Algebra_8_Questions

म.स. र ल.स. (HCF and LCM)

For Q.No.8 (a) in BLE Examination

1. म.स. पत्ता लगाउनुहोस् (Find the HCF of): \( 2ab^2 \) and \( 4a^2b \) [1A]
 
   The Given Expressions are
\( 2ab^2 = \boxed{2} \cdot \boxed{a} \cdot \boxed{b} \cdot b \)
\( 4a^2b = \boxed{2} \cdot 2 \cdot \boxed{a} \cdot a \cdot \boxed{b} \)
So, HCF is
HCF = \( 2 \cdot a \cdot b \)
orHCF =\( 2ab \) 
2. म.स. निकाल्नुहोस् (Find the HCF of): \( (x + 3)(x - 5) \), \( (x - 5)(x + 6) \) [1A]
 
   The Given Expressions are
1st Expression = \( (x + 3) \cdot \boxed{(x - 5)} \)
2nd Expression = \( \boxed{(x - 5)} \cdot (x + 6) \)
So, HCF is
HCF = \( (x - 5) \) 
3. म.स. निकाल्नुहोस् (Find the HCF of): \( a^2b \), \( ab^2c \), \( a^2b^2c \) [1A]
 
   The Given Expressions are
1st Expression = \( a^2b = \boxed{a} \cdot a \cdot \boxed{b} \)
2nd Expression = \( ab^2c = \boxed{a} \cdot \boxed{b} \cdot b \cdot c \)
3rd Expression = \( a^2b^2c = \boxed{a} \cdot a \cdot \boxed{b} \cdot b \cdot c \)
So, HCF is
HCF = \( a \cdot b =ab \) 
4. म.स. पत्ता लगाउनुहोस् (Find the HCF of): \( x^2 + 7x + 10 \), \( x^2 - x - 6 \) [2A]
 
   The Given Expressions are
1st Expression = \( x^2 + 7x + 10 \)
or1st Expression = \( x^2 + 5x + 2x + 10 \)
or1st Expression = \( x(x + 5) + 2(x + 5) \)
or1st Expression = \( (x + 5) \cdot \boxed{(x + 2)} \)
Next
2nd Expression = \( x^2 - x - 6 \)
or2nd Expression =\( x^2 - 3x + 2x - 6 \)
or2nd Expression =\( x(x - 3) + 2(x - 3) \)
or2nd Expression =\( (x - 3) \cdot \boxed{(x + 2)} \)
So, HCF is
HCF = \( x + 2 \) 
5. म.स. पत्ता लगाउनुहोस् (Find the HCF of): \( 3x^3 - 15x^2 \), \( 2x^3 - 50x \) [2A]
 
   The Given Expressions are
1st Expression = \( 3x^3 - 15x^2 \)
or1st Expression = \( 3 \cdot \boxed{x} \cdot x \cdot \boxed{(x - 5)} \)
Next
2nd Expression = \( 2x^3 - 50x \)
or2nd Expression =\( 2x(x^2 - 25) \)
or2nd Expression =\( 2 \cdot \boxed{x} \cdot (x^2 - 5^2) \)
or2nd Expression =\( 2 \cdot \boxed{x} \cdot \boxed{(x - 5)} \cdot (x + 5) \)
So, HCF is
HCF = \( x(x - 5) \) 
6. म.स. पत्ता लगाउनुहोस् (Find the HCF of): \( a^2 - 4 \), \( a^2 + 5a + 6 \) [2A]
 
   The Given Expressions are
1st Expression = \( a^2 - 4 \)
or1st Expression = \( a^2 - 2^2 \)
or1st Expression = \( (a - 2) \cdot \boxed{(a + 2)} \)
Next
2nd Expression = \( a^2 + 5a + 6 \)
or2nd Expression =\( a^2 + 3a + 2a + 6 \)
or2nd Expression =\( a(a + 3) + 2(a + 3) \)
or2nd Expression =\( (a + 3) \cdot \boxed{(a + 2)} \)
So, HCF is
HCF = \( a + 2 \) 
7. म.स. निकाल्नुहोस् (Find the HCF of): \( 5x^2 - 125 \), \( x^2 - 10x + 25 \) and \( 2x^2 - 10x \) [2A]
 
   The Given Expressions are
1st Expression = \( 5x^2 - 125 \)
or1st Expression = \( 5(x^2 - 25) \)
or1st Expression = \( 5 \cdot (x + 5) \cdot \boxed{(x - 5)} \)
Next
2nd Expression = \( x^2 - 10x + 25 \)
or2nd Expression =\( x^2 - 5x - 5x + 25 \)
or2nd Expression =\( x(x - 5) - 5(x - 5) \)
or2nd Expression =\( \boxed{(x - 5)} \cdot (x - 5) \)
Next
3rd Expression = \( 2x^2 - 10x \)
or3rd Expression =\( 2x(x - 5) \)
or3rd Expression =\( 2 \cdot x \cdot \boxed{(x - 5)} \)
So, HCF is
HCF = \( x - 5 \) 
8. म.स. निकाल्नुहोस् (Find the HCF of): \( x^3 + 7x^2 + 12x \), \( x^3 + 64 \) and \( 3x^2 + 27x + 60 \) [2A]
 
