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

### 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

a

_{1}= 2a, b_{1}= 3b, c_{1}= -(a + 2b), a_{2}= 3a, b_{2}= 2b, c_{2}= -(2a + b)Now by using cross multiplication we get

x/(b

_{1}c_{2 }– b_{2}c_{1}) = y/(c_{1}a_{2 }– c_{2}a_{1}) = 1/(a_{1}b_{2}– a_{2}b_{1})x/(-3b(2a + b) + 2b(a + 2b)) = y/(-3a(a + 2b) + 2a(2a + b)) = 1/(4ab – 9ab)

x/(b

^{2 }– 4ab) = y/(a^{2}– 4ab) = 1/-5abx/(-4ab + b

^{2}) = 1/-5abx/b(b – 4a) = 1/-5ab

x = (4a – b)/5a

and,

= -y/(-a

^{2}+ 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, a

_{1}= 5a, b_{1}= 6b, c_{1}=-28aâ‚‚= 3a, bâ‚‚ = 4b, câ‚‚ =-18

On comparing both the equation with the general form we get

a

_{1}= 5a, b_{1}= 6b, c_{1}= -28, a_{2}= 3a, b_{2}= 4b, c_{2}= -18Now by using cross multiplication we get

x/(b

_{1}c_{2 }– b_{2}c_{1}) = y/(c_{1}a_{2 }– c_{2}a_{1}) = 1/(a_{1}b_{2}– a_{2}b_{1})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

a

_{1}= a + 2b, b_{1}= 2a – b, c_{1}= -2, a_{2}= a – 2b, b_{2}= 2a + b, c_{2}= -3Now by using cross multiplication we get

x/(b

_{1}c_{2 }– b_{2}c_{1}) = y/(c_{1}a_{2 }– c_{2}a_{1}) = 1/(a_{1}b_{2}– a_{2}b_{1})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/(2a

^{2}+ 5ab + 2b^{2}– 2a^{2}+ 5ab – 2b^{2})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 = 2a^{2}

**Solution:**

Given that,

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

Or on solving we get

x((a

^{2 }+ b^{2 }– 2ab + ab)/(a – b)) = y((a^{2 }+ b^{2 }+ 2ab – ab)/(a + b))= x((a

^{2 }+ b^{2 }– 2ab + ab)/(a – b)) – y((a^{2 }+ b^{2 }+ 2ab – ab)/(a + b)) = 0and,

x + y = 2a

^{2}On comparing both the equation with the general form we get

a

_{1}= (a^{2 }+ b^{2 }– 2ab + ab)/(a – b), b_{1}= (a^{2 }+ b^{2 }+ 2ab – ab)/(a + b), c_{1}= 0,a

_{2}= 1, b_{2}= 1, c_{2}= 2a^{2}Now by using cross multiplication we get

_{1}c_{2 }– b_{2}c_{1}) = y/(c_{1}a_{2 }– c_{2}a_{1}) = 1/(a_{1}b_{2}– a_{2}b_{1})â‡’ x/(2a

^{2}((a^{2 }+ b^{2 }– 2ab + ab)/(a – b)) – 0) = y/(0 + 2a^{2}((a^{2 }+ b^{2 }+ 2ab – ab)/(a + b)))= 1/(((a

^{2 }+ bÂ²)/(a – b) + (a^{2 }+ b^{2 }+ ab)/(a + b)))â‡’ x/(2a

^{2}((a^{2 }+ b^{2 }– 2ab + ab)/(a – b))) = y/(-2a^{2})((a^{2 }+ b^{2 }+ 2ab – ab)/(a + b)) = 1/(2a^{3}/(a^{2 }– b^{2}))Now,

x/(2a

^{2}((a^{2 }+ b^{2 }– 2ab + ab)/(a – b)) = 1/(2a^{3}/(a^{2 }– b^{2}))x = (2a

^{2}(a^{2}+ ab + b^{2})(a^{2 }– b^{2})) / 2a^{3}(a + b)x = (a

^{3 }– b^{3})/aand,

y/(-2a

^{2})((a^{2 }+ b^{2 }+ 2ab – ab)/(a + b)) = 1/(2a^{3}/(a^{2 }– b^{2}))y = (2a

^{2}(a^{2}– ab + b^{2})(a^{2 }– b^{2}))/2a^{3}(a – b)y = a

^{3 }+ b^{3}/a

Hence, x = (a^{3 }– b^{3})/a and y = a^{3}+ b^{3}/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/(a

