WBBSE Solutions For Class 9 Maths Chapter 5 Linear Simultaneous Equations Exercise 5.6

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Class IX Maths Solutions WBBSEΒ  Chapter 5 Linear Simultaneous Equations Exercise 5.6

Let us solve the following linear equations in two variables by applying the cross-multiplication method :

Question 1. 8x+5y= 11, 3x – 4y = 10

Solution:

Given

8x+5y = 11 …(1)

3x-4y = 10…..(2)

From equation (1) 8x+5y-11=0
From equation (2) 3x-4y-10=0

or,\(\frac{x}{5 \times(-10)-(-11) \times(-4)}=\frac{y}{(-11) \times 3-8(-10)}=\frac{1}{8(-4)-5 \times 3}\)

or,\(\frac{x}{-50-44}=\frac{y}{-33+80}=\frac{1}{-32-15}\)

or,\(\frac{x}{-94}=\frac{y}{47}=\frac{1}{-47}\)

∴ \(\frac{x}{-94}=\frac{1}{-47} \text { and } \frac{y}{47}=\frac{1}{-47}\)

or,\(x=\frac{-94}{-47} \text { and } y=\frac{47}{-47}\)

or, X = 2Β  and y = -1

∴ y=-1 , x = 2

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Question 2. 3x4y = 1, 4x = 3y+6

Solution:

Given

3x-4y = 1 ….(1)

4x = 3y+ 6 …(2)

From equation (1) 3x-4y-1-0
From equation (2) 4x-3y-6=0

or, \(\frac{x}{(-4) \times(-6)-(-1)(-3)}=\frac{y}{(-1) \times 4-3(-6)}=\frac{1}{3(-3)-(-4) \times 4}\)

or, \(Β \quad \frac{x}{24-3}=\frac{y}{-4+18}=\frac{1}{-9+16}\)

or, \(\quad \frac{x}{21}=\frac{y}{14}=\frac{1}{7}\)

∴ \(\frac{x}{21}=\frac{1}{7} \text { and } \frac{y}{14}=\frac{1}{7}\)

or, \(x=\frac{21}{7} \text { and } y=\frac{14}{7}\)

or, x=3 and y = 2

∴ X = 3
y = 2

Class IX Maths Solutions WBBSE

Question 3. 5x + 3y = 11, 2x-7y = – 12

Solution:

Given

5x + 3y = 11….(1)

2x-7y=-12….(2)

From equation (1) 5x + 3y-11=0
From equation (2) 2x-7y+12=0.

∴ \(\frac{x}{3 \times 12-(-11) \times(-7)}=\frac{y}{-11 \times 2-5 \times 12}=\frac{1}{5 \times(-7)-3 \times 2}\)

or, \(\frac{x}{36-77}=\frac{y}{-22-60}=\frac{1}{-35-6}\)

or, \(\frac{x}{-41}=\frac{y}{-82}=\frac{1}{-41}\)

or, \(\frac{x}{41}=\frac{y}{82}=\frac{1}{41}\)

∴ \(\frac{x}{41}=\frac{1}{41} \text { and } \frac{y}{82}=\frac{1}{41}\)

or, \(x=\frac{41}{41} \text { and } y=\frac{82}{41}\)

∴ x= 1 and y = 2

Question 4. 7x-3y-31=0, 9x-5y-41=0

Solution:

Given

7x-3y31= 0 ….(1)

9x-5y-41=0 …..(2)

or, \(\frac{x}{(-3) \times(-41)-(-31)(-5)}=\frac{y}{(-31) \times 9-7 \times(-41)}=\frac{1}{7 \times(-5)-(-3) \times 9}\)

or, \(\frac{x}{123-155}=\frac{y}{-279+287}=\frac{1}{-35+27}\)

or, \(\frac{x}{-32}=\frac{y}{8}=\frac{1}{-8}\)

∴ \(\frac{x}{-32}=\frac{1}{-8} \text { and } \frac{y}{8}=\frac{1}{-8}\)

or,\(x=\frac{-32}{-8} \text { and } y=\frac{8}{-8}\)

∴ X = 4 and y = -1

Class IX Maths Solutions WBBSE

Question 5. \(\frac{x}{6}-\frac{y}{3}=\frac{x}{12}-\frac{2 y}{3}=4\)

Solution:

Given

\(\frac{x}{6}-\frac{y}{3}=4\) …(1)

\(\frac{x}{12}-\frac{2 y}{3}=4\) …(2)

Β 

From equation (1) \(\frac{x}{6}-\frac{y}{3}=4\)

or, \(\frac{x-2 y}{6}=4\)

or, x-2y = 24…(3)

