(i)
$\dfrac{1}{2x} + \dfrac{1}{3y} = 2$
$\dfrac{1}{3x} + \dfrac{1}{2y} = \dfrac{13}{6}$
(ii)
$\dfrac{2}{\sqrt{x}} + \dfrac{3}{\sqrt{y}} = 2$
$\dfrac{4}{\sqrt{x}} - \dfrac{9}{\sqrt{y}} = -1$
(iii)
$\dfrac{4}{x} + 3y = 14$
$\dfrac{3}{x} - 4y = 23$
(iv)
$\dfrac{5}{x - 1} + \dfrac{1}{y - 2} = 2$
$\dfrac{6}{x - 1} - \dfrac{3}{y - 2} = 1$
(v)
$\dfrac{7x - 2y}{xy} = 5$
$\dfrac{8x + 7y}{xy} = 15$
(vi)
$6x + 3y - 6xy = 0$
$2x + 4y - 5xy = 0$
(vii)
$\dfrac{10}{x + y} + \dfrac{2}{x - y} = 4$
$\dfrac{15}{x + y} - \dfrac{5}{x - y} = -2$
(viii)
$\dfrac{1}{3x + y} + \dfrac{1}{3x - y} = \dfrac{3}{4}$
$\dfrac{1}{2(3x + y)} - \dfrac{1}{2(3x - y)} = -\dfrac{1}{8}$
Solution
(i)
$\dfrac{1}{2x} + \dfrac{1}{3y} = 2$ -----(i)
$\dfrac{1}{3x} + \dfrac{1}{2y} = \dfrac{13}{6}$ -----(ii)
Let $\dfrac{1}{x} = p$ and $\dfrac{1}{y} = q$
Putting these values in equations (i) and (ii), we get,
$\dfrac{p}{2} + \dfrac{q}{3} = 2$ $⇒ 3p + 2q - 12 = 0$ -----(iii)
$\dfrac{p}{3} + \dfrac{q}{2} = \dfrac{13}{6}$ $⇒ 2p + 3q - 13 = 0$ -----(iv)
Using cross-multiplication method, we obtain,
$\dfrac{p}{(-26-(-36))}$ = $\dfrac{q}{(-24-(-39))}$ = $\dfrac{1}{9-4}$
$\dfrac{p}{10}$ = $\dfrac{q}{15}$ = $\dfrac{1}{5}$
$∴ \dfrac{p}{10}$ = $\dfrac{1}{5}$ and $\dfrac{q}{15}$ = $\dfrac{1}{5}$
$∴ p = 2$ and $q = 3$
$∴ \dfrac{1}{x} = 2$ and $\dfrac{1}{y} = 3$
Hence $x = \dfrac{1}{2}$ and $y = \dfrac{1}{3}$
(ii)
$\dfrac{2}{\sqrt{x}} + \dfrac{3}{\sqrt{y}} = 2$ -----(i)
$\dfrac{4}{\sqrt{x}} - \dfrac{9}{\sqrt{y}} = -1$ -----(ii)
Let $\dfrac{1}{\sqrt{x}} = p$ and $\dfrac{1}{\sqrt{y}} = q$
Putting these values in equations (i) and (ii), we get,
$2p + 3q = 2$ -----(iii)
$4p - 9q = -1$ -----(iv)
Multiplying equation (iii) by 3 and adding the result with equation (iv), we obtain,
$6p + 9q = 6$
$4p - 9q = -1$
---------------
$10p = 5$
$⇒ p = \dfrac{5}{10} = \dfrac{1}{2}$
Substituting value of p in equation (iii), we get
$2 × \dfrac{1}{2} + 3q = 2$
$⇒ 3q = 1 ⇒ q = \dfrac{1}{3}$
As $\dfrac{1}{\sqrt{x}} = p = \dfrac{1}{2}$
$⇒ \sqrt{x} = 2 ⇒ x = 4$
As $\dfrac{1}{\sqrt{y}} = q = \dfrac{1}{3}$
$⇒ \sqrt{y} = 3 ⇒ x = 9$
Hence $x = 4$ and $y = 9$
(iii)
$\dfrac{4}{x} + 3y = 14$ -----(i)
$\dfrac{3}{x} - 4y = 23$ -----(ii)
Let $\dfrac{1}{x} = p$
Putting this value in equations (i) and (ii), we get,
$4p + 3y = 14$ -----(iii)
$3p - 4y = 23$ -----(iv)
Using cross-multiplication method, we obtain,
$\dfrac{p}{(-69-56)}$ = $\dfrac{y}{(-42-(-92))}$ = $\dfrac{1}{-16-9}$