^{2 }– b^{2}))) + cy(2a/(b^{2 }– a^{2})) – 2a/(a + b) = 0On comparing both the equation with the general form we get

a

_{1}= b, b_{1}= c, c_{1}= -(a + b),a

_{2}= 2b/(a^{2 }– b^{2}), b_{2}= 2a/(b^{2 }– a^{2}), c_{2}= 2a/(a + b)Now by using cross multiplication we get

_{1}c_{2 }– b_{2}c_{1}) = y/(c_{1}a_{2 }– c_{2}a_{1}) = 1/(a_{1}b_{2}– a_{2}b_{1})â‡’ x / ((-2ac/(a + b)) + ((2ac(a + b)/(b

^{2 }– a^{2})))) = y/((-(a + b)2ab)/(a^{2 }– b^{2})) + (2ab/(a + b))= 1/((2abc/(b

^{2}– a^{2})) – (2abc/(a^{2 }– b^{2})))â‡’ x / ((-2ac/(a + b)) + ((2ac(a + b)/(b

^{2 }– a^{2})))) = y/((-(a + b)2ab)/(a^{2 }– b^{2})) + (2ab/(a + b))= 1/(-4abc/(a

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

^{2 }– b^{2}))â‡’ x/(-4a

^{2}c/(a^{2 }– b^{2})) = y/(4ab^{2}/(a^{2 }– b^{2})) = 1/1/(-4abc/(a^{2 }– b^{2}))So,

x/(-4a

^{2}c/(a^{2 }– b^{2})) = 1/(-4abc/(a^{2 }– b^{2}))x = a/b

and,

y/(4ab

^{2}/(a^{2 }– b^{2})) = 1/(-4abc/(a^{2 }– b^{2}))y = b/c

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

### Question 20. (a â€“ b) x + (a + b) y = 2a^{2 }â€“ 2b^{2 }and (a + b) (x + y) = 4ab

**Solution:**

Given that,

(a â€“ b) x + (a + b) y = 2a

^{2}â€“ 2b^{2}(a + b) (x + y) = 4ab

(a â€“ b) x + (a + b) y – 2(a

^{2 }â€“ b^{2}) = 0(a + b)x + (a + b)y – 4ab = 0

On comparing both the equation with the general form we get

a

_{1}= a – b, b_{1}= a + b, c_{1}= -2,a

_{2}= a + b, b_{2}= a + b, c_{2}= -4abNow by using cross multiplication we get

_{1}c_{2 }– b_{2}c_{1}) = y/(c_{1}a_{2 }– c_{2}a_{1}) = 1/(a_{1}b_{2}– a_{2}b_{1})â‡’ x/(-(a + b)4ab + 2(a + b) (a

^{2}– b^{2})) = y/(âˆ’ 2(a^{2}âˆ’ b^{2})(a + b) + 4ab(a â€“ b))= 1/((a âˆ’ b)(a + b) âˆ’ (a + b)(a + b))

â‡’ x/(2(a + b)(a

^{2}– b^{2}+ 2ab)) = 1/-2b(a + b)x = (2ab – a

^{2}+ b^{2})/band,

= -y/(2(a – b) (a

^{2}+ b^{2}) -2b (a + b)) = 1/ -2b(a + b)y = (a – b)(a

^{2}+ b^{2})/ b(a + b)

Hence, x = (2ab – a^{2}+ b^{2})/b and y = (a – b)(a^{2}+ b^{2})/ b(a + b)