From equation (2) \(\frac{x}{12}-\frac{2 y}{3}=4\)

or, \(\frac{x-8 y}{12}=4\)

or,X-8y = 48 ….(3)

From equation (1) x-2y-24=0
From equation (4) x-8y-48=0

∴ \(\frac{x}{(-2) \times(-48)-(-24)(-8)}=\frac{y}{(-24) \times 1-1 \times(-48)}=\frac{1}{1 \times(-8)-(-2) \times 1}\)

or,\(\frac{x}{96-192}=\frac{y}{-24+48}=\frac{1}{-8+2}\)

or, \(\frac{x}{-96}=\frac{y}{24}=\frac{1}{-6}\)

∴ \(\frac{x}{-96}=\frac{1}{-6} \text { and } \frac{y}{24}=\frac{1}{-6}\)

or, \(x=\frac{-96}{-6} \text { and } y=\frac{24}{-6}\)

or, x =16, y= -4

∴ x= 6
y = -4

Question 6. \(\frac{x}{5}+\frac{y}{3}=\frac{x}{4}-\frac{y}{3}-\frac{3}{20}=0\)

Solution:

Given

Β 

\(\frac{x}{5}+\frac{y}{3}=0\) …(1)

\(\frac{x}{4}-\frac{y}{3}-\frac{3}{20}=0\)…(2)

Class IX Maths Solutions WBBSE

From equation (1) \(\frac{x}{5}+\frac{y}{3}=0\)

or, \(\frac{3 x+5 y}{15}=0\)

or, 3x+5y =0 ….(3)

From equation (2) \(\frac{x}{4}-\frac{y}{3}-\frac{3}{20}=0\)

or, \(\frac{15 x-20 y-9}{60}=0\)

or, 15x-20y -9 =0 …(4)

From equation (3) 3x+5y+0=0
From equation (4) 15x-20y-9=0

∴ \(\frac{x}{5 \times(-9)-0 \times(-20)}=\frac{y}{0 \times 15-3 \times(-9)}=\frac{1}{3 \times(-20)-5 \times 15}\)

or, \(\frac{x}{-45-0}=\frac{y}{0+27}=\frac{1}{-60-75}\)

or,\(\frac{x}{-45}=\frac{y}{27}=\frac{1}{-13}\)

∴\( \frac{x}{-45}=\frac{1}{-135} \text { and } \frac{y}{27}=\frac{1}{-135}\)

or,\(x=\frac{-45}{-135} \text { and } y=\frac{27}{-135}\)

or,\(x=\frac{1}{3} \text { and } y=-\frac{1}{5}\)

∴ \(x=\frac{1}{3} \text { and } y=-\frac{1}{5}\)

Question 7. \(\frac{x+2}{7}+\frac{y-x}{4}=2 x-8, \quad \frac{2 y-3 x}{3}+2 y=3 x+4\)

Solution:

Given

\(\frac{x+2}{7}+\frac{y-x}{4}=2 x-8\) ….(1)

\(\frac{2 y-3 x}{3}+2 y=3 x+4\) ….(2)

Class IX Maths Solutions WBBSE

From equation(1) \(\frac{x+2}{7}+\frac{y-x}{4}=2 x-8\)

or, \(\frac{4 x+8+7 y-7 x}{28}=2 x-8\)

or, \(\frac{-3 x+7 y+8}{28}=2 x-8\)

or,56-224= -3x + 7y+8
or,56x+3x-7y = 8 +224

or,59x-7y = 232…(3)

From equation (2) \( \frac{2 y-3 x}{3}+2 y=3 x+4 \)

Or, \( \frac{2 y-3 x+6 y}{3}=3 x+4 \)

or, 9x+12=8y-3x

or, 9x + 3x – 8y = – 12
or,12x-8y=-12
or,3x-2y=-3 ….(4)

From equation (3) 59x-7y-232 = 0
From equation (4) 3x-2y+3-0

\( \frac{x}{(-7) \times 3-(-232) \times(-2)}=\frac{y}{(-232) \times 3-59 \times 3}=\frac{1}{59 \times(-2)-(-7) \times 3} \)

Class 9 Mathematics West Bengal Board

or, \( \frac{x}{-21-464}=\frac{y}{-696-177}=\frac{1}{-118+21} \)

or, \( \frac{x}{-485}=\frac{y}{-873}=\frac{1}{-97} \)