   The Given Expressions are
1st Expression = \( x^3 + 7x^2 + 12x \)
or1st Expression = \( x(x^2 + 7x + 12) \)
or1st Expression = \( x(x^2 + 4x + 3x + 12) \)
or1st Expression = \( x[x(x + 4) + 3(x + 4)] \)
or1st Expression = \( x \cdot (x + 3) \cdot \boxed{(x + 4)} \)
Next
2nd Expression = \( x^3 + 64 \)
or2nd Expression =\( x^3 + 4^3 \)
or2nd Expression =\( (x + 4)(x^2 - 4x + 4^2) \)
or2nd Expression =\( \boxed{(x + 4)} \cdot (x^2 - 4x + 16) \)
Next
3rd Expression = \( 3x^2 + 27x + 60 \)
or3rd Expression =\( 3(x^2 + 9x + 20) \)
or3rd Expression =\( 3(x^2 + 5x + 4x + 20) \)
or3rd Expression =\( 3[x(x + 5) + 4(x + 5)] \)
or3rd Expression =\( 3 \cdot (x + 5) \cdot \boxed{(x + 4)} \)
So, HCF is
HCF = \( x + 4 \) 
9. म.स. पत्ता लगाउनुहोस् (Find HCF of): \( y^2 - 5y + 6 \), \( y^3 - 8 \), \( 2y^2 - 8 \) [2A]
 
   The Given Expressions are
1st Expression = \( y^2 - 5y + 6 \)
or1st Expression = \( y^2 - 3y - 2y + 6 \)
or1st Expression = \( y(y - 3) - 2(y - 3) \)
or1st Expression = \( (y - 3) \cdot \boxed{(y - 2)} \)
Next
2nd Expression = \( y^3 - 8 \)
or2nd Expression =\( y^3 - 2^3 \)
or2nd Expression =\( (y - 2)(y^2 + 2y + 4) \)
or2nd Expression =\( \boxed{(y - 2)} \cdot (y^2 + 2y + 4) \)
Next
3rd Expression = \( 2y^2 - 8 \)
or3rd Expression =\( 2(y^2 - 4) \)
or3rd Expression =\( 2(y - 2)(y + 2) \)
or3rd Expression =\( 2 \cdot \boxed{(y - 2)} \cdot (y + 2) \)
So, HCF is
HCF = \( y - 2 \) 
10. म.स. पत्ता लगाउनुहोस् (Find the HCF of): \( m^3 - 9m \), \( m^2 + m - 6 \), \( m^4 + 27m \) [2A]
 
   The Given Expressions are
1st Expression = \( m^3 - 9m \)
or1st Expression = \( m(m^2 - 9) \)
or1st Expression = \( m \cdot (m + 3) \cdot (m - 3) \)
Next
2nd Expression = \( m^2 + m - 6 \)
or2nd Expression =\( m^2 + 3m - 2m - 6 \)
or2nd Expression =\( m(m + 3) - 2(m + 3) \)
or2nd Expression =\( \boxed{(m + 3)} \cdot (m - 2) \)
Next
3rd Expression = \( m^4 + 27m \)
or3rd Expression =\( m(m^3 + 27) \)
or3rd Expression =\( m(m^3 + 3^3) \)
or3rd Expression =\( m \cdot \boxed{(m + 3)} \cdot (m^2 - 3m + 9) \)
So, HCF is
HCF = \( m + 3 \) 
11. म.स. पत्ता लगाउनुहोस् (Find the HCF of): \( 4a^3 - 9a \), \( 6a^2 + 9a \), \( 6a^3 + 5a^2 - 6a \) [2A]
 
   The Given Expressions are
1st Expression = \( 4a^3 - 9a \)
or1st Expression = \( a(4a^2 - 9) \)
or1st Expression = \( a(2a - 3) \cdot \boxed{(2a + 3)} \)
Next
2nd Expression = \( 6a^2 + 9a \)
or2nd Expression =\( 3a(2a + 3) \)
or2nd Expression =\( 3 \cdot \boxed{a} \cdot \boxed{(2a + 3)} \)
Next
3rd Expression = \( 6a^3 + 5a^2 - 6a \)
or3rd Expression =\( a(6a^2 + 5a - 6) \)
or3rd Expression =\( a(6a^2 + 9a - 4a - 6) \)
or3rd Expression =\( a[3a(2a + 3) - 2(2a + 3)] \)
or3rd Expression =\( \boxed{a} \cdot (3a - 2) \cdot \boxed{(2a + 3)} \)
So, HCF is
HCF = \( a(2a + 3) \) 
12. म.स. पत्ता लगाउनुहोस् (Find the HCF of): \( p^3 + 2p^2 - 15p \), \( p^2 - 7p + 12 \) and \( 3p^2 - 27 \) [2A]
 
   The Given Expressions are
1st Expression = \( p^3 + 2p^2 - 15p \)
or1st Expression = \( p(p^2 + 2p - 15) \)
or1st Expression = \( p(p^2 + 5p - 3p - 15) \)
or1st Expression = \( p[p(p + 5) - 3(p + 5)] \)
or1st Expression = \( p \cdot (p + 5) \cdot \boxed{(p - 3)} \)
Next
  2nd Expression = \( p^2 - 7p + 12 \)
or2nd Expression =\( p^2 - 4p - 3p + 12 \)
or2nd Expression =\( p(p - 4) - 3(p - 4) \)
or2nd Expression =\( (p - 4) \cdot \boxed{(p - 3)} \)
Next
  3rd Expression = \( 3p^2 - 27 \)
or3rd Expression =\( 3(p^2 - 9) \)
or3rd Expression =\( 3(p^2 - 3^2) \)
or3rd Expression =\( 3 \cdot (p + 3) \cdot \boxed{(p - 3)} \)
So, HCF is
HCF = \( p - 3 \) 
13. म.स. पत्ता लगाउनुहोस् (Find the HCF of): \( x^3 + 6x^2 - 4x - 24 \), \( x^2 + 5x + 6 \) and \( x^2 - 4 \) [2A]
 
