Class 10 RD Sharma Solutions - Chapter 3 Pair of Linear Equations in Two Variables - Exercise 3.4 | Set 2

Chapter 3 of RD Sharma’s Class 10 Mathematics book delves into the topic of Pair of Linear Equations in Two Variables. This chapter is fundamental in understanding how to solve systems of linear equations and their applications. Exercise 3.4 | Set 2 provides the practice problems to strengthen students' ability to solve these equations using the various methods including the graphical, substitution, and elimination methods.

What is a Pair of Linear Equations in Two Variables?

A pair of linear equations in the two variables consists of the two equations each of which is linear in two variables. The general form of these equations is:

𝑎₁𝑥+𝑏₁𝑦=𝑐₁

a₂x+b₂y=c₂

where 𝑎₁ , 𝑏₁ , 𝑐₁, 𝑎₂, 𝑏₂ and 𝑐₂ are constants. Solving this pair of equations involves finding the values of the x and y that satisfy both equations simultaneously. This can be done using various methods such as:

• Graphical Method: Plotting both the lines on a graph to find their intersection.
• Substitution Method: Solving one equation for one variable and substituting it into the other equation.
• Elimination Method: Adding or subtracting the equations to eliminate one of the variables.

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

Question 15. 2ax + 3by = a + 2b and 3ax + 2by = 2a + b

Solution:

Given that,

2ax + 3by = a + 2b

3ax + 2by = 2a + b

On comparing both the equation with the general form we get

 a₁ = 2a, b₁ = 3b, c₁ = -(a + 2b), a₂ = 3a, b₂ = 2b, c₂ = -(2a + b)

Now by using cross multiplication we get

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

x/(-3b(2a + b) + 2b(a + 2b)) = y/(-3a(a + 2b) + 2a(2a + b)) = 1/(4ab - 9ab)

x/(b² - 4ab) = y/(a² - 4ab) = 1/-5ab

x/(-4ab + b²) = 1/-5ab

x/b(b - 4a) = 1/-5ab

x = (4a - b)/5a

and,

= -y/(-a² + 4ab) = 1/-5ab

= -y/a(-a + 4b) = 1/-5ab

 y = (4b - a)/5b

Hence, x = (4a - b)/5a and y = (4b - a)/5b

Question 16. 5ax + 6by = 28 and 3ax + 4by = 18

Solution:-

Given that,

5ax + 6by = 28

3ax + 4by = 18

Or, 5ax + 6by - 28 = 0

3ax + 4by - 18 = 0

Here, a₁ = 5a, b₁ = 6b, c₁=-28

a₂= 3a, b₂ = 4b, c₂ =-18

On comparing both the equation with the general form we get

 a₁ = 5a, b₁ = 6b, c₁ = -28, a₂ = 3a, b₂ = 4b, c₂ = -18

Now by using cross multiplication we get

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

x/4b = -y/-6a = 1/2ab

x/4b = 1/2b

x = 2/a

and,

-y/-6a = 1/2ab

y = 3/b

Hence, x = 2/a and y = 3/b

Question 17. (a + 2b)x + (2a - b)y = 2 and (a – 2b)x + (2a + b)y = 3

Solution:

Given that,

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

(a – 2b)x + (2a + b)y = 3

Or 

(a + 2b) x + (2a – b)y - 2 = 0

(a – 2b) x + (2a + b)y - 3 = 0

On comparing both the equation with the general form we get

 a₁ = a + 2b, b₁ = 2a - b, c₁ = -2, a₂ = a - 2b, b₂ = 2a + b, c₂ = -3

Now by using cross multiplication we get

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

x/(-3(2a - b)) + (2(2a + b)) = y/(-2(a - 2b)) + (3(a + 2b)) = 1/((a + 2b)(2a + b) - (a - 2b)(2a - b))

x/(5b - 2a) = y/(a + 10b) = 1/(2a² + 5ab + 2b² - 2a² + 5ab - 2b²)

x/(5b - 2a) = y/(a + 10b) = 1/10ab

So,

x/(5b - 2a) = 1/10ab

x= (5b - 2a)/10ab

and,

y/(a + 10b) = 1/10ab

y = (a + 10b)/10ab

Hence, x = (5b - 2a)/10ab and y= (a + 10b)/10ab

Question 18. x((a - b) + (ab/(a - b))) = y((a + b) - (ab/(a + b))) and x + y = 2a²

