Class 9 Math Chapter 5 WBBSE Chapter 5 Linear Simultaneous Equations Exercise 5.4
Question 1. Let us express the variable x of the equation \(\frac{x}{3}+\frac{y}{2}=8\) in term of variable y.
Solution: \(\frac{x}{3}+\frac{y}{2}=8\)
\(or, \quad \frac{x}{3}=8-\frac{y}{2}or, \frac{x}{3}=\frac{16-y}{2}
or, x=\frac{3}{2}(16-y)\)
Question 2. \(\frac{2}{x}+\frac{7}{y}=1\)=1 Express the value of y in terms of y.
Solution: \(\frac{2}{x}+\frac{7}{y}=1\)
\(or,\quad \frac{7}{y}=1-\frac{2}{x}or,\quad \frac{7}{y}=\frac{x-2}{x}
or,\quad(x-2) y=7 x
or,\quad y=\frac{7 x}{x-2}\)
Question 3. Let us solve the following equations by comparison method and check whether the solutions satisfy the equations.
1. 2(x − y) = 3, 5x + 8y = 14
Solution: 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)
From equation (2) 5x+8y = 14
or, 5x = 14-8y
or, \(x=\frac{3+2 y}{2}\) ….(4)
Comparing the value of x from equations (3) & (4),
\(\frac{3+2 y}{2}=\frac{14-8 y}{5}\)Class 9 Math Chapter 5 WBBSE
or, 15+10y=28-16y
or, 10y+ 16y=28-15
or, 26y=13
or, \(y=\frac{13}{26}\)
or, \(y=\frac{1}{2}\)
Putting the value of y in equation (3)
\(x=\frac{3+2 \times \frac{1}{2}}{2}\)\(\begin{aligned}
& x=\frac{3+1}{2} \\
& x=\frac{4}{2} \\
& x=2 \\
& x=2
\end{aligned}\)
Class 9 Math Chapter 5 WBBSE
\(y=\frac{1}{2}\)From equation (1) L. H.S. = 2(x-y)
\(\begin{aligned}& =2\left(2-\frac{1}{2}\right) \\
& =2\left(\frac{4-1}{2}\right) \\
& =3
\end{aligned}\)
= R. H. S.
Again, from equation (1), L. H. S.= 5x + 8y
= \(5 \times 2+8 \times \frac{1}{2}\)
= 10 +4
= 14
= R. H. S.
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)
From equation (1) \(\frac{3}{y}=5-2 x\)
or, \(\frac{1}{y}=\frac{5-2 x}{3}\) …..(3)
From equation (2) \(\frac{-2}{y}=3-5 x\)
or, \(\frac{2}{y}=5 x-3\)
or, \(\frac{1}{y}=\frac{5 x-3}{2}\) ….(4)
Comparing the value of \(\frac{1}{y}\) from equations (3)& (4) we get,
\(\frac{5-2 x}{3}=\frac{5 x-3}{2}=\)or, 15x – 9 = 10-4x
or, 15x+4x= 10 +9
or, 19x= 19
or, x = 19/19
or, = 1
Putting the value of x in equation (3)
or, \(\frac{5-2 x}{3}=\frac{5 x-3}{2}\)
or, \(\frac{1}{y}=\frac{3}{3}\)
or, \(\frac{1}{y}=1\)
or, y = 1
X = 1
From equation (1)L. H.S. = \(2 x+\frac{3}{y}\)
= \(2 \times 1+\frac{3}{1}\)
=2+3
= 5
= R. H. S.
From equation (2) L. H. S.= \(5 x-\frac{2}{y}\)
= \(5 \times 1-\frac{2}{1}\)
=5-2
= 3
= R. H. S.
