(i) \(\ \frac{1}{2x}+\frac{1}{3y}=2 \), \(\ \frac{1}{3x}+\frac{1}{2y}=\frac{13}{6} \)
Solution:
\(\ \frac{1}{2x}+\frac{1}{3y}=2 \) ...(1)
and \(\ \frac{1}{3x}+\frac{1}{2y}=\frac{13}{6} \) ...(2)
Let \(\ \frac{1}{x}=a \) and \(\ \frac{1}{y}=b \)
(1) ⇒ \(\ \frac{a}{2}+\frac{b}{3}=2 \)
⇒ \(\ 6.\frac{a}{2}+6.\frac{b}{3}=2×6 \)
⇒ \(\ 3a+2b-12=0 \) ...(3)
(2) ⇒ \(\ \frac{a}{3}+\frac{b}{2}=\frac{13}{6} \)
⇒ \(\ 6\frac{a}{2}+6\frac{b}{3}=\frac{13}{6} ×6 \)
⇒ \(\ 2a+3b-13=0 \) ...(4)
On cross-multiplying (3) and (4), we get-
\(\ \frac{a}{(2)(-13)-(3)(-12)}=\frac{b}{(-12)(2)-(-13)(3)}=\frac{1}{(3)(3)-(2)(2)} \)
⇒ \(\ \frac{a}{-26+36}=\frac{b}{-24+39}=\frac{1}{9-4} \)
⇒ \(\ \frac{a}{10}=\frac{b}{15}=\frac{1}{5} \)
Now, \(\ \frac{a}{10}=\frac{1}{5} ⇒a=2 \)
Again, \(\ \frac{b}{15}=\frac{1}{5} ⇒b=3 \)
⸫ \(\ a=\frac{1}{x}=2⇒x=\frac{1}{2} \)
and \(\ b=\frac{1}{y}=3⇒y=\frac{1}{3} \)
(ii) \(\ \frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2 \), \(\ \frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1 \)
Solution:
\(\ \frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2 \) ...(1)
and \(\ \frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1 \) ...(2)
Let \(\ \frac{1}{\sqrt{x}}=a \) and \(\ \frac{1}{\sqrt{y}}=b \)
(1) ⇒ \(\ 2a+3b-2=0 \) ...(3)
(2) ⇒ \(\ 4a-9b+1=0 \)...(4)
On cross-multiplying (3) and (4), we get-
\(\ \frac{a}{3(-1)-(-9)(-2)}=\frac{b}{(-2)(4)-1(2)}=\frac{1}{2(-9)-4(3)} \)
⇒ \(\ \frac{a}{3-18}=\frac{b}{-8-2}=\frac{1}{-18-12} \)
⇒ \(\ \frac{a}{-15}=\frac{b}{-10}=\frac{1}{-30} \)
Now, \(\ \frac{a}{-15}=\frac{1}{-30} ⇒a=\frac{1}{2} \)
Again, \(\ \frac{b}{-10}=\frac{1}{-30} ⇒b=\frac{1}{3} \)
⸫ \(\ a=\frac{1}{\sqrt{x}}=\frac{1}{2}⇒x=2^2=4 \)
and \(\ b=\frac{1}{\sqrt{y}}=\frac{1}{3}⇒y=3^2=9 \)
(iii) \(\ \frac{4}{x}+3y=14 \), \(\ \frac{3}{x}-4y=23 \)
Solution:
\(\ \frac{4}{x}+3y=14 \) ...(1)
and \(\ \frac{3}{x}-4y=23 \) ...(2)
Let \(\ \frac{1}{x}=a \)
(1) ⇒ \(\ 4a+3y-14=0 \)...(3)
(2) ⇒ \(\ 3a-4y-23=0 \) ...(4)
On cross-multiplying (3) and (4), we get-
\(\ \frac{a}{3(-23)-(-4)(-14)}=\frac{y}{(-14)(3)-(-23)(4)}=\frac{1}{4(-4)-3(-3)} \)
⇒ \(\ \frac{a}{-69-56}=\frac{y}{-42+92}=\frac{1}{16-9} \)
⇒ \(\ \frac{a}{-125}=\frac{y}{-50}=\frac{1}{-25} \)
Now, \(\ \frac{a}{-125}=\frac{1}{-25} ⇒a=5 \)
Again, \(\ \frac{y}{-50}=\frac{1}{-25} ⇒y=2 \)