$\dfrac{p}{-125}$ = $\dfrac{y}{50}$ = $\dfrac{1}{-25}$
$∴ \dfrac{p}{-125}$ = $\dfrac{1}{-25}$ and $\dfrac{y}{50}$ = $\dfrac{1}{-25}$
$∴ p = 5$ and $y = -2$
As $ p = \dfrac{1}{x} = 5 ⇒ x = \dfrac{1}{5}$
Hence $x = \dfrac{1}{5}$ and $y = -2$
(iv)
$\dfrac{5}{x - 1} + \dfrac{1}{y - 2} = 2$ -----(i)
$\dfrac{6}{x - 1} - \dfrac{3}{y - 2} = 1$ -----(ii)
Let $\dfrac{1}{x-1} = p$ and $\dfrac{1}{y-2} = q$
Putting these values in equations (i) and (ii), we get,
$5p + q = 2$ -----(iii)
$6p - 3q = 1$ -----(iv)
Multiplying equation (iii) by 3 and adding the result to (iv), we get
$15p + 3q = 6$
$6p - 3q = 1$
--------------
$21p = 7$ ⇒ $p = \dfrac{7}{21} = \dfrac{1}{3}$
Putting $p = \dfrac{1}{3}$ in equation (iii), we get
$5 × \dfrac{1}{3} + q = 2$ $⇒ q = 2 - \dfrac{5}{3} = \dfrac{1}{3}$
$\dfrac{1}{x-1} = p = \dfrac{1}{3}$
$ x - 1 = 3$
$ x = 4$
$\dfrac{1}{y-2} = p = \dfrac{1}{3}$
$ y - 2 = 3$
$ y = 5$
Hence $x = 4$ and $y = 5$
(v)
$\dfrac{7x - 2y}{xy} = 5$
$\dfrac{8x + 7y}{xy} = 15$
$\dfrac{7x - 2y}{xy} = 5$
$⇒ \dfrac{7x}{xy} - \dfrac{2y}{xy} = 5$
$⇒ \dfrac{7}{y} - \dfrac{2}{x} = 5$ -----(i)
$\dfrac{8x + 7y}{xy} = 15$
$⇒ \dfrac{8x}{xy} + \dfrac{7y}{xy} = 15$
$⇒ \dfrac{8}{y} + \dfrac{7}{x} = 15$ -----(ii)
Putting $\dfrac{1}{x} = p$ and $\dfrac{1}{y} = q$ in equation (i) and (ii), we get
$\dfrac{7}{y} - \dfrac{2}{x} = 5$
$⇒ -2p + 7q - 5 = 0$ -----(iii)
$\dfrac{8}{y} + \dfrac{7}{x} = 5$
$⇒ 7p + 8q -15 = 0$ -----(iv)
Using cross-multiplication method, we obtain,
$\dfrac{p}{-105-(-40)}$ = $\dfrac{q}{-35-30}$ = $\dfrac{1}{-16-49}$
$\dfrac{p}{-65}$ = $\dfrac{q}{-65}$ = $\dfrac{1}{-65}$
$∴ \dfrac{p}{-65}$ = $\dfrac{1}{-65}$ and $\dfrac{q}{-65}$ = $\dfrac{1}{-65}$
$∴ p = 1$ and $q = 1$
$\dfrac{1}{x} = p = 1$
$ x = 1$
$\dfrac{1}{y} = q = 1$
$ y = 1$
Hence $x = 1$ and $y = 1$
(vi)
$6x + 3y = 6xy$
$2x + 4y = 5xy$
By dividing both the equations by $xy$, we obtain,
$6x + 3y = 6xy$ ⇒ $\dfrac{6}{y} + \dfrac{3}{x} = 6$ -----(i)
$2x + 4y = 5xy$ ⇒ $\dfrac{2}{y} + \dfrac{4}{x} = 5$ -----(ii)
Putting $\dfrac{1}{x} = p$ and $\dfrac{1}{y} = q$ in equation (i) and (ii), we get
$\dfrac{6}{y} + \dfrac{3}{x} = 5$
$⇒ 3p + 6q - 6 = 0$ -----(iii)
$\dfrac{2}{y} + \dfrac{4}{x} = 5$
$⇒ 4p + 2q -5 = 0$ -----(iv)
Using cross-multiplication method, we obtain,
$\dfrac{p}{-30-(-12)}$ = $\dfrac{q}{-24-(-15)}$ = $\dfrac{1}{6-24}$
$\dfrac{p}{-18}$ = $\dfrac{q}{-9}$ = $\dfrac{1}{-18}$
$∴ \dfrac{p}{-18}$ = $\dfrac{1}{-18}$ and $\dfrac{q}{-9}$ = $\dfrac{1}{-18}$