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, \(x=\frac{3(2-y)}{2}\) ….(3)
From equation(2) \(\frac{x}{3}=1-\frac{y}{2}\)
or, \(\frac{x}{3}=\frac{2-y}{2}\)
or, \(x=\frac{3(2-y)}{2}\)….(4)
Comparing the value of x in equations (3) & (4),
\(\frac{2(3-y)}{3}=\frac{3(2-y)}{2}\)Class 9 Math Chapter 5 WBBSE
or, 4(3-y) 9 (2-y)
or, 12-4y=18-9y
or, 9y-4y=18-12
or, 5y = 6
or, \(y=\frac{6}{5}\)
Putting the value of y in equation (3) we get,
\(x=\frac{2\left(3-\frac{6}{5}\right)}{3}\)or, [/latex]x=\frac{2\left(\frac{15-6}{5}\right)}{3}
or, \quad x=2 \times \frac{9}{5} \times \frac{1^{-}}{3}[/latex]
or, \(\begin{aligned}
& x=\frac{6}{5} \\
& =x=\frac{6}{5}
\end{aligned}
y=\frac{6}{5}
\)
Class IX Maths Solutions WBBSE
From equation (1 ) L. H.S. =\(\frac{x}{2}+\frac{y}{3}\)
\(\begin{aligned}& =\frac{\frac{6}{5}}{2}+\frac{\frac{6}{5}}{3} \\
& =\frac{6}{5} \times \frac{1}{2}+\frac{6}{5} \times \frac{1}{3} \\
& =\frac{3}{5}+\frac{2}{5} \\
& =\frac{5}{5} \\
& =1
\end{aligned}\)
= R. H. S.
From equation (2) L. H. S. = \(\frac{x}{3}+\frac{y}{2}\)
\(\begin{aligned}& =\frac{\frac{6}{5}}{3}+\frac{\frac{6}{5}}{2} \\
& =\frac{6}{5} \times \frac{1}{3}+\frac{6}{5} \times \frac{1}{2} \\
& =\frac{2}{5}+\frac{3}{5} \\
& =\frac{2+3}{5} \\
& =\frac{5}{5}
\end{aligned}\)
Class IX Maths Solutions WBBSE
= 1
= R. H. S.
4. 4x-3y = 18, 4y – 5x = -7
Solution:
4x-3y = 18 …(1)
4y – 5x = -7 ….(2)
From equation (1) 4x = 18+ 3y
or, \(x=\frac{18+3 y}{4}\) ….(3)
From equation (2) -5x=-4y-7.
or, 5x = 4y+ 7
or, \(x=\frac{4 y+7}{5}\) ….(4)
Comparing the value of ‘x’ from equations (3) & (4),
or, \(\frac{18+3 y}{4}=\frac{4 y+7}{5}\)
or,16y+28 = 90+ 15y
or,16y 15y = 90-28
or, y = 62
Putting the value of y, we get
\(x=\frac{18+3 \times 62}{4}\)\(or, \quad x=\frac{18+186}{4}
or, x=\frac{204}{4}
or, \quad x=51\)
x =51,y= 62
From equation (1) L. H.S. = 4x-3y
= 4 x 51-3 x 62
= 204
= 18
=R. H. S.
From equation (2) L. H. S.= 4y-5x
= 4 x 62-5 x 51
=248-255
=-7
= R. H. S.
Class IX Maths Solutions WBBSE Question 4. Let us solve the equations 2x + y = 8 and 2y-3x=-5 by comparison method and justify them by solving graphically.
Solution:
2x + y = 8 ….(1)
2y-3x=-5…(2)
Equation y = 8-2x …(3)
Equation 2y=3x-5
or, \(y=\frac{3 x-5}{2}\) …(4)
From equations (3) & (4),
\(8-2 x=\frac{3 x-5}{2}\)or, 3x-5=16-4x
or, 3x+4x=16+5
or, 7x=21
or, \(x=\frac{21}{7}\)
or, x = 3
Putting this value of x in equation (3),
y=8-2×3
or,y = 2
x = 3
⇒ y = 8-2x
| x | 0 | 2 | -1 |
| y | 8 | 4 | 10 |
Class IX Maths Solutions WBBSE
\(y=\frac{3 x-5}{2}\)
| x | 1 | 3 | -7 |
| y | -1 | 2 | -13 |
Question 5. Let us solve the following equations in two variables by comparison method:
1. 3x-2y= 2, 7x + 3y = 43
Solution:
3x-2y=2 …(1)
7x + 3y = 43 …(2)
From equation (1) 3x=2+2y
or, \(x=\frac{2+2 y}{3}\) …(3)
From equation (2) 7x=43-3y
or, \(x=\frac{43-3 y}{7}\) ….(4)
Comparing the value of x from equations (3) & (4)
\(\frac{2+2 y}{3}=\frac{43-3}{7}\)
or,14+ 14y = 129-9y
or, 14y+9y=129-14
or, 23y=115
or, \(y=\frac{115}{23}\)
or,y = 5
Putting the value of y in equation (3),
\(x=\frac{2+2 \times 5}{3}\)\(x=\frac{12}{3}\)
X = 4
∴x=4&y=5
2. 2x-3y=8, \(\frac{x+y}{x-y}=\frac{7}{3}\)
Solution: 2x-3y=8 …(1)
\(\frac{x+y}{x-y}=\frac{7}{3}\)….(2)
Class IX Maths Solutions WBBSE
From equation (1) 2x=8+3y
or, \(x=\frac{8+3 y}{2}\) ….(3)
From equation (2) 7x-7y = 3x + 3y
or, 7x-3x=3y+ 7y.