   The Given Expressions are
1st Expression = \( x^3 + 6x^2 - 4x - 24 \)
or1st Expression = \( x^2(x + 6) - 4(x + 6) \)
or1st Expression = \( (x^2 - 4)(x + 6) \)
or1st Expression = \( (x - 2) \cdot \boxed{(x + 2)} \cdot (x + 6) \)
Next
  2nd Expression = \( x^2 + 5x + 6 \)
or2nd Expression =\( x^2 + 3x + 2x + 6 \)
or2nd Expression =\( x(x + 3) + 2(x + 3) \)
or2nd Expression =\( (x + 3) \cdot \boxed{(x + 2)} \)
Next
  3rd Expression = \( x^2 - 4 \)
or3rd Expression =\( x^2 - 2^2 \)
or3rd Expression =\( (x - 2) \cdot \boxed{(x + 2)} \)
So, HCF is
HCF = \( x + 2 \) 
14. म.स. पत्ता लगाउनुहोस् (Find the HCF of): \( a^2 - 25 \), \( a^2 - 6a + 5 \) and \( (a - 5)^2 \) [2A]
 
   The Given Expressions are
1st Expression = \( a^2 - 25 \)
or1st Expression = \( a^2 - 5^2 \)
or1st Expression = \( (a + 5) \cdot \boxed{(a - 5)} \)
Next
  2nd Expression = \( a^2 - 6a + 5 \)
or2nd Expression =\( a^2 - 5a - a + 5 \)
or2nd Expression =\( a(a - 5) - 1(a - 5) \)
or2nd Expression =\( (a - 1) \cdot \boxed{(a - 5)} \)
Next
  3rd Expression = \( (a - 5)^2 \)
or3rd Expression =\( \boxed{(a - 5)} \cdot (a - 5) \)
So, HCF is
HCF = \( a - 5 \) 
15. म.स. पत्ता लगाउनुहोस् (Find the HCF of): \( 2x^2 - 8 \), \( x^2 - 4x + 4 \) and \( x^2 - 3x + 2 \) [2A]
 
   The Given Expressions are
1st Expression = \( 2x^2 - 8 \)
or1st Expression = \( 2(x^2 - 4) \)
or1st Expression = \( 2(x^2 - 2^2) \)
or1st Expression = \( 2 \cdot (x + 2) \cdot \boxed{(x - 2)} \)
Next
  2nd Expression = \( x^2 - 4x + 4 \)
or2nd Expression =\( x^2 - 2x - 2x + 4 \)
or2nd Expression =\( x(x - 2) - 2(x - 2) \)
or2nd Expression =\( \boxed{(x - 2)} \cdot (x - 2) \)
Next
  3rd Expression = \( x^2 - 3x + 2 \)
or3rd Expression =\( x^2 - 2x - x + 2 \)
or3rd Expression =\( x(x - 2) - 1(x - 2) \)
or3rd Expression =\( (x - 1) \cdot \boxed{(x - 2)} \)
So, HCF is
HCF = \( x - 2 \) 
16. ल.स. पत्ता लगाउनुहोस् (Find the LCM of): \( 6ab^2 \) and \( 3ab \) [1A]
 
   The Given Expressions are
1st Expression = \( 6ab^2 = 2 \cdot 3 \cdot a \cdot b^2 \)
2nd Expression = \( 3ab = 3 \cdot a \cdot b \)
To find the LCM, we take the highest power of all factors.
So, LCM is
LCM = \( 2 \cdot 3 \cdot a \cdot b^2 = 6ab^2 \) 
17. ल.स. पत्ता लगाउनुहोस् (Find the LCM of): \( a^3b^2 \) and \( a^2b^3 \) [1A]
 
   The Given Expressions are
1st Expression = \( a^3b^2 = \boxed{a^3} \cdot b^2 \)
2nd Expression = \( a^2b^3 = a^2 \cdot \boxed{b^3} \)
To find the LCM, we take the highest power of all factors.
So, LCM is
LCM = \( \boxed{a^3} \cdot \boxed{b^3} = a^3b^3 \) 
18. ल.स. निकाल्नुहोस् (Find the LCM of): \( a + b \), \( a^2 - b^2 \) [1A]
 
   The Given Expressions are
1st Expression = \( \boxed{(a + b)} \)
2nd Expression = \( a^2 - b^2 = (a + b) \cdot \boxed{(a - b)} \)
To find the LCM, we take all the factors with their highest powers.
So, LCM is
LCM = \( (a + b)(a - b)=a^2 - b^2 \) 
19. ल.स. पत्ता लगाउनुहोस् (Find the LCM of): \( a^2 - 1 \), \( a^2 + a - 2 \) [2A]
 
   The Given Expressions are
1st Expression = \( a^2 - 1 \)
or1st Expression = \( a^2 - 1^2 \)
or1st Expression = \( \boxed{(a + 1)} \cdot \boxed{(a - 1)} \)
Next
2nd Expression = \( a^2 + a - 2 \)
or2nd Expression =\( a^2 + 2a - a - 2 \)
or2nd Expression =\( a(a + 2) - 1(a + 2) \)
or2nd Expression =\( (a - 1) \cdot \boxed{(a + 2)} \)
So, LCM is
LCM = \( (a + 1)(a - 1)(a + 2) \)
orLCM = \( (a^2 - 1)(a + 2) \) 
20. ल.स. पत्ता लगाउनुहोस् (Find the LCM of): \( x^2 + x - 20 \), \( x^2 - 25 \) [2A]
 
   The Given Expressions are
1st Expression = \( x^2 + x - 20 \)
or1st Expression = \( x^2 + 5x - 4x - 20 \)
or1st Expression = \( x(x + 5) - 4(x + 5) \)
or1st Expression = \( \boxed{(x + 5)} \cdot \boxed{(x - 4)} \)
Next
  2nd Expression = \( x^2 - 25 \)
or2nd Expression =\( x^2 - 5^2 \)
or2nd Expression =\( (x + 5) \cdot \boxed{(x - 5)} \)
So, LCM is
LCM = \( (x + 5)(x - 4)(x - 5) \)
orLCM = \( (x^2 - 25)(x - 4) \) 
21. ल.स. पत्ता लगाउनुहोस् (Find the LCM of): \( 4a - 24 \), \( a^2 - 36 \) and \( a^2 - 3a - 18 \) [3A]
 
