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

### 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

a

_{1}= 1, b_{1}= 2, c_{1}= 1, a_{2}= 2, b_{2}= −3, c_{2}= -12Now 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/(-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

a

_{1}= 3, b_{1}= 2, c_{1}= 25, a_{2}= 2, b_{2}= 1, c_{2}= 10Now 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/(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

a

_{1}= 2, b_{1}= 1, c_{1}= -35, a_{2}= 3, b_{2}= 4, c_{2}= -65Now 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/(-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

a

_{1}= 2, b_{1}= -1, c_{1}= -6, a_{2}= 1, b_{2}= -1, c_{2}= -2Now 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/(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

a

_{1}= 1, b_{1}= 1, c_{1}= -2, a_{2}= 1, b_{2}= -1, c_{2}= -6Now 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)/(-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

a

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

^{2}– b^{2})x/(-ab – b

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

x/-(-a

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

and, y/(a

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

a

_{1}= 1, b_{1}= a, c_{1}= – b, a_{2}= a, b_{2}= -b, c_{2}= -cNow 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/(-ac – b

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

^{2}) = 1/(-b – a^{2})x = (ac + b

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

y/(- ab + c) = 1/(-b – a

^{2})y =(ab – c)/(a

^{2}+ b)

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

### Question 8. ax + by = a^{2 }and bx + ay = b^{2}

**Solution:**

Given that,

ax + by = a

^{2}bx + ay = b

^{2}Or, ax + by – a

^{2}= 0bx + ay – b

^{2}= 0On comparing both the equation with the geneal form we get

a

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

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

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

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

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

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

y/(-ab(a – b)) = 1/(a

^{2 }– b^{2})(a – b)(a

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

y= -ab/(a + b)

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

a

_{1}= 5, b_{1}= -2, c_{1}= – 1, a_{2}= 15, b_{2}= 7, c_{2}= -10Now 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/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

a

_{1}= 2, b_{1}= 3, c_{1}= – 13, a_{2}= 5, b_{2}= -4, c_{2}= 2Now 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)/(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

a

_{1}= 57, b_{1}= 6, c_{1}= – 5, a_{2}= 38, b_{2}= 21, c_{2}= -9Now 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/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 = a^{2} – b^{2}

**Solution:**

Given that,

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

ax – by = a

^{2 }– b^{2}(x/a) + (y/b) – 2 = 0

ax – by – (a

^{2 }– b^{2}) = 0On comparing both the equation with the geneal form we get

a

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

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

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

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

x/(-(a

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

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

and,

y/(-(a

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

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

Hence, x = a and y = b

**Question 13. (x/a) + (y/b) = a + b and (x/a**^{2}**) + (y/b**^{2}**) = 2**

**Solution:**

Given that,

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

(x/a

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

(x/a

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

a

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

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

x/((a – b)b

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

^{2}and,

-y/((-a – b)/a

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

^{2}

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

**Question 14. x/a = y/b and ax + by = a**^{2}** + b**^{2}

**Solution:**

Given that,

x/a = y/b

ax + by = a

^{2 }+ b^{2}On comparing both the equation with the geneal form we get

a

_{1}= 1/a, b_{1}= 1/b, c_{1}= 0, a_{2}= a, b_{2}= b, c_{2}= -(a^{2 }+ b^{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/((a

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

x/((a

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

and,

y/((a

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

Hence, x = a and y = b

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