Class IX Maths Solutions WBBSE Chapter 5 Linear Simultaneous Equations Exercise 5.5
Question 1. Let us express the value of the variable x from the equation \(\frac{2}{x}+\frac{3}{y}=1\) in terms of the variable y.
Solution: \(\frac{2}{x}+\frac{3}{y}=1\)
\(\frac{2}{x}=1-\frac{3}{y}\) \(\frac{2}{x}=\frac{y-3}{y}\)Read and Learn More WBBSE Solutions For Class 9 Maths
or, (y-3)x = 2y
or, \(x=\frac{2 y}{y-3}\)
Question 2. Let us write the value of x by putting \(\frac{7-4 x}{-5}\) -4x -5 instead of y in the equation 2x + 3y = 9.
Solution: 2x+3y=9
\(or, \quad 2 x+\frac{3(7-4 x)}{-5}=9or, \quad 2 x-\frac{3(7-4 x)}{5}=9
or, \quad \frac{10 x-21+12 x}{5}=9
or, \quad 22 x-21=45\)
Class IX Maths Solutions WBBSE
or, 22x = 45 +21
or,22x=66
or, \(x=\frac{66}{22}\)
∴ x = 3.
Question 3. Let us solve the following equations in two variables by substitution method and check them graphically.
1. 3x – y = 7, 2x + 4y = 0
Solutioin:
Given
3x – y = 7 …(1)
2x + 4y = 0…(2)
From equation (1)-y=-3x+7
or, y = 3x-7 …..(3)
Putting the value of y in equation (2),
2x + 4(3x-7)= 0
or, 2x+12x-28=0
or, 14x= 28
or, \(x=\frac{28}{14}\)
x = 2
Putting the value of y in equation (4),
y=3×2-7
or, y = 6-7
or,y = -1
∴ x = 2
y=-1
From equation (3)
y = 3x-7
| x | 2 | 4 | -1 |
| y | -1 | 5 | -10 |
From equation (2) 4y = -2x
or, \(y=\frac{-x}{2}\)
| x | 2 | 6 | -2 |
| y | -1 | -3 | 1 |
2. \(\frac{x}{2}+\frac{y}{3}=2=\frac{x}{4}+\frac{y}{2}\)
Solution:
\(\frac{x}{2}+\frac{y}{3}=2\) …(1)
\(\frac{x}{4}+\frac{y}{2}= \) …(2)
Class IX Maths Solutions WBBSE
From equation (1) \(\frac{x}{2}=2-\frac{y}{3}\)
or, \(\frac{x}{2}=\frac{6-y}{3}\)
or, \(x=\frac{12-2 y}{3}\) …(3)
Putting the value of y in equation (4),
\(\frac{\frac{12-2 y}{3}}{4}+\frac{y}{2}=2\)
or, \(\frac{2(6-y)}{3} \times \frac{1}{4}+\frac{y}{2}=2\)
or, \(\frac{6-y}{6}+\frac{y}{2}=2\)
or, \(\frac{6-y+3 y}{6}=2\)
or, 6+ 2y = 12
or, 2y=12-6
or, y = 6/2
∴ y = 3
Putting the value of y in equation (4),
\(x=\frac{12-2 \times 3}{3}\)
or, \(x=\frac{12-6}{3}\)
or, \(x=\frac{6}{3}\)
∴ x = 2, y=2
⇒ y = 3
From equation (3)
\(x=\frac{12-2 y}{3}\)
| x | 4 | 2 | -6 |
| y | 0 | 3 | -3 |
From equation (2)
\(\frac{x}{4}=2-\frac{y}{2}\)
or, x = 8- 2y
| x | 4 | 0 | 12 | 2 |
| y | 2 | 4 | -2 | 3 |
Class IX Maths Solutions WBBSE
Question 4. Let us solve the following equations in two variables by substitution method and check whether the solutions satisfy the equations:
1. \(2 x+\frac{3}{y}=1,5 x-\frac{2}{y}=\frac{11}{12}\)
Solution:
Given
\(2 x+\frac{3}{y}=1\) …(1)
\(5 x-\frac{2}{y}=\frac{11}{12}\) …..(2)
From equation (1) \(\frac{3}{y}=1-2 x\)
or, (1-2x)y = 3
or, \(y=\frac{3}{1-2 x}\)
Putting the value of y in equation (2),
\(5 x-\frac{2(1-2 x)}{3}=\frac{11}{12}\)
or, \(\frac{15 x-2+4 x}{3}=\frac{11}{12}\)
or, \(\quad 19 x-2=\frac{11}{4}\)
or, \(19 x=\frac{11}{4}+2\)
or, \(19 x=\frac{11+8}{4}\)
or, \(x=\frac{19}{19 \times 4}\)
or, \(x=\frac{1}{4}\)
Putting the value of x in equation (3)
\(y=\frac{3}{1-2 \times \frac{1}{4}}\)
or, \(y=\frac{3}{1-\frac{1}{2}}\)
or, y = 6
⇒ \(x=\frac{1}{4}\)
y = 6
From equation (1) L. H.S. = \(2 x+\frac{3}{y}\)
\(\begin{aligned}& =2 \times \frac{1}{4}+\frac{3}{6} \\
& =\frac{1}{2}+\frac{1}{2} \\
& =\frac{1+1}{2} \\
& =\frac{2}{2}
\end{aligned}\)
Class IX Maths Solutions WBBSE
= 1
= R. H. S.