   The Given Expressions are
1st Expression = \( 4a - 24 \)
or1st Expression = \( 4(a - 6) = \boxed{2^2} \cdot \boxed{(a - 6)} \)
Next
2nd Expression = \( a^2 - 36 \)
or2nd Expression =\( a^2 - 6^2 \)
or2nd Expression =\( (a - 6) \cdot \boxed{(a + 6)} \)
Next
3rd Expression = \( a^2 - 3a - 18 \)
or3rd Expression =\( a^2 - 6a + 3a - 18 \)
or3rd Expression =\( a(a - 6) + 3(a - 6) \)
or3rd Expression =\( (a - 6) \cdot \boxed{(a + 3)} \)
So, LCM is
LCM = \( 2^2(a - 6)(a + 6)(a + 3) \)
orLCM = \( 4(a^2 - 36)(a + 3) \) 
22. ल.स. पत्ता लगाउनुहोस् (Find the LCM of): \( (x + 2)^2 \), \( x^2 + 6x + 8 \) and \( x^2 + 7x + 10 \) [3A]
 
   The Given Expressions are
1st Expression = \( (x + 2)^2 = \boxed{(x + 2)^2} \)
Next
  2nd Expression = \( x^2 + 6x + 8 \)
or2nd Expression =\( x^2 + 4x + 2x + 8 \)
or2nd Expression =\( x(x + 4) + 2(x + 4) \)
or2nd Expression =\( \boxed{(x + 4)} \cdot (x + 2) \)
Next
  3rd Expression = \( x^2 + 7x + 10 \)
or3rd Expression =\( x^2 + 5x + 2x + 10 \)
or3rd Expression =\( x(x + 5) + 2(x + 5) \)
or3rd Expression =\( \boxed{(x + 5)} \cdot (x + 2) \)
So, LCM is
LCM = \( (x + 2)^2(x + 4)(x + 5) \) 
23. ल.स. पत्ता लगाउनुहोस् (Find the LCM of): \( x^2 + 3x - 10 \), \( x^2 - 6x + 8 \) and \( x^2 + x - 20 \) [3A]
 
   The Given Expressions are
1st Expression = \( x^2 + 3x - 10 \)
or1st Expression = \( x^2 + 5x - 2x - 10 \)
or1st Expression = \( x(x + 5) - 2(x + 5) \)
or1st Expression = \( \boxed{(x - 2)} \cdot \boxed{(x + 5)} \)
Next
2nd Expression = \( x^2 - 6x + 8 \)
or2nd Expression =\( x^2 - 4x - 2x + 8 \)
or2nd Expression =\( x(x - 4) - 2(x - 4) \)
or2nd Expression =\((x - 2) \cdot \boxed{(x - 4)} \)
Next
3rd Expression = \( x^2 + x - 20 \)
or3rd Expression =\( x^2 + 5x - 4x - 20 \)
or3rd Expression =\( x(x + 5) - 4(x + 5) \)
or3rd Expression =\( (x + 5) \cdot (x - 4) \)
So, LCM is
LCM = \( (x - 2)(x + 5)(x - 4) \) 
24. ल.स. पत्ता लगाउनुहोस् (Find the LCM of): \( a^2 + 6a + 8 \), \( a^2 + 9a + 20 \) and \( a^2 + 7a + 10 \) [3A]
 
   The Given Expressions are
1st Expression = \( a^2 + 6a + 8 \)
or1st Expression = \( a^2 + 4a + 2a + 8 \)
or1st Expression = \( a(a + 4) + 2(a + 4) \)
or1st Expression = \( \boxed{(a + 4)} \cdot \boxed{(a + 2)} \)
Next
2nd Expression = \( a^2 + 9a + 20 \)
or2nd Expression =\( a^2 + 5a + 4a + 20 \)
or2nd Expression =\( a(a + 5) + 4(a + 5) \)
or2nd Expression =\( \boxed{(a + 5)} \cdot (a + 4) \)
Next
3rd Expression = \( a^2 + 7a + 10 \)
or3rd Expression =\( a^2 + 5a + 2a + 10 \)
or3rd Expression =\( a(a + 5) + 2(a + 5) \)
or3rd Expression =\( (a + 5) \cdot(a + 2) \)
So, LCM is
LCM = \( (a + 4)(a + 2)(a + 5) \) 
25. म.स. र ल.स. पत्ता लगाउनुहोस् (Find the HCF and LCM of): \( x^2 - 5x + 6 \) and \( x^3 - 4x \) [3A]
 
   The Given Expressions are
1st Expression = \( x^2 - 5x + 6 \)
or1st Expression = \( x^2 - 3x - 2x + 6 \)
or1st Expression = \( x(x - 3) - 2(x - 3) \)
or1st Expression = \( \boxed{(x - 2)} \cdot (x - 3) \)
Next
2nd Expression = \( x^3 - 4x \)
or2nd Expression = \( x(x^2 - 4) \)
or2nd Expression = \( x \cdot (x^2 - 2^2) \)
or2nd Expression = \( x \cdot \boxed{(x - 2)} \cdot (x + 2) \)
So, HCF is
HCF = \( x - 2 \)
Next, LCM is
LCM = \( (x - 2)(x - 3) x(x + 2) \)
orLCM = \( x(x^2 - 4)(x - 3) \) 
26. म.स. र ल.स. पत्ता लगाउनुहोस् (Find the HCF and LCM of): \( x^2 - 25 \) and \( x^2 - 9x + 20 \) [3A]
 
