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Class 10 RD Sharma Solutions – Chapter 3 Pair of Linear Equations in Two Variables – Exercise 3.4 | Set 1

  • Last Updated : 30 Apr, 2021

Solve each of the following systems of equations by the method of cross multiplication.

Question 1. x + 2y + 1 = 0 and 2x – 3y – 12 = 0 

Solution:

Given that,

x + 2y + 1 = 0

2x – 3y – 12 = 0

On comparing both the equation with the geneal form we get



 a1 = 1, b1 = 2, c1 = 1, a2 = 2, b2 = −3, c2 = -12

Now by using cross multiplication we get

x/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)

x/(-24 – (-3)) = y/(-2 – (-12)) = 1/(-3 – 4)

x/(-24 + 32) = y/(2 + 12) = 1/ (-3 – 4)

x = -21/-7

So, x = 3

and 



y = 14/-7

y = -2

Hence, x = 3 and y = -2

Question 2. 3x + 2y + 25 = 0 and 2x + y + 10 = 0

Solution:

Given that,

3x +2y + 250

2x + y + 10 = 0

On comparing both the equation with the geneal form we get

 a1 = 3, b1 = 2, c1 = 25, a2 = 2, b2 = 1, c2 = 10

Now by using cross multiplication we get

x/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)

x/(20 – 25) = y/(50 – 30) = 1/(3 – 4)

x/-5 = y/20 = 1/-1

x = -5 × (-1) 

x = 5

x = -5 × (-1) = 5

y/20 = -1

y = 20 × (-1) = −20

Hence, x = 5y and y = −2

Question 3. 2x + y = 35 and 3x + 4y = 65

Solution:



Given that,

2x + y = 35

3x + 4y = 65

Or 2x + y – 35 = 0

3x + 4y – 65 = 0

On comparing both the equation with the geneal form we get

 a1 = 2, b1 = 1, c1 = -35, a2 = 3, b2 = 4, c2 = -65

Now by using cross multiplication we get

x/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)

x/(-65 – (-140)) = y/(-105-(-65 × 2)) = 1/( 8 – 3)

x/(-65 + 140) = y/( -105 + 130) = 1/5

x = 75/5

x = 15

y = 25/5

y = 5

Hence, x = 15 and y = 5

Question 4. 2x – y = 6 and x – y = 2

Solution:

Given that,

2x – y = 6

x – y = 2

Or 2x – y – 6 = 0

x – y – 2 = 0

On comparing both the equation with the geneal form we get

 a1 = 2, b1 = -1, c1 = -6, a2 = 1, b2 = -1, c2 = -2

Now by using cross multiplication we get

x/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)

x/(2 – 6) = y/(-6 + 4) = 1/(-2 + 1)

So, x = 4

and y = 2

Hence, x = 4 and y = 2



Question 5. (x + y)/xy = 2 and (x − y)/xy = 6

Solution:

Given that,

(x + y)/xy = 2,

(x − y)/xy = 6

Or, 1/x + 1/y – 2 = 0

and,

1/x – 1/y – 6 = 0

On comparing both the equation with the geneal form we get

 a1 = 1, b1 = 1, c1 = -2, a2 = 1, b2 = -1, c2 = -6

Now by using cross multiplication we get

x/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)

(1/x)/(-6 + 2) = (1/y)/(2 + 6) = 1/(1 + 1)

So, x = -1/2

and y = 1/4

Hence, x = -1/2 and y = 1/4

Question 6. ax + by = a – b and bx – ay = a + b

Solution:

Given that,

ax + by = a – b

bx – ay = a + b

Or, ax + by – (a – b) = 0

bx – ay – (a + b) = 0

On comparing both the equation with the geneal form we get

 a1 = a, b1 = b, c1 = -(a – b), a2 = b, b2 = -a, c2 = (a + b)

Now by using cross multiplication we get

x/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)

x/(−b(a + b) − (-a)(-(a − b))) = y/((-(a − b) × b) – (-(a + b) × a) = 1/ (-a2 – b2)

x/(-ab – b2 – a2 + ab) = y/(-ab – b2 + a2 + ab) = 1/-(-a2 + b2)