Again from equation (2) L. H. S. = \(5 x-\frac{2}{y}\)
\(\begin{aligned}& =5 \times \frac{1}{4}-\frac{2}{6} \\
& =\frac{5}{4}-\frac{1}{3} \\
& =\frac{15-4}{12}
\end{aligned}\)
= \(\frac{11}{12}\)
= R. H. S.
∴ \(x=\frac{1}{4}\) & y = 6 satify the equations (1) & (2).
2. \(\frac{2}{x}+\frac{3}{y}=2, \frac{5}{x}+\frac{10}{y}=5 \frac{5}{6}\)
Solution:
Given
\(\frac{2}{x}+\frac{3}{y}=2\) …(1)
\(\frac{5}{x}+\frac{10}{y}=5 \frac{5}{6}\)….(2)
From equation (1) \(\frac{3}{y}=2-\frac{2}{x}\)
or, \(\frac{3}{y}=\frac{2 x-2}{x}\)
or,(2x-2)y = 3 x
or, \(y=\frac{3 x}{2 x-2}\) ….(3)
Putting the value of x in equation (2),
\(\frac{5}{x}+\frac{10(2 x-2)}{3 x}=\frac{35}{6}\)Class 9 Mathematics West Bengal Board
or, \(\frac{15+20 x-20}{3 x}=\frac{35}{6}\)
or, \(\frac{20 x-5}{x}=\frac{35}{2}\)
or, 40x-10=35x
or, 40x-35x=10
or, 5x = 10
or, \(x=\frac{10}{5}\)
or, x = 2
Putting the value of x in equation (3),
or, \(y=\frac{3 \times 2}{2 \times 2-2}\)
or, \(y=\frac{6}{4-2}\)
or, \(y=\frac{6}{2}\)
∴ y = 3, x = 2
From equation (1) L. H.S. = \(\frac{2}{x}+\frac{3}{y}\)
= \(\frac{2}{2}+\frac{3}{3}\)
= 1 + 1 = 2
= R. H. S.
Again, from equation (2) L. H. S. \(\frac{5}{x}+\frac{10}{y}\)
\(\begin{aligned}& =\frac{5}{2}+\frac{10}{3} \\
& =\frac{15+20}{6} \\
& =\frac{35}{6} \\
& =5 \frac{5}{6}
\end{aligned}\)
= R. H. S.