Solution:

Given that,

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

Or on solving we get

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

= x((a² + b² - 2ab + ab)/(a - b)) - y((a² + b² + 2ab - ab)/(a + b)) = 0

and,

x + y = 2a²

On comparing both the equation with the general form we get

 a₁ = (a² + b² - 2ab + ab)/(a - b), b₁ = (a² + b² + 2ab - ab)/(a + b), c₁ = 0, 

a₂ = 1, b₂ = 1, c₂ = 2a²

Now by using cross multiplication we get

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

⇒ x/(2a²((a² + b² - 2ab + ab)/(a - b)) - 0) = y/(0 + 2a²((a² + b² + 2ab - ab)/(a + b))) 

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

⇒ x/(2a²((a² + b² - 2ab + ab)/(a - b))) = y/(-2a²)((a² + b² + 2ab - ab)/(a + b)) = 1/(2a³/(a² - b²))

Now,

x/(2a²((a² + b² - 2ab + ab)/(a - b)) = 1/(2a³/(a² - b²))

x = (2a²(a² + ab + b²)(a² - b²)) / 2a³(a + b)

x = (a³ - b³)/a

and,

 y/(-2a²)((a² + b² + 2ab - ab)/(a + b)) = 1/(2a³/(a² - b²))

y = (2a²(a² - ab + b²)(a² - b²))/2a³(a - b)

y = a³ + b³/a

Hence, x = (a³ - b³)/a and y = a³+ b³/a

Question 19. bx + cy = a + b and ax[(1/(a - b)) - (1/(a + b))] + cy[(1/(b - a)) - (1/(b + a))] = 2a/(a + b)

Solution:

Given that,

bx + cy = a + b

ax[(1/(a - b)) - (1/(a + b))] + cy[(1/(b - a)) - (1/(b + a))] = 2a/(a + b)

Or

bx + cy -(a + b) = 0

ax((1/(a - b)) - (1/(a + b))) + cy(((1/(b - a)) - (1/(b + a))) - 2a/(a + b) = 0

ax(2b/(a² - b²))) + cy(2a/(b² - a²)) - 2a/(a + b) = 0

On comparing both the equation with the general form we get

 a₁ = b, b₁ = c, c₁ = -(a + b),

a₂ = 2b/(a² - b²), b₂ = 2a/(b² - a²), c₂ = 2a/(a + b) 

Now by using cross multiplication we get

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

⇒ x / ((-2ac/(a + b)) + ((2ac(a + b)/(b² - a²)))) = y/((-(a + b)2ab)/(a² - b²)) + (2ab/(a + b)) 

= 1/((2abc/(b²- a²)) - (2abc/(a² - b²)))

⇒ x / ((-2ac/(a + b)) + ((2ac(a + b)/(b² - a²)))) = y/((-(a + b)2ab)/(a² - b²)) + (2ab/(a + b)) 

= 1/(-4abc/(a² - b²))

⇒ x/(-2ac((1/(a + b)) + (1/(a - b))) = y/(2ab((-1/(a - b)) + (1/(a + b))) = 1/(-4abc/(a² - b²))

⇒ x/(-4a²c/(a² - b²)) = y/(4ab²/(a² - b²)) = 1/1/(-4abc/(a² - b²))

So,

x/(-4a²c/(a² - b²)) = 1/(-4abc/(a² - b²))

x = a/b

and,

y/(4ab²/(a² - b²)) = 1/(-4abc/(a² - b²))

y = b/c

Hence, x = a/b, y = b/c

Question 20. (a – b) x + (a + b) y = 2a² – 2b² and (a + b) (x + y) = 4ab

Solution:

Given that,

(a – b) x + (a + b) y = 2a² – 2b²

(a + b) (x + y) = 4ab

(a – b) x + (a + b) y - 2(a² – b²) = 0

(a + b)x +  (a + b)y - 4ab = 0

On comparing both the equation with the general form we get

 a₁ = a - b, b₁ = a + b, c₁ = -2,

a₂ = a + b, b₂ = a + b, c₂ = -4ab

Now by using cross multiplication we get

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

⇒ x/(-(a + b)4ab + 2(a + b) (a² - b²)) = y/(− 2(a² − b²)(a + b) + 4ab(a – b)) 