$∴ p = 1$ and $q = \dfrac{1}{2}$
$\dfrac{1}{x} = p = 1$ ⇒ $ x = 1$
$\dfrac{1}{y} = q = \dfrac{1}{2}$ ⇒ $ y = 2$
Hence $x = 1$ and $y = 2$
(vii)
$\dfrac{10}{x + y} + \dfrac{2}{x - y} = 4$ -----(i)
$\dfrac{15}{x + y} - \dfrac{5}{x - y} = -2$ -----(ii)
Substituting $\dfrac{1}{x + y}$ by $p$ and $\dfrac{1}{x - y}$ by $q$ in equations (i) and (ii), we get
$10p + 2q = 4$
$10p + 2q - 4 = 0$-----(iii)
$15p - 5q = -2$
$15p - 5q + 2 = 0$-----(iv)
Using cross-multiplication method, we obtain,
$\dfrac{p}{2 - 10}$ = $\dfrac{q}{-30-10}$ = $\dfrac{1}{-25 - 15}$
$\dfrac{p}{-8}$ = $\dfrac{q}{-40}$ = $\dfrac{1}{-40}$
$∴ \dfrac{p}{-8}$ = $\dfrac{1}{-40}$ and $\dfrac{q}{-40}$ = $\dfrac{1}{-40}$
$∴ p = \dfrac{1}{5}$ and $q = 1$
As $\dfrac{1}{x+y} = p = \dfrac{1}{5}$ ⇒ $ x+y = 5$-----(v)
$\dfrac{1}{x-y} = q = 1$ ⇒ $ x-y = 1$ -----(vi)
Adding (v) and (vi), we get $2x = 6$ ⇒ $x = 3$
Substituting value of $x$ in equation (v), $ 3 + y = 5$, we get $y = 2$
Hence $x = 3$ and $y = 2$
(viii)
$\dfrac{1}{3x + y} + \dfrac{1}{3x - y} = \dfrac{3}{4}$ -----(i)
$\dfrac{1}{2(3x + y)} - \dfrac{1}{2(3x - y)} = -\dfrac{1}{8}$ -----(ii)
Substituting $\dfrac{1}{3x + y}$ by $p$ and $\dfrac{1}{3x - y}$ by $q$ in equations (i) and (ii), we get
$p + q = \dfrac{3}{4}$ -----(iii)
$p - q = -\dfrac{1}{4}$-----(iv)
Adding (iii) and (iv), we get
$2p = \dfrac{3}{4} - \dfrac{1}{4}$
$2p = \dfrac{1}{2}$
$p = \dfrac{1}{4}$
Substituting value of $p$ in equation (iv), $ \dfrac{1}{4} - q = -\dfrac{1}{4}$
$q = \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2}$
As $\dfrac{1}{3x+y} = p = \dfrac{1}{4}$ ⇒ $ 3x+y = 4$-----(v)
$\dfrac{1}{3x - y} = q = \dfrac{1}{2}$ ⇒ $ 3x - y = 2$ -----(vi)
Adding (v) and (vi), we get $6x = 6$ ⇒ $x = 1$
Substituting value of $x$ in equation (v), we get $ 3 + y = 4$ ⇒ $y = 1$
Hence $x = 1$ and $y = 1$
2
Formulate the following problems as a pair of equations, and hence find their solutions:
(i) Ritu can row downstream 20 km in 2 hours, and upstream 4km in 2 hours. Find her speed of rowing in still water
and the speed of the current.
(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in
3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels
60 km by train and remaining by bus. If she travels 100 km by train and the remaining by bus, she takes
10 minutes longer. Find the speed of the train and the bus separately.
Solution
(i)
Let the Ritu’s speed of rowing in still water and the speed of stream be $x$ km/h and $y$ km/h respectively.