⸫ \(\ a=\frac{1}{x}=5⇒x=\frac{1}{5} \)
and \(\ y=2 \)
(iv) \(\ \frac{5}{x-1}+\frac{1}{y-2}=2 \), \(\ \frac{6}{x-1}-\frac{3}{y-2}=1 \)
Solution:
\(\ \frac{5}{x-1}+\frac{1}{y-2}=2 \) ...(1)
and \(\ \frac{6}{x-1}-\frac{3}{y-2}=1 \) ...(2)
Let \(\ \frac{1}{x-1}=a \) and \(\ \frac{1}{y-2}=b \)
(1) ⇒ \(\ 5a+b-2=0 \) ...(3)
(2) ⇒ \(\ 6a-3b-1=0 \) ...(4)
On cross-multiplying (3) and (4), we get-
\(\ \frac{a}{2(-1)-(-3)(-2)}=\frac{b}{(-2)(6)-(-1)(5)}=\frac{1}{5(-3)-6(1)} \)
⇒ \(\ \frac{a}{-1-6}=\frac{b}{-12+5}=\frac{1}{-15-6} \)
⇒ \(\ \frac{a}{-7}=\frac{b}{-7}=\frac{1}{-21} \)
Now, \(\ \frac{a}{-7}=\frac{1}{-21} ⇒a=\frac{1}{3} \)
Again, \(\ \frac{b}{-7}=\frac{1}{-21} ⇒b=\frac{1}{3} \)
⸫ \(\ x-1=\frac{1}{a}=3⇒x=3+1=4 \)
and \(\ y-2=\frac{1}{b}=3⇒y=3+2=5 \)
(v) \(\ \frac{7x-2y}{xy}=5 \), \(\ \frac{8x+7y}{xy}=15 \)
Solution:
\(\ \frac{7x-2y}{xy}=5 \) ⇒ \(\ \frac{7x}{xy}-\frac{2y}{xy}=5 \)
⇒ \(\ \frac{7}{y}-\frac{2}{x}=5 \) ...(1)
and \(\ \frac{8x+7y}{xy}=15 \) ⇒ \(\ \frac{8x}{xy}+\frac{7y}{xy}=15 \)
⇒ \(\ \frac{8}{y}+\frac{7}{x}=15 \) ...(2)
Let \(\ \frac{1}{y}=a \) and \(\ \frac{1}{x}=b \)
(1) ⇒ \(\ 7a-2b-5=0 \) ...(3)
(2) ⇒ \(\ 8a+7b-15=0 \) ...(4)
On cross-multiplying (3) and (4), we get-
\(\ \frac{a}{(-2)(-15)-(7)(-5)}=\frac{b}{(-5)(8)-(-15)(7)}=\frac{1}{(7)(7)-(8)(-2)} \)
⇒ \(\ \frac{a}{30+35}=\frac{b}{-40+105}=\frac{1}{49+16} \)
⇒ \(\ \frac{a}{65}=\frac{b}{65}=\frac{1}{65} \)
Now, \(\ \frac{a}{65}=\frac{1}{65} ⇒a=1 \)
Again, \(\ \frac{b}{65}=\frac{1}{65} ⇒b=1 \)
⸫ \(\ a=\frac{1}{y}=1⇒y=1 \)
and \(\ b=\frac{1}{x}=1⇒x=1 \)
(vi) \(\ 6x+3y=6xy \)
\(\ 2x+4y=5xy \)
Solution:
\(\ 6x+3y=6xy \) ⇒ \(\ \frac{6x+3y}{xy}=6 \)
⇒ \(\ \frac{6x}{xy}+\frac{3y}{xy}=6 \) ⇒ \(\ \frac{6}{y}+\frac{3}{x}=6 \) ...(1)
and \(\ 2x+4y=5xy \) ⇒ \(\ \frac{2x+4y}{xy}=5 \)
⇒ \(\ \frac{2x}{xy}+\frac{4y}{xy}=5 \) ⇒ \(\ \frac{2}{y}+\frac{4}{x}=5 \) ...(2)
Let \(\ \frac{1}{y}=a \) and \(\ \frac{1}{x}=b \)
(1) ⇒ \(\ 6a+3b-6=0 \) ...(3)
(2) ⇒ \(\ 2a+4b-5=0 \) ...(4)
On cross-multiplying (3) and (4), we get-
\(\ \frac{a}{(3)(-5)-(4)(-6)}=\frac{b}{(-6)(2)-(-5)(6)}=\frac{1}{(6)(4)-(2)(3)} \)
⇒ \(\ \frac{a}{-15+24}=\frac{b}{-12+30}=\frac{1}{24-6} \)
⇒ \(\ \frac{a}{9}=\frac{b}{18}=\frac{1}{18} \)
Now, \(\ \frac{a}{9}=\frac{1}{18} ⇒a=\frac{1}{2} \)
Again, \(\ \frac{b}{18}=\frac{1}{18} ⇒b=1 \)