= 1/((a − b)(a + b) − (a + b)(a + b))

⇒ x/(2(a + b)(a² - b² + 2ab)) = 1/-2b(a + b)

x = (2ab - a² + b²)/b

and,

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

y = (a - b)(a² + b²)/ b(a + b)

Hence, x = (2ab - a² + b²)/b and y = (a - b)(a² + b²)/ b(a + b)

Question 21. a²x + b²y = c² and b²x + a²y = d²

Solution:

Given that,

a²x + b²y = c²

b²x + a²y = d²

Or

a²x + b²y  - c² = 0

b²x + a²y - d² = 0

On comparing both the equation with the general form we get

 a₁ = a², b₁ = b², c₁ = -c²,

a₂ = b², b₂ = a², c₂ = -d²

Now by using cross multiplication we get

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

x/(-b²d² + a²c²) = y/(-c²b² + a²d²) = 1/(a⁴-b⁴)

 x/(a²c² - b²d²) = y/(a²d² - c²b²) = 1/(a⁴-b⁴)

Therefore,

 = x/(a²c² - b²d²) = 1/(a⁴ - b⁴)

x = (a²c² - b²d²)/(a⁴ - b⁴)

and,

= y/(a²d² - c²b²) = 1/(a⁴-b⁴)

y = (a²c² - b²d²) / (a⁴-b⁴)

Hence, x = (a²c² - b²d²)/(a⁴ - b⁴),  y = (a²c² - b²d²) / (a⁴ - b⁴)

Question 22. ax + by = (a + b)/2 and 3x + 5y = 4

Solution:

Given that,

ax + by = (a+b)/2

3x + 5y = 4

Or 

ax + by - (a + b)/2 = 0

3x + 5y - 4 = 0

On comparing both the equation with the general form we get

 a₁ = a, b₁ = b, c₁ = -(a + b)/2,

a₂ = 3, b₂ = 5, c₂ = -4

Now by using cross multiplication we get

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

= x/(-4b + 5((a + b)/2)) = y/(-3((a + b)/2) + 4a) = 1/(5a - 3b)

= x/((5a - 3b)/2) = y/((5a - 3b)/2) = 1/(5a - 3b)

Now,

x/((5a - 3b)/2) = 1/(5a - 3b)

x = (5a - 3b)/(2(5a - 3b))

x = 1/2

and,

y/((5a - 3b)/2) = 1/(5a - 3b)

y = (5a - 3b)/(2(5a - 3b))

y = 1/2

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

Question 23. 2 (ax – by) + a + 4b = 0 and 2 (bx + ay) + b – 4a = 0

Solution:

Given that,

2 (ax – by) + (a + 4b) = 0

2 (bx + ay) + (b – 4a) = 0

On comparing both the equation with the general form we get

 a₁ = 2a, b₁ = -2b, c₁ = a + 4b,

a₂ = 2b, b₂ = 2a, c₂ = b - 4a

Now by using cross multiplication we get

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

= x/((-2b(a + 4b)) - (2a(b - 4a ))) = y/((2b(a + 4b)) - (2a(b - 4a))) = 1/(4a² + 4b²)

= x/(-2b² + 8ab - 2ab + 8a²) = y/(2ab + 8b² - 2ab + 8a²) = 1/4(a² + b²)

= x/-2(a² + b²) = y/8(a² + b²) = 1/4(a² + b²)

So, 

= x/-2(a² + b²) = 1/4(a² + b²)

x = -1/2

and,

= y/8(a² + b²) = 1/4(a² + b²)

y = 2

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

Question 24. 6 (ax + by) = 3a + 2b and 6 (bx – ay) = 3b – 2a 

Solution:

given that,

6 (ax + by) = 3a + 2b

6 (bx – ay) = 3b – 2a 

6 (ax + by) -(3a + 2b)=0....       (1)

6 (bx – ay) -(3b – 2a) =0.....       (2)

On comparing both the equation with the general form we get

 a₁ = 6a, b₁ = 6b, c₁ = -(3a - 2b),

a₂ = 6b, b₂ = 66a, c₂ = -(3b - 2a)