   The Given Expressions are
1st Expression = \( x^2 - 25 \)
or1st Expression = \( x^2 - 5^2 \)
or1st Expression = \( (x + 5) \cdot \boxed{(x - 5)} \)
Next
2nd Expression = \( x^2 - 9x + 20 \)
or2nd Expression = \( x^2 - 5x - 4x + 20 \)
or2nd Expression = \( x(x - 5) - 4(x - 5) \)
or2nd Expression = \( (x - 4) \cdot \boxed{(x - 5)} \)
  So, HCF is
HCF = \( x - 5 \)
Next, LCM is
LCM = \( (x + 5) \cdot (x - 5) \cdot (x - 4) \)
orLCM = \( (x^2 - 25)(x - 4) \) 
27. म.स. र ल.स. पत्ता लगाउनुहोस् (Find the HCF and LCM of): \( x^2 + x - 6 \) and \( x^2 - 9 \) [3A]
 
   The Given Expressions are
1st Expression = \( x^2 + x - 6 \)
or1st Expression = \( x^2 + 3x - 2x - 6 \)
or1st Expression = \( x(x + 3) - 2(x + 3) \)
or1st Expression = \( \boxed{(x + 3)} \cdot (x - 2) \)
Next
2nd Expression = \( x^2 - 9 \)
or2nd Expression = \( x^2 - 3^2 \)
or2nd Expression = \( \boxed{(x + 3)} \cdot (x - 3) \)
  So, HCF is
HCF = \( x + 3 \)
Next, LCM is
LCM = \( (x + 3)(x - 2)(x - 3) \)
orLCM = \( (x - 2)(x^2 - 9) \) 
28. म.स. र ल.स. पत्ता लगाउनुहोस् (Find the HCF and LCM of): \( x^2 - z^2 + y^2 + 2xy \) and \( x^2 - y^2 + z^2 - 2xz \) [3A]
 
   The Given Expressions are
1st Expression = \( x^2 - z^2 + y^2 + 2xy \)
or1st Expression = \( (x^2 + 2xy + y^2) - z^2 \)
or1st Expression = \( (x + y)^2 - z^2 \)
or1st Expression = \( \boxed{(x + y + z)} \cdot (x + y - z) \)
Next
2nd Expression = \( x^2 - y^2 + z^2 - 2xz \)
or2nd Expression = \( (x^2 - 2xz + z^2) - y^2 \)
or2nd Expression = \( (x - z)^2 - y^2 \)
or2nd Expression = \( [(x - z) + y] \cdot [(x - z) - y] \)
or2nd Expression = \( \boxed{(x + y - z)} \cdot (x - y - z) \)
  So, HCF is
HCF = \( x + y - z \)
Next, LCM is
LCM = \( (x + y - z) \cdot (x + y + z) \cdot (x - y - z) \) 

आनुपातिक बीजीय भिन्न (Rational Algebraic Fraction)

For Q.No.6 (b) in BLE Examination

1. सरल गर्नुहोस् (Simplify): \( \dfrac{1}{a - 2} - \dfrac{4}{a^2 - 4} \) [2U]
 
   The simplification is
\( \dfrac{1}{a - 2} - \dfrac{4}{a^2 - 4} \)
or \( \dfrac{1}{a - 2} - \dfrac{4}{(a - 2)(a + 2)} \)
or \( \dfrac{1 \cdot (a + 2) - 4}{(a - 2)(a + 2)} \)
or \( \dfrac{a + 2 - 4}{(a - 2)(a + 2)} \)
or \( \dfrac{(a - 2)}{(a - 2)(a + 2)} \)
or \( \dfrac{1}{a + 2} \)
2. सरल गर्नुहोस् (Simplify): \( \dfrac{a^2 + b^2}{a^2 - b^2} - \dfrac{a - b}{a + b} \) [2U]
 
   The simplification is
\( \dfrac{a^2 + b^2}{a^2 - b^2} - \dfrac{a - b}{a + b} \)
or \( \dfrac{a^2 + b^2}{(a - b)(a + b)} - \dfrac{a - b}{a + b} \)
or \( \dfrac{(a^2 + b^2) - (a - b)(a - b)}{(a - b)(a + b)} \)
or \( \dfrac{a^2 + b^2 - (a - b)^2}{a^2 - b^2} \)
or \( \dfrac{a^2 + b^2 - (a^2 - 2ab + b^2)}{a^2 - b^2} \)
or \( \dfrac{a^2 + b^2 - a^2 + 2ab - b^2}{a^2 - b^2} \)
or \( \dfrac{2ab}{a^2 - b^2} \)
3. सरल गर्नुहोस् (Simplify): \( \dfrac{x}{x^2 + 3x + 2} - \dfrac{2}{x^2 - 1} \) [2U]
 
   The simplification is
\( \dfrac{x}{x^2 + 3x + 2} - \dfrac{2}{x^2 - 1} \)
or\( \dfrac{x}{(x + 2)(x + 1)} - \dfrac{2}{(x - 1)(x + 1)} \)
or \( \dfrac{x(x - 1) - 2(x + 2)}{(x + 2)(x + 1)(x - 1)} \)
or \( \dfrac{x^2 - x - 2x - 4}{(x + 2)(x^2 - 1)} \)
or \( \dfrac{x^2 - 3x - 4}{(x + 2)(x^2 - 1)} \)
or \( \dfrac{(x - 4)(x + 1)}{(x + 2)(x + 1)(x - 1)} \)
or \( \dfrac{x - 4}{(x + 2)(x - 1)} \)

4. सरल गर्नुहोस् (Simplify): \( \dfrac{2x}{x - 2} - \dfrac{4}{x - 2} \) [2U]
 
   The simplification is
\( \dfrac{2x}{x - 2} - \dfrac{4}{x - 2} \)
or \( \dfrac{2x - 4}{x - 2} \)
or \( \dfrac{2(x - 2)}{(x - 2)} \)
or \( 2 \)
5. सरल गर्नुहोस् (Simplify): \( \dfrac{3}{a^2 - 4} + \dfrac{1}{(a - 2)^2} \) [2U]
 