### Question 21. a^{2}x + b^{2}y = c^{2 }and b^{2}x + a^{2}y = d^{2}

**Solution:**

Given that,

a

^{2}x + b^{2}y = c^{2}b

^{2}x + a^{2}y = d^{2}Or

a

^{2}x + b^{2}y – c^{2 }= 0b

^{2}x + a^{2}y – d^{2 }= 0On comparing both the equation with the general form we get

a

_{1}= a^{2}, b_{1}= b^{2}, c_{1}= -c^{2},a

_{2}= b^{2}, b_{2}= a^{2}, c_{2}= -d^{2}Now by using cross multiplication we get

_{1}c_{2 }– b_{2}c_{1}) = y/(c_{1}a_{2 }– c_{2}a_{1}) = 1/(a_{1}b_{2}– a_{2}b_{1})x/(-b

^{2}d^{2}+ a^{2}c^{2}) = y/(-c^{2}b^{2}+ a^{2}d^{2}) = 1/(a^{4}-b^{4})x/(a

^{2}c^{2}– b^{2}d^{2}) = y/(a^{2}d^{2 }– c^{2}b^{2}) = 1/(a^{4}-b^{4})Therefore,

= x/(a

^{2}c^{2}– b^{2}d^{2}) = 1/(a^{4}– b^{4})x = (a

^{2}c^{2 }– b^{2}d^{2})/(a^{4 }– b^{4})and,

= y/(a

^{2}d^{2 }– c^{2}b^{2}) = 1/(a^{4}-b^{4})y = (a

^{2}c^{2 }– b^{2}d^{2}) / (a^{4}-b^{4})

Hence, x = (a^{2}c^{2 }– b^{2}d^{2})/(a^{4 }– b^{4}), y = (a^{2}c^{2 }– b^{2}d^{2}) / (a^{4 }– b^{4})

### 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

a

_{1}= a, b_{1}= b, c_{1}= -(a + b)/2,a

_{2}= 3, b_{2}= 5, c_{2}= -4Now by using cross multiplication we get

_{1}c_{2 }– b_{2}c_{1}) = y/(c_{1}a_{2 }– c_{2}a_{1}) = 1/(a_{1}b_{2}– a_{2}b_{1})= 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

a

_{1}= 2a, b_{1}= -2b, c_{1}= a + 4b,a

_{2}= 2b, b_{2}= 2a, c_{2}= b – 4aNow by using cross multiplication we get

_{1}c_{2 }– b_{2}c_{1}) = y/(c_{1}a_{2 }– c_{2}a_{1}) = 1/(a_{1}b_{2}– a_{2}b_{1})= x/((-2b(a + 4b)) – (2a(b – 4a ))) = y/((2b(a + 4b)) – (2a(b – 4a))) = 1/(4a

^{2}+ 4b^{2})= x/(-2b

^{2}+ 8ab – 2ab + 8a^{2}) = y/(2ab + 8b^{2}– 2ab + 8a^{2}) = 1/4(a^{2}+ b^{2})= x/-2(a

^{2}+ b^{2}) = y/8(a^{2}+ b^{2}) = 1/4(a^{2}+ b^{2})So,

= x/-2(a

^{2}+ b^{2}) = 1/4(a^{2}+ b^{2})x = -1/2

and,

= y/8(a

^{2}+ b^{2}) = 1/4(a^{2}+ b^{2})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

a

_{1}= 6a, b_{1}= 6b, c_{1}= -(3a – 2b),a

_{2}= 6b, b_{2}= 66a, c_{2}= -(3b – 2a)Now by using cross multiplication we get

_{1}c_{2 }– b_{2}c_{1}) = y/(c_{1}a_{2 }– c_{2}a_{1}) = 1/(a_{1}b_{2}– a_{2}b_{1})= x/(-6b(3b – 2a) – 6a(3a – 2b)) = y/(-6b(3a – 2b) + 6a(3b – 2a)) = 1/(-36a