Ritu’s speed of rowing;
Upstream = $(x - y)$ km/h
Downstream = $(x + y)$ km/h
Ritu can row downstream 20 km in 2 hours
$2(x + y) = 20$
$x + y = 10$ -----(i)
Ritu can row upstream 4 km in 2 hours
$2(x - y) = 4$
$x - y = 2$ -----(ii)
Adding (i) and (ii), we get,
$2x = 12 ⇒ x = 6$
Putting value of $x$ in (ii), we get $6 - y = 2$ ⇒ $y = 4$
Hence, Ritu’s speed of rowing in still water is 6 km/h and the speed of the current is 4 km/h
(ii)
Let the number of days taken by a woman and a man to finish the work be $x$ and $y$ respectively.
Therefore,
work done by a woman in 1 day \( = \dfrac{1}{x}\) and
work done by a man in 1 day \( = \dfrac{1}{y}\)
2 women and 5 men can together finish an embroidery work in 4 days
$∴ \dfrac{2}{x} + \dfrac{5}{y} = \dfrac{1}{4}$ -----(i)
3 women and 6 men can finish it in 3 days
$∴ \dfrac{3}{x} + \dfrac{6}{y} = \dfrac{1}{3}$ -----(ii)
Substituting $\dfrac{1}{x} = p$ and $\dfrac{1}{y} = q $ in equations (i) and (ii), we obtain
$\dfrac{2}{x} + \dfrac{5}{y} = \dfrac{1}{4}$
⇒ $2p + 5q = \dfrac{1}{4}$
⇒ $8p + 20q - 1 = 0 $ -----(iii)
$\dfrac{3}{x} + \dfrac{6}{y} = \dfrac{1}{3}$
⇒ $3p + 6q = \dfrac{1}{3}$
⇒ $9p + 18q - 1 = 0 $ -----(iv)
Using cross-multiplication method, we obtain,
$\dfrac{p}{-20 - (-18)}$ = $\dfrac{q}{-9-(-18))}$ = $\dfrac{1}{144 - 180}$
$\dfrac{p}{-2}$ = $\dfrac{q}{-1}$ = $\dfrac{1}{-36}$
$∴ \dfrac{p}{-2}$ = $\dfrac{1}{-36}$ and $\dfrac{q}{-1}$ = $\dfrac{1}{-36}$
$∴ p = \dfrac{1}{18}$ and $q = \dfrac{1}{36}$
As $\dfrac{1}{x} = p = \dfrac{1}{18}$ ⇒ $ x = 18$-----(v)
$\dfrac{1}{y} = q = \dfrac{1}{36}$ ⇒ $ y = 36$ -----(vi)
Hence number of days taken by a woman is 18 and by a man is 36
(iii)
Let the speed of train and bus be $u$ km/h and $v$ km/h respectively.
Roohi travels 300 km and takes 4 hours if she travels 60 km by train and remaining by bus
$\dfrac{{60}}{u} + \dfrac{{240}}{v} = 4$ -----(i)
If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer
$\dfrac{{100}}{u} + \dfrac{{200}}{v} = \dfrac{{25}}{6}$ -----(ii)
Substituting $\dfrac{1}{u}=p$ and $\dfrac{1}{v}=q$ in equations (i) and (ii), we obtain
$\dfrac{{60}}{{u}} + \dfrac{{240}}{v} = 4$
$⇒ 60p + 240q = 4$ -----(iii)
$\dfrac{{100}}{u} + \dfrac{{200}}{v} = \dfrac{{25}}{6}$
$⇒ 100p + 200q = \dfrac{{25}}{6} $
$⇒ 600p + 1200q = 25 $-----(iv)
Multiplying equation (iii) by 10, we obtain
$600p + 2400q = 40 $-----(v)
Subtracting equation (iv) from (v), we obtain
$1200q = 15$
$⇒q = \dfrac{{15}}{{1200}}$
$⇒q = \dfrac{1}{{80}}$
Substituting $q = \dfrac{1}{{80}} $ in equation (iii), we obtain
$60p + 240 × \dfrac{1}{{80}} = 4$
$60p = 4 - 3$
$p = \dfrac{1}{{60}}$
Therefore, $p = \dfrac{1}{u} = \dfrac{1}{{60}}$
$u = 60$
And $q = \dfrac{1}{v} = \dfrac{1}{{80}}$
$⇒ v = 80$
Hence, speed of the train = 60 km/h and speed of the bus = 80 km/h