or, 4x = 10y
or, \(x=\frac{10 y}{4}\)
or, \(x=\frac{5 y}{2}\) …(4)
Companing the value of x from equations (3) & (4),
\(\frac{8+3 y}{2}=\frac{5 y}{2}\)
or, 8+3y = 5y
or, 8+ 3y = 5y
or, 3y-5y=-8
or,-2y=-8
or, \(y=\frac{-8}{-2}\)
y = 4
Putting the value of y in equation (4),
\(x=\frac{5 \times 4}{2}\)or, X = 10
∴ X = 10
y = 4
3. \(\frac{1}{3}(x-y)=\frac{1}{4}(y-1), \frac{1}{7}(4 x-5 y)=x-7\)
Solution:
\(\frac{1}{3}(x-y)=\frac{1}{4}(y-1)\) ….(1)
\(\frac{1}{7}(4 x-5 y)=x-7\) …(2)
Class 9 Mathematics West Bengal Board
From equation (1) \(\frac{x-y}{3}=\frac{y-1}{4}\)
or,4x-4y=3y-3
or,4x 3y+4y-3
or, \(x=\frac{7 y-3}{4}\) ….(3)
From equation (2) 7(x-7)=4x-5y
or, 7x-49=4x-5y
or, 7x-4x=49-5y
or, 3x=49-5y
or, \(x=\frac{49-5 y}{3}\) ……(4)
Comparing the value of x from equations (3) & (4),
\(\frac{7 y-3}{4}=\frac{49-5 y}{3}\)
or, 21y-9-196-20y
or, 21y+20y 196 +9
or, 41y=205
or, \(y=\frac{205}{4}\)
or, y = 5
Putting the value of y in equation (3),
\(x=\frac{7 \times 5-3}{4}\)Class 9 Mathematics West Bengal Board
\(or, x=\frac{35-3}{4}or, \quad x=\frac{32}{4}
or, x=8 \)
Required solution x = 8,y = 5
4. \(\frac{x+1}{y+1}=\frac{4}{5}, \frac{x-5}{y-5}=\frac{1}{2}\)
Solution:
\(\frac{x+1}{y+1}=\frac{4}{5}\) …….(1)
\(\frac{x-5}{y-5}=\frac{1}{2}\) ….(2)
From equation (1) 5x+5=4y+4
or, 5x = 4y+ 4-5
or, 5x= 4y-1
or, \(x=\frac{4 y-1}{5}\) ….(3)
From equation (2) 2x-10-y-5
or, 2x=y-5+10
or, 2x = y +5
or, \(x=\frac{y+5}{2}\) …(4)
Comparing the value of x from equations (2) & (4),
\(\frac{4 y-1}{5}=\frac{y+5}{2}\)or, 8y-2=5y+25
or, 8y – 5y = 25+ 2
or, 3y=27
or, \(y=\frac{27}{3}\)
or,y=9
Putting the value of y in equation (4),
\(\begin{aligned}&x=\frac{9+5}{2}\\
&x=\frac{14}{2}
\end{aligned}\)
Class 9 Mathematics West Bengal Board
x =7,y =9
5. x + y = 11, y+2= \(\frac{1}{8}\) (10y+x)
Solution: x + y = 11……(1)
y+2= \(\frac{1}{8}\) (10y+x) …..(2)
From equation (1) x = 11-y ….(3)
From equation (2) \(\frac{(y+2)}{1}=\frac{(10 y+x)}{8}\)
or, 10y + x = 8y + 16
or, x = 8y + 16-10y
or, x = -2y+16 ….(4)
Comparing the value of x from equations (3) & (4),