^{2 }– 36b^{2})= x/(-18(a

^{2 }+ b^{2})) = y/(-12(a^{2 }+ b^{2})) = 1/(-36(a^{2 }+ b^{2}))Therefore,

x/(-18(a

^{2 }+ b^{2})) = 1/(-36(a^{2 }+ b^{2}))x = 1/2

and,

y/(-12(a

^{2 }+ b^{2})) = 1/(-36(a^{2 }+ b^{2}))y = 1/3

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

### Question 25. (a^{2}/x) âˆ’ (b^{2}/y) = 0 and (a^{2}b/x) âˆ’ (b^{2}a/y) = a + b, x, y â‰ 0

**Solution:**

Given that,

(a

^{2}/x) âˆ’ (b^{2}/y) = 0(a

^{2}b/x) âˆ’ (b^{2}a/y) = a + bOr

(a

^{2}b/x) âˆ’ (b^{2}a/y) – (a + b) = 0On comparing both the equation with the general form we get

a

_{1}= a^{2}, b_{1}= -b^{2}, c_{1}= 0,a

_{2}= a^{2}b, b_{2}= b^{2}a, c_{2}= -(a + b)Now by using cross multiplication we get

_{1}c_{2 }– b_{2}c_{1}) = y/(c_{1}a_{2 }– c_{2}a_{1}) = 1/(a_{1}b_{2}– a_{2}b_{1})= (1/x)/(b

^{2}(a + b) – 0) = (1/y)/(0 + (a^{2}(a + b))) = 1/(a^{3}b^{2 }– a^{2}b^{3})= (1/x)/(b

^{2}(a + b)) = (1/y)/(a^{2}(a + b)) = 1/a^{2}b^{2}(a + b)So,

= (1/x)/(b

^{2}(a + b)) = 1/a^{2}b^{2}(a + b)x = a

^{2}and,

= (1/y)/(a

^{2}(a + b)) = 1/a^{2}b^{2}(a + b)y = b

^{2}

Hence, x = a^{2 }and y = b^{2}

### Question 26. mx â€“ ny = m^{2} + n^{2} and x + y = 2m

**Solution:**

Given that,

mx â€“ ny = m

^{2}+ n^{2}x + y = 2m

Or

mx â€“ ny -(m

^{2}+ n^{2}) = 0x + y – 2m = 0

On comparing both the equation with the general form we get

a

_{1}= m, b_{1}= -n, c_{1}= -(m^{2}+ n^{2}),a

_{2}= 1, b_{2}= 1, c_{2}= -2mNow by using cross multiplication we get

_{1}c_{2 }– b_{2}c_{1}) = y/(c_{1}a_{2 }– c_{2}a_{1}) = 1/(a_{1}b_{2}– a_{2}b_{1})= x/(2mn + (m

^{2}+ n^{2})) = y/(-(m^{2 }+ n^{2}) + 2m^{2}) = 1/(m + n)= x/(m + n)

^{2}= y/(m^{2 }– n^{2}) = 1/(m + n)Therefore,

x/(m + n)

^{2}= 1/(m + n)x = m + n

and,

y/(m

^{2}– n^{2}) = 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

a

_{1}= a/b, b_{1}= -b/a, c_{1}= -(a + b),a

_{2}= a, b_{2}= b, c_{2}= -2abNow by using cross multiplication we get

_{1}c_{2 }– b_{2}c_{1}) = y/(c_{1}a_{2 }– c_{2}a_{1}) = 1/(a_{1}b_{2}– a_{2}b_{1})= 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 = a^{2} + b^{2} and x + y = 2ab

**Solution:**

Given that,

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

^{2 }+ b^{2}x + y = 2ab

Or

(b/a)x + (a/b)y – (a

^{2 }+ b^{2}) = 0x + y – 2ab = 0

On comparing both the equation with the general form we get

a

_{1}= b/a, b_{1}= a/b, c_{1}= -(a^{2 }+ b^{2}),a

_{2}= 1, b_{2}= 1, c_{2}= -2abNow by using cross multiplication we get

_{1}c_{2 }– b_{2}c_{1}) = y/(c_{1}a_{2 }– c_{2}a_{1}) = 1/(a_{1}b_{2}– a_{2}b_{1})= x/(b

^{2 }– a^{2}) = y/(-b^{2 }+ a^{2}) = 1/((b^{2 }– a^{2})/ab)Therefore,

x/(b

^{2 }– a^{2}) = 1/((b^{2 }– a^{2})/ab)x = ab

y/(-b

^{2 }+ a^{2}) = 1/((b^{2 }– a^{2})/ab)y = ab

Hence, x = ab, y = ab

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