⸫ \(\ y=\frac{1}{a}=\frac{1}{2}⇒y=2 \)
and \(\ x=\frac{1}{b}=\frac{1}{1}⇒x=1 \)
(vii) \(\ \frac{10}{x+y}+\frac{2}{x-y}=4 \), \(\ \frac{15}{x+y}-\frac{5}{x-y}=-2 \)
Solution:
\(\ \frac{10}{x+y}+\frac{2}{x-y}=4 \) ...(1)
and \(\ \frac{15}{x+y}-\frac{5}{x-y}=-2 \) ...(2)
Let \(\ \frac{1}{x+y}=a \) and \(\ \frac{1}{x-y}=b \)
(1) ⇒ \(\ 10a+2b-4=0 \) ...(3)
(2) ⇒ \(\ 15a-5b+2=0 \) ...(4)
On cross-multiplying (3) and (4), we get-
\(\ \frac{a}{2(2)-(-5)(-4)}=\frac{b}{(4)(15)-(2)(10)}=\frac{1}{10(-5)-15(2)} \)
⇒ \(\ \frac{a}{4-20}=\frac{b}{-60-20}=\frac{1}{-50-30} \)
⇒ \(\ \frac{a}{-16}=\frac{b}{-80}=\frac{1}{-80} \)
Now, \(\ \frac{a}{-16}=\frac{1}{-80} ⇒a=\frac{1}{5} \)
Again, \(\ \frac{b}{-80}=\frac{1}{-80} ⇒b=1 \)
\(\ a=\frac{1}{x+y}=\frac{1}{5}⇒ x+y=5 ⇒ x=5-y \) ...(5)
\(\ b=\frac{1}{x-y}=1⇒ x-y=1 ⇒ x=y+1 \) ...(6)
Comparing (5)&(6), we get,
\(\ 5-y=y+1⇒2y=5-1⇒y=2 \) ...(7)
(6)⇒ \(\ x= 2+1=3 \)
(viii) \(\ \frac{1}{3x+y}+\frac{1}{3x-y}=\frac{3}{4} \), \(\ \frac{1}{2(3x+y)}-\frac{1}{2(3x-y)}=-\frac{1}{8} \)
Solution:
\(\ \frac{1}{3x+y}+\frac{1}{3x-y}=\frac{3}{4} \) ...(1)
\(\ \frac{1}{2(3x+y)}-\frac{1}{2(3x-y)}=-\frac{1}{8} \) ...(2)
Let \(\ \frac{1}{3x+y}=a \) and \(\ \frac{1}{3x-y}=b \)
(1) ⇒ \(\ a+b=\frac{3}{4}⇒ 4a+4b=3 \) ...(3)
(2) ⇒ \(\ \frac{a}{2}-\frac{b}{2}=-\frac{1}{8}⇒-2a-2b=-\frac{1}{2} \)
\(\ ⇒4b-4a=-1 \) ...(4)
(3)+(4)⇒ \(\ 4a+4b+4a-4b=3-1 \)
⇒ \(\ 8a=2 ⇒ a=\frac{1}{4} \) ...(5)
Now, (3)⇒ \(\ 4\left(\frac{1}{4}\right)+4b=3 \) [using (5)]
⇒ \(\ 1+4b=3 ⇒ b=\frac{3-1}{4}⇒b=\frac{1}{2} \) ...(6)
Now, \(\ \frac{1}{3x+y}=\frac{1}{4}⇒ 3x+y=4 \) ...(7)
\(\ \frac{1}{3x-y}=\frac{1}{2}⇒ 3x-y=2 \) ...(8)
Again (7)+(8)⇒ \(\ 3x+y+3x-y=4+2 \)
\(\ ⇒ 6x=6 ⇒ x=1 \) ...(9)
(8)⇒ \(\ y=3(1)-2 \) [using (9)]
⇒ \(\ y = 3-2 ⇒ y=1 \)
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 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
Solution:
Let the speed of Ritu be \(\ x \) km/h and that of the current be \(\ y \) km/h
For downstream, speed = \(\ (x+y) \) km/h
For upstream, speed = \(\ (x-y) \) km/h
A/q, \(\ x+y = \frac{20}{2} ⇒ x+y=10 \) …(1)
Again, \(\ x-y = \frac{4}{2} ⇒ x-y=2 \) …(2)
Now, (1)+(2) ⇒ \(\ x+y+x-y=10+2 \)
\(\ ⇒ 2x = 12 ⇒ y = 6 \) ...(3)
Now, (2)⇒ \(\ x=2+6 \) [using (3)]
\(\ ⇒x=8 \)
⸫ The required speed of Ritu is 8 km/h and that of the current is 6 km/h.