11 – y = -2y+ 16
or, -y+2y= 16-11
or,y = 5
Putting the value of y in equation (3),
X=11-5
or,X = 6
∴ X = 6
y = 5
6. \(\frac{x}{3}+\frac{y}{4}=1\), 2x + 4y = 11
Solution: \(\frac{x}{3}+\frac{y}{4}=1\)…..(1)
2x + 4y = 11 …(2)
From equation (1) \(\frac{x}{3}+\frac{y}{4}=1\),
or, \(\frac{4 x+3 y}{12}=1\)
or, 4x + 3y = 12
or, 4x=12-3y
or, \(x=\frac{12-3 y}{4}\)…(3)
From equation (2) 2x = 11-4y
or, \(x=\frac{11-4 y}{2}\) ….(4)
Comparing the value of x from equations (3) & (4),
or, \(\frac{12-3 y}{4}=\frac{11-4 y}{2}\)
or, 24 – 6y = 44 – 16y
or, 16y – 6y = 44-24
or,10y=20
or, \(y=\frac{20}{10}\)
∴ y = 2
\(x=\frac{12-3 \times 2}{4}\)Class 9 Mathematics West Bengal Board
or, \(x=\frac{6}{4}\)
∴ \(x=\frac{3}{2}\), y = 2
7. \(x+\frac{2}{y}=7,2 x-\frac{6}{y}=9\)
Solution:
\(x+\frac{2}{y}=7\) …..(1)
\(2 x-\frac{6}{y}=9\) …(2)
From equation (1) \(x=7-\frac{2}{y}\)
or, \(x=\frac{7 y-2}{y}\)
From equation (2) \(2 x=9+\frac{6}{y}\)
or, \(2 x=\frac{9 y+6}{y}\)
or, \(x=\frac{9 y+6}{2 y}\) …..(4)
Comparing the value of x from equations (3) & (4),
\(\begin{aligned}& \frac{7 y-2}{y}=\frac{9 y+6}{2 y} \\
& \frac{7 y-2}{1}=\frac{9 y+6}{2}
\end{aligned}\)
or, 14y- 4= 9y+6
or,14y-9y=6+4
or, 5y = 10
or, \(y=\frac{10}{5}\),
∴ y = 2
Putting the value of y in equation (3),
\(\begin{aligned}& x=\frac{7 \times 2-2}{2} \\
& x=\frac{14-2}{2} \\
& x=\frac{12}{2}
\end{aligned}\)
Class 9 Maths WB Board
8. \(\frac{1}{x}+\frac{1}{y}=\frac{5}{6}, \frac{1}{x}-\frac{1}{y}=\frac{1}{6}\)
Solution: \(\frac{1}{x}+\frac{1}{y}=\frac{5}{6}\)…(1)
\(\frac{1}{x}-\frac{1}{y}=\frac{1}{6}\) ….(2)
From equation (1) \(\frac{1}{x}=\frac{5}{6}-\frac{1}{y}\) …(3)
From equation (2) \(\frac{1}{x}=\frac{1}{6}+\frac{1}{y}\) …(4)
Comparing the value of x from equations (3) & (4),
\(\begin{aligned}& \frac{5}{6}-\frac{1}{y}=\frac{1}{6}+\frac{1}{y} \\
& \frac{5}{6}-\frac{1}{6}=\frac{1}{y}+\frac{1}{y} \\
& \frac{5-1}{6}=\frac{1+1}{y} \\
& \frac{4}{6}=\frac{2}{y} \\
& 4 y=12 \\
& y=\frac{12}{4} \\
& y=3
\end{aligned}\)
Putting the value of y in equation (3).