\( \frac{x}{-485}=\frac{1}{-97} \text { and } \frac{y}{-873}=\frac{1}{-97} \)

or, \( x=\frac{-485}{-97} \text { and } y=\frac{-873}{-97} \)

or, x = 5 and y = 9
x = 5,y=9

Question 8. x+5y = 36, \( \frac{x+y}{x-y}=\frac{5}{3} \)

Solution:

Given

x+5y = 36 ….(1)

\( \frac{x+y}{x-y}=\frac{5}{3} \) …(2)

Β 

or, 5x-5y=3x+3y
or, 5x-3x-5y-3y=0
or, 2x – 8y = 0
or,X-4y=0 …(4)

From equation (1) x+5y-36=0
From equation (3) x-4y+0=0

\( \frac{x}{5 \times 0-(-36) \times(-4)}=\frac{y}{(-36) \times 1-1 \times 0}=\frac{1}{1 \times(-4)-5 \times 1} \)

Β 

or, \( \frac{x}{0-144}=\frac{y}{-36-0}=\frac{1}{-4-5} \)

or, \( \frac{x}{-144}=\frac{y}{-36}=\frac{1}{-9} \)

or, \( \frac{x}{144}=\frac{y}{36}=\frac{1}{9} \)

\( \frac{x}{144}=\frac{1}{9} \text { and } \frac{y}{36}=\frac{1}{9} \)

Class 9 Mathematics West Bengal Board

or, \( x=\frac{144}{9} \text { and } y=\frac{36}{9} \)

or,Β  x=16 and y = 4

Question 9. 13x12y+15=0, 8x-7y = 0

Solution:

Given

13x-12y+15=0 …(1)

8x-7y+0=0…(2)

\( \frac{x}{-12 \times 0-15 \times(-7)}=\frac{y}{15 \times 8-13 \times 0}=\frac{1}{13 \times(-7)-(-12) \times 8} \)

Β 

Or, \( \frac{x}{0+105}=\frac{y}{120-0}=\frac{1}{-91+96} \)

Or, \( \frac{x}{105}=\frac{y}{120}=\frac{1}{5} \)

\( \frac{x}{105}=\frac{1}{5} \text { and } \frac{y}{120}=\frac{1}{5} \) \( x=\frac{105}{5} \text { and } y=\frac{120}{5} \)

Or, x=21 and y = 24
x = 21,
y = 24

Question 10. x + y = 2b, x – y = 2a

Solution:

Given

x + y = 2b …(1)

x – y = 2a …(2)

From equation (1) x+y-2b=0
From equation (2) xy- 2a =0

\( \frac{x}{1 \times(-2 a)-(2 b)(-1)}=\frac{y}{(-2 b) \times 1-1 \times(-2 a)}=\frac{1}{1 \times(-1)-1 \times 1} \)

Class 9 Mathematics West Bengal Board

Or, \( \frac{x}{-2 a-2 b}=\frac{y}{-2 b+2 a}=\frac{1}{-1-1} \)

Or, \( \frac{x}{-2(a+b)}=\frac{y}{-2(b-a)}=\frac{1}{-2} \)

Or, \( \frac{x}{2(a+b)}=\frac{y}{2(b-a)}=\frac{1}{2} \)

\( \frac{x}{2(a-b)}=\frac{1}{2} \text { and } \frac{y}{2(b-a)}=\frac{1}{2} \)

Β 

Or,\( x=\frac{2(a-b)}{2} \text { and } y=\frac{2(b-a}{2} \)

X= a-b, y= b-a

Question 11. x – y = 2a, ax + by = a2 + b2

Solution:

Given

x – y = 2a …(1)
ax + by = a2 + b2 …(2)

From equation (1) x-y-2a=0
From equation (2) ax + by-(a2+b2)=0

\( \frac{x}{(-1)\left\{-\left(a^2+b^2\right)\right\}-(-2 a) \times b}=\frac{y}{-2 a \times a-1 \times\left\{-\left(a^2+b^2\right)\right\}}=\frac{1}{1 \times b-(-1) \times a} \)

Class 9 Maths WB Board

Or, \( \frac{x}{a^2+b^2+2 a b}=\frac{y}{-2 a^2+a^2+b^2}=\frac{1}{b+a} \)

Or, \( \frac{x}{(a+b)^2}=\frac{y}{b^2-a^2}=\frac{1}{b+a} \)

Or, \( \frac{x}{(a+b)(a+b)}=\frac{y}{(b+a)(b-a)}=\frac{1}{b+a} \)

\( \frac{x}{(a+b)(a+b)}=\frac{1}{b+a} \text { and } \frac{y}{(b+a)(b-a)}=\frac{1}{b+a} \)

