\;2$
$2{{\text{x}}^2} - 23{\text{x}} + 63 = 0$
$2{{\text{x}}^2} - 14{\text{x}} - 9{\text{x}} + 63 = 0$
$2x(x - 7) - 9(x - 7) = 0$
$(2x - 9)(x - 7) = 0$
$\mathrm x=7,\dfrac{\;9}2$
These are the values of x, and the required answer.

Note: We can solve the equation using quadratic formula as well but it is a calculative method. It increases the chances of making an error. We should use it only when splitting the middle term is not possible.