3. \(\frac{x+y}{x y}=3, \frac{x-y}{x y}=1\)
Solution:
Given
\(\frac{x+y}{x y}=3\) …(1)
\(\frac{x-y}{x y}=1\) ….(2)
From equation (1) \(\frac{x}{x y}+\frac{y}{x y}=3\)
\(or, \frac{1}{y}+\frac{1}{x}=3or, \frac{1}{y}=3-\frac{1}{x}
or, \quad \frac{1}{y}=\frac{3 x-1}{x}\)
Class 9 Mathematics West Bengal Board
or, (3x-1)y = x
or, \(y=\frac{x}{3 x-1}\) ….(3)
From equation (1) \(\frac{x}{x y}-\frac{y}{x y}=1\)
or, \(\frac{1}{y}-\frac{1}{x}=1\) …..(4)
Putting the value \(y=\frac{x}{3 x-1}\) in equation (4)
or, \(\frac{1}{\frac{x}{3 x-1}}-\frac{1}{x}=1\)
or, \(\frac{3 x-1}{x}-\frac{1}{x}=1\)
or, \(\frac{3 x-1-1}{x}=1\)
or, 3x-2=x
or,3x-x=2
or,2x = 2
or, \(x=\frac{2}{2}\)
or, X = 1
Putting the value of x in equation (3),
or, \(y=\frac{1}{3 \times 1-1}\)
or, \(y=\frac{1}{3-1}\)
or, \(y=\frac{1}{2}\)
∴ x =1
\(y=\frac{1}{2}\)From equation (1) L. H.S. = \(\frac{x+y}{x y}\)
= \(=\frac{1+\frac{1}{2}}{1 \times \frac{1}{2}}\)
\(\begin{aligned}& \frac{2+1}{2} \\
= & \frac{1}{2} \\
= & \frac{3}{2} \times \frac{2}{1}
\end{aligned}\)
Class 9 Mathematics West Bengal Board
= 3
= R. H. S.
Again, from equation (2) L. H. S. = \(\frac{x-y}{x y}\)
\(\begin{aligned}& =\frac{1-\frac{1}{2}}{1 \times \frac{1}{2}} \\
& =\frac{\frac{2-1}{2}}{\frac{1}{2}} \\
& =\frac{1}{2} \times \frac{2}{1}
\end{aligned}\)
= 1
= R. H. S.
4. \(\frac{x+y}{x-y}=\frac{7}{3}, \quad x+y=\frac{7}{10}\)
Solution:
Given
\(\frac{x+y}{x-y}=\frac{7}{3}\) ….(1)
\(x+y=\frac{7}{10}\) ……(2)
From equation (1) \(\frac{x+y}{x-y}=\frac{7}{3}\)
or, 7x-7y=3x+3y
or, 7x-3x=7y+ 3y
or, 4x = 10y
or, \(x=\frac{10 y}{4}\)
or, \(x=\frac{5 y}{2}\) …..(3)
Putting the value of y in equation (2),
\(or,\frac{5 y}{2}+y=\frac{7}{10}or,\quad\frac{5 y+2 y}{2}=\frac{7}{10}
or,\frac{7 y}{1}=\frac{7}{5}
or,\quad y=\frac{7}{5 \times 7}
or, \quad y=\frac{1}{5}\)
Class 9 Math Chapter 5 WBBSE
Putting the value of y in equation (2),
\(or, \quad x+\frac{1}{5}=\frac{7}{10}or,\quad x=\frac{7}{10}-\frac{1}{5}
or,\quad x=\frac{7-2}{10}
or,\quad x=\frac{5}{10}
or,\quad x=\frac{1}{2}
therefore x=\frac{1}{2}
y=\frac{1}{5}\)
From equation (1) L. H.S. \(=\frac{x+y}{x-y}\)
= \(=\frac{\frac{1}{2}+\frac{1}{5}}{\frac{1}{2}-\frac{1}{5}}\)
= \(=\frac{\frac{7}{10}}{\frac{3}{10}}\)
= \(=\frac{7}{3}\)
= R. H. S.
Again, from equation (1) L. H. S. = x + y
\(\begin{aligned}& =\frac{1}{2}+\frac{1}{5} \\
& =\frac{5+2}{10} \\
& =\frac{7}{10}
\end{aligned}\)
= R. H. S.