Now by using cross multiplication we get

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

= x/(-6b(3b - 2a) - 6a(3a - 2b)) = y/(-6b(3a - 2b) + 6a(3b - 2a)) = 1/(-36a² - 36b²)

= x/(-18(a² + b²)) = y/(-12(a² + b²)) = 1/(-36(a² + b²))

Therefore,

x/(-18(a² + b²)) = 1/(-36(a² + b²))

x = 1/2

and,

y/(-12(a² + b²)) = 1/(-36(a² + b²))

y = 1/3

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

Question 25. (a²/x) − (b²/y) = 0 and (a²b/x) − (b²a/y) = a + b, x, y ≠ 0

Solution:

Given that,

(a²/x) − (b²/y) = 0

(a²b/x) − (b²a/y) = a + b 

Or

(a²b/x) − (b²a/y) - (a + b) = 0

On comparing both the equation with the general form we get

 a₁ = a², b₁ = -b², c₁ = 0,

a₂ = a²b, b₂ = b²a, c₂ = -(a + b)

Now by using cross multiplication we get

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

= (1/x)/(b²(a + b) - 0) = (1/y)/(0 + (a²(a + b))) = 1/(a³b² - a²b³)

= (1/x)/(b²(a + b)) = (1/y)/(a²(a + b)) = 1/a²b²(a + b)

So,

= (1/x)/(b²(a + b)) = 1/a²b²(a + b)

x = a²

and,

= (1/y)/(a²(a + b)) = 1/a²b²(a + b)

y = b²

Hence, x = a² and y = b²

Question 26. mx – ny = m² + n² and x + y = 2m

Solution:

Given that,

mx – ny = m² + n²

x + y = 2m

Or

mx – ny -(m² + n²) = 0

x + y - 2m = 0

On comparing both the equation with the general form we get

 a₁ = m, b₁ = -n, c₁ = -(m² + n²),

a₂ = 1, b₂ = 1, c₂ = -2m

Now by using cross multiplication we get

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

= x/(2mn + (m² + n²)) = y/(-(m² + n²) + 2m²) = 1/(m + n)

= x/(m + n)² = y/(m² - n²) = 1/(m + n)

Therefore,

 x/(m + n)² = 1/(m + n)

x = m + n

and,

y/(m² - n²) = 1/(m + n)

y = m - n

Hence, x = m + n, y = m - n

Question 27. (ax/b) - (by/a) = a + b and ax - by = 2ab

Solution:

Given that,

(ax/b) - (by/a) = a + b

ax - by = 2ab

Or

(ax/b) - (by/a) - (a + b) = 0

ax - by - 2ab = 0

On comparing both the equation with the general form we get

 a₁ = a/b, b₁ = -b/a, c₁ = -(a + b),

a₂ = a, b₂ = b, c₂ = -2ab

Now by using cross multiplication we get

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

= x/b(b - a) = -y/a(-a + b) = 1/(b - a)

So,

x/b(b - a) = 1/(b - a)

x = b

and,

-y/a(-a + b) = 1/(b - a)

y = -a

Hence, x = b, y = -a

Question 28. (b/a)x + (a/b)y = a² + b² and x + y = 2ab

Solution:

Given that,

(b/a)x + (a/b)y = a² + b²

x + y = 2ab

Or

(b/a)x + (a/b)y - (a² + b²) = 0

x + y - 2ab = 0

On comparing both the equation with the general form we get

 a₁ = b/a, b₁ = a/b, c₁ = -(a² + b²),

a₂ = 1, b₂ = 1, c₂ = -2ab

Now by using cross multiplication we get

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

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

Therefore,

 x/(b² - a²) = 1/((b² - a²)/ab)

x = ab

y/(-b² + a²) = 1/((b² - a²)/ab)

y = ab

Hence, x = ab, y = ab

Conclusion

Mastering the concepts of solving pairs of linear equations in the two variables is crucial for the tackling more complex algebraic problems. Exercise 3.4 | Set 2 of RD Sharma's textbook offers valuable practice in the applying different solving techniques. By solving these exercises, students will enhance their problem-solving skills and build a strong foundation in algebra.