\(or, \frac{1}{x}=\frac{5-2}{6}or, \frac{1}{x}=\frac{3}{6}
or, \quad 3 x=6
or, \quad x=\frac{6}{3}
or, \quad x=2
therefore x=2 \)
Class 9 Maths WB Board
9. \(\frac{x+y}{x y}=2, \frac{x-y}{x y}=1\)
Solution: \(\frac{x+y}{x y}=2\) …(1)
\(\frac{x-y}{x y} = 1\) …(2)
From equation (1) \(\frac{x}{x y}+\frac{y}{x y}=2\)
or,\(\frac{1}{y}+\frac{1}{x}=2\)
or, \(\frac{1}{y}=2-\frac{1}{x}\) ….(3)
From equation (2)=\(\frac{x}{x y}-\frac{y}{x y}=1\)
or, \(\frac{1}{y}-\frac{1}{x}=1\)
or, \(\frac{1}{y}=1+\frac{1}{x}\)
Comparing the value of x from equations (3) & (4),
\(or, \quad 2-\frac{1}{x}=1+\frac{1}{x}or, \quad 2-1=\frac{1}{x}+\frac{1}{x}
or, \quad 1=\frac{1+1}{x}
or, 1=\frac{2}{x}
or, \quad x=2\)
Putting the value of y in equation (3),
\(or, \frac{1}{y}=\frac{4-1}{2}or, \quad \frac{1}{y}=\frac{3}{2}
or, 3 y=2
or, \quad y=\frac{2}{3}\)
Class 9 Maths WB Board
∴ x =2, \(y=\frac{2}{3}\)
10. \(\frac{x+y}{5}+\frac{x-y}{4}=5, \frac{x+y}{4}+\frac{x-y}{5}=5 \frac{4}{5}\)
Solution: \(\frac{x+y}{5}+\frac{x-y}{4}=5\) …(1)
\(\frac{x+y}{4}+\frac{x-y}{5}=5 \frac{4}{5}\) …(2)
From equation (1)=\(\frac{x+y}{5}+\frac{x-y}{4}=5\)
\(or, \quad \frac{4 x+4 y+5 x-5 y}{20}=5or, \quad 9 x-v=100
or, \quad 9 x=100+y\)
or, \(x=\frac{100+y}{9}\) …(3)
From equation (2) \(\frac{x+y}{4}+\frac{x-y}{5}=\frac{29}{5}\)
\(or, \quad \frac{5 x+5 y+4 x-4 y}{20}=\frac{29}{5}or, \quad 9 x+y=\frac{29}{5} \times 20
or, \quad 9 x+y=116
or, \quad 9 x=116-y\)
\(x=\frac{16-y}{9}\)….(4)
Class 9 Maths WB Board
Comparing the value of x from equations (3) & (4),
\(or, \quad \frac{100+y}{9}=\frac{116-y}{9}or, \quad 100+y=116-y
or, y+y=116-100
or, 2 y=16
or, y=\frac{16}{2}
or, y=8\)
Putting the value of y in equation (3),
\(\begin{aligned}& x=\frac{100+8}{9} \\
& x=\frac{108}{9}
\end{aligned}\)
∴ x = 12
y = 8
11. \(\frac{4}{x}-\frac{y}{2}=-1, \frac{8}{x}+2 y=10\)
Solution: \(\frac{4}{x}-\frac{y}{2}=-1\)….(1)
\(\frac{8}{x}+2y=10\) …(2)
From equation (1)\(-\frac{y}{2}=-\frac{4}{x}-1\)
or, \(\frac{y}{2}=\frac{4}{x}+1\)
\(y=2\left(\frac{4}{x}+1\right)\)…(3)
From equation (2) 2y = \(=10-\frac{8}{x}\)
or, \(y=\frac{10}{2}-\frac{8}{2 x}\)
or, \(y=5-\frac{4}{x}\) …(4)
Comparing the value of x from equations (3) & (4),
\(2\left(\frac{4}{x}+1\right)=5-\frac{4}{x}\)\(or, \frac{8}{x}+2=5-\frac{4}{x}
or, \frac{8}{x}+\frac{4}{x}=5-2\)
\(or, \quad \frac{8+4}{x}=3
or, \quad \frac{12}{x}=3\)
or, 3x = 12
\(x=\frac{12}{3}\)
or, x = 4
Putting the value of y in equation (4),
\(y=5-\frac{4}{4}\)or, y=5-1
y = 4
∴ x= 4 , y = 4
12. 2-2(3x-y)=10(4-y)- 5x = 4(y-x)
Solution: 2- 2(3x-y) = 4(y-x) …..(1)
10(4-y)-5x = 4(y-x) …(2)
From equation (1) 2-6x+2y= 4y – 4x
or, – 6x+4x=4y-2y-2
or, -2x =-2y-2
or, 2x=2-2y
or, \(x=\frac{2-2 y}{2}\) …(3)
From equation (2) 40-10y-5x = 4y – 4x
or, – 5x + 4x = 4y+ 10y – 40
or, -x=14y-40
or, X = 40-14y
Comparing the value of x from equations (3) & (4),
\(\frac{2-2 y}{2}=40-14 y\)
or, \(\frac{2}{2}-\frac{2 y}{2}=40-14y\)
or, 1-y=40-14y
or, -y+14y=40-1
or, 13y=39
or, y = \(\frac{39}{13}\)
y = 3
Putting the value of y in equation (4)
x=40-14 x 3
or, X = 40-42
∴ x=-2,y =3