   The simplification is
\( \dfrac{3}{a^2 - 4} + \dfrac{1}{(a - 2)^2} \)
or \( \dfrac{3}{(a - 2)(a + 2)} + \dfrac{1}{(a - 2)(a - 2)} \)
or \( \dfrac{3(a - 2) + 1(a + 2)}{(a - 2)^2 (a + 2)} \)
or \( \dfrac{3a - 6 + a + 2}{(a - 2)^2 (a + 2)} \)
or \( \dfrac{4a - 4}{(a - 2)^2 (a + 2)} \)
or \( \dfrac{4(a - 1)}{(a - 2)^2 (a + 2)} \)
6. सरल गर्नुहोस् (Simplify): \( \dfrac{3a^2 - 6a}{a^2 - 1} + \dfrac{3}{a^2 - 1} \) [2U]
 
   The simplification is
\( \dfrac{3a^2 - 6a}{a^2 - 1} + \dfrac{3}{a^2 - 1} \)
or \( \dfrac{3a^2 - 6a + 3}{a^2 - 1} \)
or \( \dfrac{3(a^2 - 2a + 1)}{a^2 - 1} \)
or \( \dfrac{3(a - 1)^2}{(a - 1)(a + 1)} \)
or \( \dfrac{3 \cdot (a - 1) \cdot (a - 1)}{(a - 1) \cdot (a + 1)} \)
or \( \dfrac{3(a - 1)}{a + 1} \)
7. सरल गर्नुहोस् (Simplify): \( \dfrac{2x^2 + x}{2x + 1} - \dfrac{2xy + y}{2x + 1} \) [2U]
 
   The simplification is
\( \dfrac{2x^2 + x}{2x + 1} - \dfrac{2xy + y}{2x + 1} \)
or \( \dfrac{(2x^2 + x) - (2xy + y)}{2x + 1} \)
or \( \dfrac{x(2x + 1) - y(2x + 1)}{2x + 1} \)
or \( \dfrac{(2x + 1)(x - y)}{(2x + 1)} \)
or \( x - y \)
8. सरल गर्नुहोस् (Simplify): \( \dfrac{x^2 - 2xy + y^2}{x^2 - y^2} \times \dfrac{x + y}{x - y} \) [2U]
 
   The simplification is
\( \dfrac{x^2 - 2xy + y^2}{x^2 - y^2} \times \dfrac{x + y}{x - y} \)
or \( \dfrac{(x - y)^2}{(x - y)(x + y)} \times \dfrac{x + y}{x - y} \)
or \( 1 \)
9. सरल गर्नुहोस् (Simplify): \( \dfrac{a^2 - 4a}{a^2 - 4} - \dfrac{4}{4 - a^2} \) [2U]
 
   The simplification is
\( \dfrac{a^2 - 4a}{a^2 - 4} - \dfrac{4}{4 - a^2} \)
or \( \dfrac{a^2 - 4a}{a^2 - 4} - \dfrac{4}{-(a^2 - 4)} \)
or \( \dfrac{a^2 - 4a}{a^2 - 4} + \dfrac{4}{a^2 - 4} \)
or \( \dfrac{a^2 - 4a + 4}{a^2 - 4} \)
or \( \dfrac{(a - 2)^2}{(a - 2)(a + 2)} \)
or \( \dfrac{a - 2}{a + 2} \)
10. सरल गर्नुहोस् (Simplify): \( \dfrac{x^2 + y^2}{x - y} + \dfrac{2xy}{y - x} \) [2U]
 
   The simplification is
\( \dfrac{x^2 + y^2}{x - y} + \dfrac{2xy}{y - x} \)
or \( \dfrac{x^2 + y^2}{x - y} + \dfrac{2xy}{-(x - y)} \)
or \( \dfrac{x^2 + y^2}{x - y} - \dfrac{2xy}{x - y} \)
or \( \dfrac{x^2 + y^2 - 2xy}{x - y} \)
Factorizing the numerator:
or \( \dfrac{(x - y)^2}{x - y} \)
or \( x - y \)
11. सरल गर्नुहोस् (Simplify): \( \dfrac{p^2}{p - q} + \dfrac{q^2}{q - p} \) [2U]
 
   The simplification is
\( \dfrac{p^2}{p - q} + \dfrac{q^2}{q - p} \)
or \( \dfrac{p^2}{p - q} + \dfrac{q^2}{-(p - q)} \)
or \( \dfrac{p^2}{p - q} - \dfrac{q^2}{p - q} \)
The denominators are the same, so we subtract the numerators.
or \( \dfrac{p^2 - q^2}{p - q} \)
or \( \dfrac{(p + q) \cdot (p - q)}{(p - q)} \)
or \( p + q \)
12. सरल गर्नुहोस् (Simplify): \( \dfrac{x - 15}{x^2 - 9} + \dfrac{18}{x^2 - 9} \) [2U]
 
   The simplification is
\( \dfrac{x - 15}{x^2 - 9} + \dfrac{18}{x^2 - 9} \)
or \( \dfrac{x - 15 + 18}{x^2 - 9} \)
or \( \dfrac{x + 3}{x^2 - 3^2} \)
or \( \dfrac{x + 3}{(x - 3) \cdot (x + 3)} \)
or \( \dfrac{1}{x - 3} \)
13. सरल गर्नुहोस् (Simplify): \( \dfrac{1}{x - 1} + \dfrac{2x}{1 - x^2} \) [2U]
 
   The simplification is
\( \dfrac{1}{x - 1} + \dfrac{2x}{1 - x^2} \)
or \( \dfrac{1}{x - 1} + \dfrac{2x}{-(x^2 - 1)} \)
or \( \dfrac{1}{x - 1} - \dfrac{2x}{(x - 1)(x + 1)} \)
or \( \dfrac{1 \cdot (x + 1) - 2x}{(x - 1)(x + 1)} \)
or \( \dfrac{x + 1 - 2x}{x^2 - 1} \)
or \( \dfrac{1 - x}{x^2 - 1} \)
or \( \dfrac{-(x - 1)}{(x - 1)(x + 1)} \)
or \( \dfrac{-1}{x + 1} \)
14. सरल गर्नुहोस् (Simplify): \( \dfrac{2}{a + 1} + \dfrac{2a}{a - 1} - \dfrac{a^2 + 3}{a^2 - 1} \) [2U]
 