(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.
Solution:
Let the time taken by 1 woman to finish the work be \(\ x \) days and that of by 1 man be \(\ y \) days.
work done by 1 woman in 1 day = \(\ \frac{1}{x} \)
work done by 1 man in 1 day = \(\ \frac{1}{y} \)
A/q, \(\ 4\left(\frac{2}{x}+\frac{5}{y}\right)=1 ⇒ \frac{2}{x}+\frac{5}{y}=\frac{1}{4} \) …(1)
Again \(\ 3\left(\frac{3}{x}+\frac{6}{y}\right)=1 ⇒ \frac{3}{x}+\frac{6}{y}=\frac{1}{3} \) …(2)
Let, \(\ \frac{1}{x}=a \) and \(\ \frac{1}{y}=b \)
Now, (1) ⇒ \(\ 2a+5b=\frac{1}{4} ⇒ 8a+20b=1 \) …(3)
(2) ⇒ \(\ 3a+6b=\frac{1}{3} ⇒ 9a+18b=1 \) …(4)
Now,
(3)×9-(4)×8 ⇒
\(\ 72a+180b-(72a+144b)=9-8 \)
\(\ ⇒ 72a+180b-72a-144b=1 \)
\(\ ⇒ 180b-144b=1 \)
\(\ ⇒ b=\frac{1}{36} \)
(4)⇒ \(\ 9a+18\left(\frac{1}{36}\right)=1 \)
\(\ ⇒ 9a+\frac{1}{2}=1\)
\(\ ⇒ 9a=1-\frac{1}{2}\)
\(\ ⇒ 9a=\frac{1}{2}\)
\(\ ⇒ a=\frac{1}{18}\)
\(\ \frac{1}{x}=\frac{1}{18}\ ⇒ x=18 \)
and \(\ \frac{1}{y}=\frac{1}{36}\ ⇒ y = 36 \)
⸫ The time taken by 1 woman to finish the work is 18 days and that of by 1 man is 36 days.
(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:
Let the speed of the train be \(\ x \) km/h and that of the bus be \(\ y \) km/h.
Given, total distance she travels = 300km
A/q, \(\ \frac{60}{x}+\frac{240}{y}=4 \)
\(\ ⇒ 4\left(\frac{15}{x}+\frac{60}{y}\right)=4 \)
\(\ ⇒ \frac{15}{x}+\frac{60}{y}=1 …(1) \)
Again \(\ \frac{100}{x}+\frac{200}{y}=4+\frac{10}{60} \)
\(\ ⇒ \frac{100}{x}+\frac{200}{y}=\frac{25}{6} \)
\(\ ⇒ \frac{600}{x}+\frac{1200}{y}=25\)
\(\ ⇒ 25\left(\frac{24}{x}+\frac{48}{y}\right)=25 \)
\(\ ⇒ \frac{24}{x}+\frac{48}{y}=1 \) …(2)
Let, \(\ \frac{1}{x}=a \) and \(\ \frac{1}{y}=b \)
Now, (1) ⇒ \(\ 15a+60b=1 \) …(3)
(2) ⇒ \(\ 24a+48b=1 \) …(4)
Now,
(3)×8-(4)×5 ⇒
\(\ 120a+480b-(120a+240b)=8-5 \)
\(\ ⇒ 120a+480b-120a-240b=3 \)
\(\ ⇒ 480b-240b=3 \)
\(\ ⇒ b=\frac{3}{240}=\frac{1}{80} \)
(3)⇒ \(\ 15a+60\left(\frac{1}{80}\right)=1 \)
\(\ ⇒ 15a+\frac{3}{4}=1 \)
\(\ ⇒ 15a=1-\frac{3}{4} \)
\(\ ⇒ a=\frac{1}{4×15} \)
\(\ ⇒ a=\frac{1}{60} \)
Now, \(\ \frac{1}{x}=\frac{1}{60} ⇒x=60 \) and \(\ \frac{1}{y}=\frac{1}{80} ⇒ y = 80 \)
⸫ The speed of the train is 60 km/h and that of the bus is 80 km/h.
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