Question

In the system of given equations $\dfrac{{3x}}{{x - 1}} + \dfrac{{4y + 20}}{{y - 3}} = 32$ and $\dfrac{{5x}}{{x - 1}} - \dfrac{{y + 5}}{{y - 3}} = 15$ , values of $x$ and $y$ which satisfies the given equations are.
A) $(5,\dfrac{4}{3})$
B) $(\dfrac{4}{3},5)$
C) $(\dfrac{5}{3},4)$
D) $(4,\dfrac{5}{4})$

Answer 1

Hint: Here in this question there are two equations from which two simplified equations will be made in which there will be two variables $x$ and $y$ .Further after using elimination method there will be one more equation containing variables $x$ and $y$ . This relation will get putted into the main equation by which only one variable quadratic equation is made and then by applying sridharacharya formula and can find values of $x$ and $y$

Complete step-by-step answer:
$\dfrac{{3x}}{{x - 1}} + \dfrac{{4y + 20}}{{y - 3}} = 32$
Taking L.C.M of this equation and further simplifying.
$ \Rightarrow \dfrac{{3x(y - 3) + (4y + 20)(x - 1)}}{{(x - 1)(y - 3)}} = 32$
$ \Rightarrow \dfrac{{3xy - 9 + 4xy + 20x - 4y - 20}}{{xy - y + 3 - 3x}} = 32$ (Opening the brackets and further simplifying) $ \Rightarrow \dfrac{{7xy + 20x - 4y - 29}}{{xy - y + 3 - 3x}} = 32$
$ \Rightarrow 7xy + 20x - 4y - 29 = 32(xy - y + 3 - 3x)$ (Cross multiplying and further simplifying) $ \Rightarrow 7xy + 20x - 4y - 29 = 32xy - 32y + 96 - 96x$
$ \Rightarrow 25xy - 28y - 116x = ( - 125)$ (Taking variables on one side)
$25xy - 28y - 116x = ( - 125)$ Equation (i)
$\dfrac{{5x}}{{x - 1}} - \dfrac{{y + 5}}{{y - 3}} = 15$
Taking L.C.M of this equation and further simplifying.
\[ \Rightarrow \dfrac{{5x(y - 3) - (x - 1)(y + 5)}}{{(x - 1)(y - 3)}} = 15\]
\[\dfrac{{ \Rightarrow 5xy - 15 - (xy - y - 5 + 5x)}}{{xy - 3x - y + 3}} = 15\]
\[ \Rightarrow \dfrac{{5xy - 15 - xy + y + 5 - 5x}}{{xy - 3x - y + 3}} = 15\] (Opening the brackets and further simplifying) \[ \Rightarrow \dfrac{{4xy - 10 + y - 5x}}{{xy - 3x - y + 3}} = 15\]
\[ \Rightarrow 4xy - 10 + y - 5x = 15xy - 45x - 15y + 45\] (Cross multiplying and further simplifying) \[ \Rightarrow 11xy - 16y - 40x = ( - 55)\] (Taking variables on one side) \[11xy - 16y - 40x = ( - 55)\] Equation (ii)
Now multiplying equation (i) with $11$ and equation (ii) with $25$ $ \Rightarrow 275xy - 308y - 1276x = ( - 1375)$ Equation (iii) $ \Rightarrow 275xy - 400y - 1000x = ( - 1375)$ Equation (iv)
Subtracting (iii) from (iv)
$ \Rightarrow - 400y + 308y - 1000x + 1276x = 0$
$ \Rightarrow - 92y + 276x = 0$
$ \Rightarrow 276x = 92y$
$ \Rightarrow \dfrac{x}{y} = \dfrac{{92}}{{276}}$
$ \Rightarrow \dfrac{x}{y} = \dfrac{1}{3}$
$ \Rightarrow y = 3x$
Putting $y = 3x$ in the main equation
$ \Rightarrow \dfrac{{3x}}{{x - 1}} + \dfrac{{4(3x) + 20}}{{3x - 3}} = 32$
\[ \Rightarrow \dfrac{{9{x^2} - 9x + 12x + 20}}{{3{x^2} - 6x + 3}} = 32\] (Taking LCM and then further solving) \[ \Rightarrow 9{x^2} + 3x + 20 = 96{x^2} - 192x + 96\] (Cross multiplying and then further simplifying) \[ \Rightarrow 87{x^2} - 189x + 76 = 0\] (Forming quadratic equation)
\[ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] (Using formula of sridharacharya) \[\therefore x = \dfrac{{189 \pm \sqrt {{{(189)}^2} - 4(87)(76)} }}{{2(87)}}\]
\[\therefore {x_1} = 0.53,{x_2} = 1.63\] (Putting the value from the equation and further solving)
Putting in y we will get ${y_1} = 1.59,{y_2} = 4.89$

Approximately option (3) is satisfying so the correct option is $(\dfrac{5}{3},4)$

Note: Sridharacharya formula: - It is a formula used to find the roots of quadratic equations and thus also known as quadratic formula. For example there is a quadratic equation given below:- $a{x^2} + bx + c = 0$ and $a,b,c$ are the constants so the sridharacharya formula can be written as ${x_1},{x_2} = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ where ${x_1},{x_2}$ are the two roots of the quadratic equation. Also, there is $D = \sqrt {{b^2} - 4ac} $ where ‘D’ is the discriminant. Don’t forget to take root after solving these common mistakes of most of the students.