Question 5. Let us solve the following equations in two variables by substitution method:
1. 2(x-y) = 3, 5x + 8y = 14
Solution:
Given
2(x-y)=3 ….(1)
5x + 8y = 14 …..(2)
From equation (1) 2x-2y=3
or, 2x = 3 + 2y
or, \(x=\frac{3+2 y}{2}\) ….(3)
Putting the value of x in equation (2),
\(\frac{5(3+2 y)}{2}+8 y=14\)Class 9 Math Chapter 5 WBBSE
or, \(\frac{15+10 y+16 y}{2}=14\)
or, 15+26y=28
or, 26y=28-15
or, 26y = 13
or, \quad y=\frac{1}{2}\)
Putting the value of y in equation (3),
\(x=\frac{3+2 \times \frac{1}{2}}{2}\)
or, \(x=\frac{3+1}{2}\)
or, \(x=\frac{4}{2}\)
or, x =2
∴ x =2 , y= 1/2
2. \(2 x+\frac{3}{y}=5,5 x-\frac{2}{y}=3\)
Solution:
\(2 x+\frac{3}{y}=5 \) …..(1)
\(5 x-\frac{2}{y}=3\) …..(2)
Class 9 Math Chapter 5 WBBSE
From equation (1) \(2 x=5-\frac{3}{y}\)
or, \(2 x=\frac{5 y-3}{y}\)
or, \(x=\frac{5 y-3}{2 y}\) ….(3)
Putting the value of x in equation (2),
\(\frac{5(5 y-3)}{2 y}-\frac{2}{y}=3\)
or, \(\frac{25 y-15-4}{2 y}=3\)
or, 25y-19 = 6y
or,25y-6y 19
or,19y= 19
or, y = \(\frac{19}{19}\)
or, y = 1
Putting the value of y in equation (3),
\(x=\frac{5-3}{2}\)
or, \(x=\frac{5-3}{2}\)
or, \(x=\frac{2}{2}\)
or, x = 1
∴ x = , y = 1
3. \(\frac{x}{2}+\frac{y}{3}=1, \frac{x}{3}+\frac{y}{2}=1\)
Solution: \(\frac{x}{2}+\frac{y}{3}=1\) ….(1)
\(\frac{x}{3}+\frac{y}{2}=1\) ….(2)
From equation (1) \(\frac{x}{2}=1-\frac{y}{3}\)
or,\(\frac{x}{2}=\frac{3-y}{3}\)
or, 3x = 6-2y
or, \(x=\frac{6-2 y}{3}\) …(3)
Putting the value of y in equation (3),
\(\frac{\frac{6-2 y}{3}}{3}+\frac{y}{2}=1\)Class 9 Math Chapter 5 WBBSE
or, \(\frac{6-2 y}{3} \times \frac{1}{3}+\frac{y}{2}=1\)
or, \(\frac{12-4 y+9 y}{18}=1\)
or, 12 + 5y = 18
or, 5y = 18-12
or, \(y=\frac{6}{5}\)
Putting the value of y in equation (3),
\(x=\frac{6-2 \times \frac{6}{5}}{3}\)\(or, x=\frac{\frac{30-12}{5}}{3}
or, \quad x=\frac{18}{5} \times \frac{1}{3}
or, \quad x=\frac{6}{5}\)
∴ \(x=\frac{6}{5}\)
∴ \(y=\frac{6}{5}\)
4. \(\frac{x}{3}=\frac{y}{4}\), 7x-5y = 2
Solution:
Given
\(\frac{x}{3}=\frac{y}{4}\) ….(1)
7x-5y – 2 ….(2)
From equation(1) 4x = 3y
or, \(x=\frac{3 y}{4}\) …..(3)
Putting the value of x in equation. (2),
\(7 \times \frac{3 y}{4}-5 y=2\)
or, \(\frac{21 y-20 y}{4}=2\)
or, y = 8
Putting the value of y in equation (3),
\(x=\frac{3 \times 8}{4}\)Class 9 Math Solution WBBSE In English
∴ x = 6, y = 8
5. \(\frac{2}{x}+\frac{5}{y}=1, \frac{3}{x}+\frac{2}{y}=\frac{19}{20}\)
Solution:
Given
\(\frac{2}{x}+\frac{5}{y}=1\) …(1)
\(\frac{3}{x}+\frac{2}{y}=\frac{19}{20}\)……..(2)
From equation (1) \(\frac{2}{x}=1-\frac{5}{y}\)
or, \(\frac{2}{x}=\frac{y-5}{y}\)