Or, \( x=\frac{(a+b)(a+b)}{(a+b)} \text { and } y=\frac{(b+a)(b-a)}{(b+a)} \)

or, x = a + b and y = b-a
x = a + b
y = b-a

Question 12. \( \frac{x}{a}+\frac{y}{b}=2 \), ax-by=a2-b2

Solution:

Given

\( \frac{x}{a}+\frac{y}{b}=2 \) ….(1)
ax-by=a2-b2 …(2)

From equation (1) \( \frac{x}{a}+\frac{y}{b}=2 \)

Or, \( \frac{b x+a y}{a b}=2 \)

Or, bx+ay = 2ab ….(3)

From equation (3) bx+ay-2ab = 0
From equation (4) ax + by – (a2 + b2) = 0

\( \frac{x}{a\left(b^2-a^2\right)-(-2 a b) \times x(-b)}=\frac{y}{(-2 a b) \times a-b\left(b^2-a^2\right)}=\frac{1}{b \times(-b)-a \times a} \)

Β 

Or, \( \frac{x}{a b^2-a^3-2 a b^2}=\frac{y}{-2 a^2 b-b^3+a^2 b}=\frac{1}{-b^2-a^2} \)

Or, \( \frac{x}{-a^3-a b^2}=\frac{y}{-a^2 b-b^3}=\frac{1}{-a^2-b^2} \)

Or, \( \frac{x}{-a\left(a^2+b^2\right)}=\frac{y}{-b\left(a^2+b^2\right)}=\frac{1}{-\left(a^2+b^2\right)} \)

Or,\( \frac{x}{a\left(a^2+b^2\right)}=\frac{y}{b\left(a^2+b^2\right)}=\frac{1}{\left(a^2+b^2\right)} \)

\( \frac{x}{a\left(a^2+b^2\right)}=\frac{1}{a^2+b^2} \text { and } \frac{y}{b\left(a^2+b^2\right)}=\frac{1}{a^2+b^2} \)

Class 9 Maths WB Board

Or, \( x=\frac{a\left(a^2+b^2\right)}{\left(a^2+b^2\right)} \text { and } y=\frac{b\left(a^2+b^2\right)}{\left(a^2+b^2\right)} \)

X=4

Y= b

Question 13. ax + by = 1, \( b x+a y=\frac{2 a b}{a^2+b^2} \)

Solution: ax + by = 1 ….(1)

\( b x+a y=\frac{2 a b}{a^2+b^2} \)…..(2)

Β 

From equation (1) ax + by – 1 = 0
From equation (2) b(a2+ b2)x + a(a2+ b2)y – 2ab = 0

\( \frac{x}{b(-2 a b)-(-1) a\left(a^2+b^2\right)}=\frac{y}{(-1) b\left(a^2+b^2\right)-(a)(-2 a b)}=\frac{1}{a \cdot a\left(a^2+b^2\right)-b \cdot b\left(a^2+b^2\right)} \)

Class 9 Maths WB Board

Or,\( \frac{x}{-2 a b^2+a^3+a b^2}=\frac{y}{-a^2 b-b^3+2 a^2 b}=\frac{1}{a^4+a^2 b^2-a^2 b^2-b^4} \)

Or,\( \frac{x}{a\left(a^2-b^2\right)}=\frac{y}{b\left(a^2-b^2\right)}=\frac{1}{\left(a^2+b^2\right)\left(a^2-b^2\right)} \)

Or,\( \frac{x}{a\left(a^2-b^2\right)}=\frac{y}{b\left(a^2-b^2\right)}=\frac{1}{\left(a^2+b^2\right)\left(a^2-b^2\right)} \)

\( \frac{x}{a\left(a^2-b^2\right)}=\frac{1}{\left(a^2+b^2\right)\left(a^2-b^2\right)} \text { and } \frac{y}{b\left(a^2-b^2\right)}=\frac{1}{\left(a^2+b^2\right)\left(a^2-b^2\right)} \)

Β 

Or, \( x=\frac{a\left(a^2-b^2\right)}{\left(a^2+b^2\right)\left(a^2-b^2\right)} \text { and } y=\frac{b\left(a^2-b^2\right)}{\left(a^2+b^2\right)\left(a^2-b^2\right)} \)

Or,\( x=\frac{a}{a^2+b^2} \text { and } y=\frac{b}{a^2+b^2} \)

\( \begin{aligned}
& therefore x=\frac{a}{a^2+b^2} \\
& y=\frac{b}{a^2+b^2}
\end{aligned} \)

Β 

Β 

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