   The simplification is
\( \dfrac{2}{a + 1} + \dfrac{2a}{a - 1} - \dfrac{a^2 + 3}{a^2 - 1} \)
or \( \dfrac{2}{a + 1} + \dfrac{2a}{a - 1} - \dfrac{a^2 + 3}{(a - 1)(a + 1)} \)
or \( \dfrac{2(a - 1) + 2a(a + 1) - (a^2 + 3)}{(a - 1)(a + 1)} \)
or \( \dfrac{2a - 2 + 2a^2 + 2a - a^2 - 3}{a^2 - 1} \)
or \( \dfrac{(2a^2 - a^2) + (2a + 2a) + (- 2 - 3)}{a^2 - 1} \)
or \( \dfrac{a^2 + 4a - 5}{a^2 - 1} \)
or \( \dfrac{(a + 5) \cdot (a - 1)}{(a + 1) \cdot (a - 1)} \)
or \( \dfrac{a + 5}{a + 1} \)
15. सरल गर्नुहोस् (Simplify): \( \dfrac{1}{x + y} + \dfrac{1}{x - y} + \dfrac{2y}{x^2 - y^2} \) [2U]
 
   The simplification is
\( \dfrac{1}{x + y} + \dfrac{1}{x - y} + \dfrac{2y}{x^2 - y^2} \)
or \( \dfrac{1}{x + y} + \dfrac{1}{x - y} + \dfrac{2y}{(x + y)(x - y)} \)
or \( \dfrac{1 \cdot (x - y) + 1 \cdot (x + y) + 2y}{(x + y)(x - y)} \)
or \( \dfrac{x - y + x + y + 2y}{x^2 - y^2} \)
or \( \dfrac{(x + x) + (-y + y + 2y)}{x^2 - y^2} \)
or \( \dfrac{2x + 2y}{x^2 - y^2} \)
or \( \dfrac{2(x + y)}{(x - y)(x + y)} \)
or \( \dfrac{2}{x - y} \)
16. सरल गर्नुहोस् (Simplify):
    \( \dfrac{x + y}{(y - z)(z - x)} - \dfrac{y + z}{(x - z)(x - y)} + \dfrac{z + x}{(z - y)(y - x)} \) [3U]
 
   The simplification is
\( \dfrac{x + y}{(y - z)(z - x)} + \dfrac{y + z}{(z - x)(x - y)} + \dfrac{z + x}{(y - z)(x - y)} \)
or \( \dfrac{(x + y)(x - y) + (y + z)(y - z) + (z + x)(z - x)}{(x - y)(y - z)(z - x)} \)
\( \dfrac{(x^2 - y^2) + (y^2 - z^2) + (z^2 - x^2)}{(x - y)(y - z)(z - x)} \)
\( \dfrac{\cancel{x^2} - \cancel{y^2} + \cancel{y^2} - \cancel{z^2} + \cancel{z^2} - \cancel{x^2}}{(x - y)(y - z)(z - x)} \)
or \( \dfrac{0}{(x - y)(y - z)(z - x)} \)
or \( 0 \)
17. सरल गर्नुहोस् (Simplify): \( \dfrac{2}{a + 1} - \dfrac{2}{a - 1} + \dfrac{4}{a^2 + 1} \) [2U]
 
   The simplification is
\( \dfrac{2}{a + 1} - \dfrac{2}{a - 1} + \dfrac{4}{a^2 + 1} \)
or \( \dfrac{2(a - 1) - 2(a + 1)}{(a + 1)(a - 1)} + \dfrac{4}{a^2 + 1} \)
or \( \dfrac{2a - 2 - 2a - 2}{a^2 - 1} + \dfrac{4}{a^2 + 1} \)
or \( \dfrac{- 4}{a^2 - 1} + \dfrac{4}{a^2 + 1} \)
or \( \dfrac{- 4(a^2 + 1) + 4(a^2 - 1)}{(a^2 - 1)(a^2 + 1)} \)
or \( \dfrac{- 4a^2 - 4 + 4a^2 - 4}{(a^2)^2 - 1^2} \)
or \( \dfrac{\cancel{- 4a^2} - 4 + \cancel{4a^2} - 4}{a^4 - 1} \)
or \( \dfrac{- 8}{a^4 - 1} \)
or \( - \dfrac{8}{a^4 - 1} \)
18. सरल गर्नुहोस् (Simplify):
    \( \dfrac{1}{(x - 3)(x - 4)} + \dfrac{1}{(x - 4)(x - 5)} + \dfrac{1}{(x - 5)(x - 3)} \) [3U]
 
   The simplification is
\( \dfrac{1}{(x - 3)(x - 4)} + \dfrac{1}{(x - 4)(x - 5)} + \dfrac{1}{(x - 5)(x - 3)} \)
or \( \dfrac{1 \cdot (x - 5) + 1 \cdot (x - 3) + 1 \cdot (x - 4)}{(x - 3)(x - 4)(x - 5)} \)
or \( \dfrac{x - 5 + x - 3 + x - 4}{(x - 3)(x - 4)(x - 5)} \)
or \( \dfrac{(x + x + x) + (- 5 - 3 - 4)}{(x - 3)(x - 4)(x - 5)} \)
or \( \dfrac{3x - 12}{(x - 3)(x - 4)(x - 5)} \)
or \( \dfrac{3 \cdot \boxed{(x - 4)}}{(x - 3) \cdot \boxed{(x - 4)} \cdot (x - 5)} \)
or \( \dfrac{3}{(x - 3)(x - 5)} \)
19. सरल गर्नुहोस् (Simplify):
    \( \dfrac{a}{(a - b)(a - c)} + \dfrac{b}{(b - a)(b - c)} + \dfrac{c}{(c - a)(c - b)} \) [3U]
 
