Question

Solve for x:
$\dfrac{{x - 3}}{{x - 4}} + \dfrac{{x - 5}}{{x - 6}} = \dfrac{{10}}{3};x \ne 4,6$

Answer 1

Hint: Use the lowest common factor to solve the question and simplify the equation. Start solving the equation by taking the L.C.M of $x - 4$ and $x - 6$. Do the calculation carefully use the quadratic formula for finding the value of \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].

Complete step-by-step answer:
We have given equation is
$\dfrac{{x - 3}}{{x - 4}} + \dfrac{{x - 5}}{{x - 6}}$$ = \dfrac{{10}}{3}$
Take the L.C.M of $(x - 4)(x - 6)$
$\dfrac{{(x - 3)(x - 6) + (x - 5)(x - 4)}}{{(x - 4)(x - 6)}} = \dfrac{{10}}{3}$
Open the brackets
$\dfrac{{({x^2} - 6x - 3x + 18) + ({x^2} - 4x - 5x + 20)}}{{{x^2} - 4x - 6x + 24}} = \dfrac{{10}}{3}$
Solve the equation of the denominator and numerator
$\dfrac{{({x^2} - 9x + 18) + ({x^2} - 9x + 20)}}{{{x^2} - 10x + 24}} = \dfrac{{10}}{3}$
Open the brackets and solve the equation
$\dfrac{{{x^2} - 9x + 18 + {x^2} - 9x + 20}}{{{x^2} - 10x + 24}} = \dfrac{{10}}{3}$
Add the variables and constants
$\dfrac{{2{x^2} - 18x + 38}}{{{x^2} - 10x + 24}} = \dfrac{{10}}{3}$
Evaluate the equation
$(2{x^2} - 18x + 38).3 = ({x^2} - 10x + 24).10$
Multiply the left-hand side by 3 and right-hand side by 10
$6{x^2} - 54x + 114 = 10{x^2} - 100x + 240$
Add the variable and constant that having same power
$6{x^2} - 10{x^2} - 54x + 100x + 114 - 240 = 0$
Add and subtract
$ - 4{x^2} + 46x - 126 = 0$
Multiply the equation by -1
$ - 1( - 4{x^2} + 46x - 126) = 0$
We get
 $4{x^2} - 46x + 126 = 0$
Divide the equation by 2 to simplify it
\[\dfrac{{4{x^2} - 46x + 126}}{2} = \dfrac{0}{2}\]
We get
$2{x^2} - 23x + 63 = 0$
Here we have our equation in the form of $a{x^2} + bx + c = 0$
We can find the value of x by the formula
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Here we have $a = 2,b = - 23,c = 63$
Put the values in the formula
$x = \dfrac{{ - ( - 23) \pm \sqrt { {(-23)^2} - 4.2.63} }}{{2.2}}$
Solve the number that is under the root first
$x = \dfrac{{23 \pm \sqrt {529 - 504} }}{4}$
Subtract 504 by 529
$x = \dfrac{{23 \pm \sqrt {25} }}{4}$
We know 25 is a perfect square of 5
So, here we have two values of x
\[x = \dfrac{{23 + 5}}{4},\dfrac{{23 - 5}}{4}\]
Add and subtract
\[x = \dfrac{{28}}{4},\dfrac{{18}}{4}\]
Divide both by 4
$x = 7,\dfrac{9}{2}$
Hence we have the values of $x = 7,\dfrac{9}{2}$ that is satisfy the equation $x \ne 4,6$

Note: The most important thing to solve this question is calculation. Always make the equation simple by dividing by their common factor and try to make the digit always positive, that is with the ${x^2}$. Use the quadratic formula of finding x. Students mostly make the mistake in the calculation in that type of question.