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• RD Sharma Class 10 Solutions

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

### 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

aâ‚‚= 3a, bâ‚‚ = 4b, câ‚‚ =-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 â€“ 2b)x + (2a + b)y = 3

Solution:

Given that,

(a + 2b)x + (2a – b)y = 2

(a â€“ 2b)x + (2a + b)y = 3

Or

(a + 2b) x + (2a â€“ b)y – 2 = 0

(a â€“ 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)

â‡’ x/(2a2((a2 + b2 – 2ab + ab)/(a – b)) – 0) = y/(0 + 2a2((a2 + b2 + 2ab – ab)/(a + b)))

= 1/(((a2 + bÂ²)/(a – b) + (a2 + b2 + ab)/(a + b)))

â‡’ 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)

â‡’ 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)))

â‡’ x / ((-2ac/(a + b)) + ((2ac(a + b)/(b2 – a2)))) = y/((-(a + b)2ab)/(a2 – b2)) + (2ab/(a + b))

= 1/(-4abc/(a2 – b2))

â‡’ x/(-2ac((1/(a + b)) + (1/(a – b))) = y/(2ab((-1/(a – b)) + (1/(a + b))) = 1/(-4abc/(a2 – b2))

â‡’ 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 â€“ b) x + (a + b) y = 2a2 â€“ 2b2 and (a + b) (x + y) = 4ab

Solution:

Given that,

(a â€“ b) x + (a + b) y = 2a2 â€“ 2b2

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

(a â€“ b) x + (a + b) y – 2(a2 â€“ 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)

â‡’ x/(-(a + b)4ab + 2(a + b) (a2 – b2)) = y/(âˆ’ 2(a2 âˆ’ b2)(a + b) + 4ab(a â€“ b))

= 1/((a âˆ’ b)(a + b) âˆ’ (a + b)(a + b))

â‡’ 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 â€“ by) + a + 4b = 0 and 2 (bx + ay) + b â€“ 4a = 0

Solution:

Given that,

2 (ax â€“ by) + (a + 4b) = 0

2 (bx + ay) + (b â€“ 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 â€“ ay) = 3b â€“ 2a

Solution:

given that,

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

6 (bx â€“ ay) = 3b â€“ 2a

6 (ax + by) -(3a + 2b)=0….       (1)

6 (bx â€“ ay) -(3b â€“ 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) âˆ’ (b2/y) = 0 and (a2b/x) âˆ’ (b2a/y) = a + b, x, y â‰  0

Solution:

Given that,

(a2/x) âˆ’ (b2/y) = 0

(a2b/x) âˆ’ (b2a/y) = a + b

Or

(a2b/x) âˆ’ (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 â€“ ny = m2 + n2 and x + y = 2m

Solution:

Given that,

mx â€“ ny = m2 + n2

x + y = 2m

Or

mx â€“ 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

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