So, 

x/-(-a2 + b2) = 1/-(-a2 + b2

x = 1



and, y/(a2 + b2) = 1/-(-a2 + b2)

y = -1

Hence, x = 1 and y = -1

Question 7. x + ay = b and ax – by = c

Solution:

Given that,

x + ay = b

ax – by = c

Or, x + ay – b = 0

ax – by – c = 0

On comparing both the equation with the geneal form we get

a1 = 1, b1 = a, c1 = – b, a2 = a, b2 = -b, c2 = -c

Now by using cross multiplication we get

x/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)

x/(-ac – b2) = y/(-ab + c) = 1/(-b – a2)

x/(- ac – b2) = 1/(-b – a2

x = (ac + b2)/(a2 + b)

and,

y/(- ab + c) = 1/(-b – a2)

y =(ab – c)/(a2 + b)

Hence, x = (ac + b2)/(a2 +b) and y =(ab – c)/(a2 + b)

Question 8. ax + by = a2 and bx + ay = b2

Solution:

Given that,

ax + by = a2

bx + ay = b2

Or, ax + by – a2 = 0

bx + ay – b2 = 0

On comparing both the equation with the geneal form we get

a1 = a, b1 = b, c1 = – a2, a2 = b, b2 = a, c2 = -b2

Now by using cross multiplication we get

x/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)

x/(-b3 + a3) = y/(-a2b + ab2) = 1/(a2 – b2)

x/(a3 – b3) = y/(-ab(a – b)) = 1/(a2 – b2)

x= (a3 – b3)/(a2 – b2)

x = ((a – b)(a2 + ab + b2))/(a + b)(a – b)

x = (a2 + ab + b2)/a + b

and,

y/(-ab(a – b)) = 1/(a2 – b2)

(a – b)(a2 + ab + b2) (a + ba – b)

 y = -ab(a – b)/(a + b)(a – b)

y= -ab/(a + b)



Hence, x = (a2 + ab + b2)/(a + b) and y = -ab/(a + b)

Question 9. (5/(x + y)) – (2/(x – y)) = -1 and (15/(x + y)) + (7/(x – y)) = 10

Solution:

Given that,

(5/(x + y)) – (2/(x – y)) = -1

(15/(x + y)) + (7/(x – y)) = 10

Let, us assume

x + y = a, and x – y = b, then

(5/a) – (2/b) + 1 = 0

(15/a) – (7/b) + 10 = 0

On comparing both the equation with the geneal form we get

a1 = 5, b1 = -2, c1 = – 1, a2 = 15, b2 = 7, c2 = -10

Now by using cross multiplication we get

x/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)

((1/a)/(20 – 7)) = ((1/b)(15 + 50)) = 1/(35 + 30)

((1/a)/13) = ((1/b)/65) = 1/65

So,

1/a = 13/65

a = 5

and,

1/b = 65/65

b =1

= x + y = 5……       (1)

= x – y = 1…..      (2)

Now, by adding eq(1) and (2) then we get,

x = 3 and y = 2

Hence, x = 3 and y = 2

Question 10. (2/x) + (3/y) = 13 and (5/x) – (4/y) = -2

Solution:

Given that,

(2/x) +(3/y) = 13

(5/x) – (4/y) = -2

Or, (2/x) + (3/y) – 13 = 0

(5/x) – (4/y) + 2 = 0

On comparing both the equation with the geneal form we get

a1 = 2, b1 = 3, c1 = – 13, a2 = 5, b2 = -4, c2 = 2

Now by using cross multiplication we get

x/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)

 = ((1/x)/(6 – 52)) = ((1/y)(-65 – 4)) = 1/(-8 – 5)

= ((1/x)/-46) = ((1/y)/(-69)) = 1/(-23)

Now,

1/x = -46/(-23)

x = 1/2

and,

1/y = -69/-23

y = 1/3

Hence, x = 1/2 and y = 1/3

Question 11. (11x + y)) + (6/(x – y)) = 5 and (38/(x + y)) + (21(x – y)) = 9

Solution:

Given that,

(57/(x + y)) + (6/(x – y)) = 5

(38/(x + y)) + (21(x – y)) = 9

Let us considered



x + y = p

x – y = q

So,

(57/p) + (6/q) – 5 = 0

(38/p) + (21/q) – 9 = 0

On comparing both the equation with the geneal form we get

a1 = 57, b1 = 6, c1 = – 5, a2 = 38, b2 = 21, c2 = -9

Now by using cross multiplication we get

x/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)

((1/p)/(-54 + 105)) = ((1/q)/(-190 + 513)) = 1/(1197 – 228)

((1/p)/51) = ((1/q)/323) = 1/969

So,

((1/p)/51) = 1/969

1/p = 51/969

p = 19

and,

((1/q)/323) = 1/969

1/q = 323/969

q = 3

Now, adding eq x + y = 19 and x – y = 3 then we get:

x = 11 and y = 8

Hence, x = 11 and y = 8

Question 12. (x/a) + (y/b) = 2 and ax – by =  a2 – b2

Solution:

Given that,

(x/a) + (y/b) = 2

ax – by = a2 – b2

(x/a) + (y/b) – 2 = 0

ax – by – (a2 – b2) = 0

On comparing both the equation with the geneal form we get

a1 = 1/a, b1 = 1/b, c1 = – 2, a2 = a, b2 = -b, c2 = -(a2 – b2)

Now by using cross multiplication we get

x/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)

x/((-1/b)(a2 + b2)-2b) = y/((-2a) + (1/a)(a2 – b2)) = 1/((-b/a) – (a/b))

x/((-a2/b) – b) = y/(-a – (b2/a)) = 1/-(a2 + b2)/ab

x/(-(a2 + b2)/b) = y/(-(a2 + b2)/a)) = 1/-(a2 + b2)/ab

So,

 x/(-(a2 + b2)/b) = 1/-( a2 + b2)/ab

x = -(a2 + b2)ab/-b(a2 + b2

x = a

and,

 y/(-(a2 + b2)/a)) = 1/-(a2 + b2)/ab

y = -(a2 + b2)ab/-a(a2 + b2

y = b

Hence, x = a and y = b

Question 13. (x/a) + (y/b) = a + b and (x/a2) + (y/b2) = 2

Solution:

Given that,

(x/a) + (y/b) = a + b

(x/a2) + (y/b2) = 2

(x/a) + (y/b) – a + b = 0

(x/a2) + (y/b2) – 2 = 0



On comparing both the equation with the geneal form we get

a1 = 1/a, b1 = 1/b, c1 = -(a + b), a2 = 1/ a2, b2 = 1/ b2, c2 = -2

Now by using cross multiplication we get

x/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)

(x/ (-2/b) + (a/b2) + (1/b)) = (-y/(-2/a) + (1/a) + (b/ a2)) = (1/(-1/ ab2) – (-1/ a2b))

Now,

x/((a – b)b2) = 1/(-1/ab2) – (-1/a2b)

x = a2

and,

-y/((-a – b)/a2) + (1/a) + (b/b2) = 1/(-1/ab2) – (-1/a2b)

y = b2

Hence, x = a2 and y = b2

Question 14. x/a = y/b and ax + by = a2 + b2

Solution:

Given that,

x/a = y/b

ax + by = a2 + b2

On comparing both the equation with the geneal form we get

a1 = 1/a, b1 = 1/b, c1 = 0, a2 = a, b2 = b, c2 = -(a2 + b2)

Now by using cross multiplication we get

x/(b1c2 – b2c1) = y/(c1a2 – c2a1) = 1/(a1b2 – a2b1)

x/((a2 + b2)/b) = y/((a2 + b2)/a) = 1/((a/b) + (b/a))

Now,

x/((a2 + b2)/b) = 1/((a/b) + (b/a))

x = a

and,

 y/((a2 + b2)/a) = 1/((a/b) + (b/a))

y = b

Hence, x = a and y = b

Attention reader! Don’t stop learning now. Participate in the Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students.




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