or, (y-5)x = 2y
or, \(x=\frac{2 y}{y-5}\) ….(3)
Putting the value of x in equation (2),
\(\frac{3}{\frac{2 y}{y-5}}+\frac{2}{y}=\frac{19}{20}\)
or, \(\frac{3(y-5)}{2 y}+\frac{2}{y}=\frac{19}{20}\)
or, \(\frac{3 y-15+4}{2 y}=\frac{19}{20}\)
or, \(\frac{3 y-11}{y}=\frac{19}{10}\)
or, 30y-110 = 19y
or,30y-19y=110
or,11y=110
or, \(y=\frac{110}{11}\)
or,y = 10
Putting the value of y in equation (3),
\(x=\frac{2 \times 10}{10-5}\)Class 9 Math Solution WBBSE In English
or,\(x=\frac{20}{5}\)
or,X = 4
∴ X=4
y = 10
6. \(\frac{1}{3}\)(x − y) = \(\frac{1}{3}\) (y-1), \(\frac{1}{3}\)(4x-5y) = x – 7
Solution:
Given
\(\frac{1}{3}\)(x − y) = \(\frac{1}{3}\) (y-1) ….(1)
\(\frac{1}{3}\)(4x-5y) = x – 7 …..(2)
From equation (1) \(\frac{x-y}{3}=\frac{y-1}{4}\)
or, 4x-4y=3y-3
or, 4x=3y – 3+ 4y
or, 4x = 7y-3
or, \(x=\frac{7 y-3}{4}\) ….(3)
Putting the value of x in equation (2),
\(\frac{1}{7}\left\{4 \frac{(7 y-3)}{4}-5 y\right\}=\frac{7 y-3}{4}-7\)Class 9 Math Solution WBBSE In English
or, \(\frac{1}{7}(7 y-3-5 y)=\frac{7 y-3-28}{4}\)
or, \(\frac{2 y-3}{7}=\frac{7 y-31}{4}\)
or, 49y-217 = 8y-12
or, 49y-8y= -12+217
or, 41y=205
or, \(y=\frac{205}{41}\)
or, y = 5
Putting the value of y in equation (3),
\(x=\frac{7 \times 5-3}{4}\)\(or, x=\frac{35-3}{4} or, x=\frac{32}{4}\)
or, x =8
∴ x = 8, y =5
7. \(\frac{x}{14}+\frac{y}{18}=1, \frac{x+y}{2}+\frac{3 x-5 y}{4}=2\)
Solution:
\(\frac{x}{14}+\frac{y}{18}=1\) ….(1)
\(\frac{x+y}{2}+\frac{3 x-5 y}{4}=2\) ….(2)
Class 9 Math Solution WBBSE In English
From equation, (1) \(\frac{x}{14}+\frac{y}{18}=1\)
or, \(\frac{9 x+7 y}{126}=1[/latex
or, 9x + 7y = 126
or, 9x=126-7y
or, [latex]x=\frac{126-7 y}{9}\) ….(3)
From equation (1) \(\frac{x+y}{2}+\frac{3 x-5 y}{4}=2\)
or, \(\frac{2 x+2 y+3 x-5 y}{4}=2\)
or, 5x – 3y = 8 (4)
From equation (4) \(x=\frac{126-7 y}{9}\)
\(\frac{5(126-7 y)}{9}-3 y=8\)
or, \(\frac{630-35 y-27 y}{9}=8\)
or, – 62y + 630 = 72
or,-62y=72-630
or, -62y=-558
or, 62y=558
or, \(y=\frac{558}{62}\)
or, y=9
Putting the value of y in equation (3),
\(x=\frac{126-7 \times 9}{9}\)Class 9 Math Solution WBBSE In English
\(or, \quad x=\frac{126-63}{9}or, x=\frac{63}{9}
or, \quad x=7\)
∴ x = 7, y=9
8. p (x+y)= q(xy) =2pq
Solution:
Given
P(x+y)=2pq ….(1)
q (x – y) = 2pq…..(2)
From equation (1) p (x+y)=2pq
or, \(x+y=\frac{2 p q}{p}\)
or, x + y = 2q
or, x = 2q-y ….(3)
Putting the value of x in equation(2)
q(2q – y-y) = 2pq
or, \(2 q-2 y=\frac{2 p q}{q}\)
or,-2y=2p-2q
or, \(y=\frac{2(p-q)}{-2}\)
or,y=q-p
Putting the value of y in equation (3),
x = 2q – (q.-p)
or, x=2q-q+p
or, x=p+q
=x=p+q
y=q-p