   The simplification is
\( \dfrac{a}{(a - b)(a - c)} + \dfrac{b}{-(a - b)(b - c)} + \dfrac{c}{-(a - c) \cdot [-(b - c)]} \)
or\( \dfrac{a}{(a - b)(a - c)} - \dfrac{b}{(a - b)(b - c)} + \dfrac{c}{(a - c)(b - c)} \)
or\( \dfrac{a \cdot (b - c) - b \cdot (a - c) + c \cdot (a - b)}{(a - b)(b - c)(a - c)} \)
or\( \dfrac{ab - ac - ab + bc + ac - bc}{(a - b)(b - c)(a - c)} \)
or\( \dfrac{\cancel{ab} - \cancel{ac} - \cancel{ab} + \cancel{bc} + \cancel{ac} - \cancel{bc}}{(a - b)(b - c)(a - c)} \)
or \( \dfrac{0}{(a - b)(b - c)(a - c)} \)
or \( 0 \)
20. सरल गर्नुहोस् (Simplify): \( \dfrac{1}{1 + x} - \dfrac{1}{x - 1} + \dfrac{2}{1 + x^2} \) [2U]
 
   The simplification is
\( \dfrac{1}{1 + x} - \dfrac{1}{x - 1} + \dfrac{2}{1 + x^2} \)
or \( \dfrac{1}{1 + x} + \dfrac{1}{1 - x} + \dfrac{2}{1 + x^2} \)
or \( \dfrac{1 \cdot (1 - x) + 1 \cdot (1 + x)}{(1 + x)(1 - x)} + \dfrac{2}{1 + x^2} \)
or \( \dfrac{1 - \cancel{x} + 1 + \cancel{x}}{1 - x^2} + \dfrac{2}{1 + x^2} \)
or \( \dfrac{2}{1 - x^2} + \dfrac{2}{1 + x^2} \)
or \( \dfrac{2(1 + x^2) + 2(1 - x^2)}{(1 - x^2)(1 + x^2)} \)
or \( \dfrac{2 + \cancel{2x^2} + 2 - \cancel{2x^2}}{1 - x^4} \)
or \( \dfrac{4}{1 - x^4} \)
21. सरल गर्नुहोस् (Simplify):
    \( \dfrac{x}{x - 2y} + \dfrac{x}{x + 2y} + \dfrac{2x^2}{x^2 + 4y^2} \) [2U]
 
   The simplification is
\( \dfrac{x}{x - 2y} + \dfrac{x}{x + 2y} + \dfrac{2x^2}{x^2 + 4y^2} \)
or \( \dfrac{x(x + 2y) + x(x - 2y)}{(x - 2y)(x + 2y)} + \dfrac{2x^2}{x^2 + 4y^2} \)
or \( \dfrac{x^2 + \cancel{2xy} + x^2 - \cancel{2xy}}{x^2 - 4y^2} + \dfrac{2x^2}{x^2 + 4y^2} \)
or \( \dfrac{2x^2}{x^2 - 4y^2} + \dfrac{2x^2}{x^2 + 4y^2} \)
or \( \dfrac{2x^2(x^2 + 4y^2) + 2x^2(x^2 - 4y^2)}{(x^2 - 4y^2)(x^2 + 4y^2)} \)
or \( \dfrac{2x^4 + \cancel{8x^2y^2} + 2x^4 - \cancel{8x^2y^2}}{(x^2)^2 - (4y^2)^2} \)
or \( \dfrac{4x^4}{x^4 - 16y^4} \)
22. सरल गर्नुहोस् (Simplify): \( \dfrac{x}{x + y} + \dfrac{y}{x - y} - \dfrac{2xy}{x^2 - y^2} \) [2U]
 
   The simplification is
\( \dfrac{x}{x + y} + \dfrac{y}{x - y} - \dfrac{2xy}{x^2 - y^2} \)
or\( \dfrac{x}{x + y} + \dfrac{y}{x - y} - \dfrac{2xy}{(x + y)(x - y)} \)
or \( \dfrac{x(x - y) + y(x + y) - 2xy}{(x + y)(x - y)} \)
or \( \dfrac{x^2 - xy + xy + y^2 - 2xy}{x^2 - y^2} \)
or \( \dfrac{x^2 + y^2 + \cancel{xy} - \cancel{xy} - 2xy}{x^2 - y^2} \)
or \( \dfrac{x^2 - 2xy + y^2}{x^2 - y^2} \)
or \( \dfrac{(x - y)^2}{(x + y)(x - y)} \)
or \( \dfrac{x - y}{x + y} \)
23. सरल गर्नुहोस् (Simplify): \( \dfrac{1}{a^2 - 3a + 2} + \dfrac{1}{a^2 - 5a + 6} \) [2U]
 
   The simplification is
\( \dfrac{1}{a^2 - 3a + 2} + \dfrac{1}{a^2 - 5a + 6} \)
or \( \dfrac{1}{(a - 1)(a - 2)} + \dfrac{1}{(a - 2)(a - 3)} \)
or \( \dfrac{1 \cdot (a - 3) + 1 \cdot (a - 1)}{(a - 1)(a - 2)(a - 3)} \)
or \( \dfrac{a - 3 + a - 1}{(a - 1)(a - 2)(a - 3)} \)
or \( \dfrac{2a - 4}{(a - 1)(a - 2)(a - 3)} \)
or \( \dfrac{2 \cdot (a - 2)}{(a - 1) \cdot (a - 2) \cdot (a - 3)} \)
or \( \dfrac{2}{(a - 1)(a - 3)} \)
24. सरल गर्नुहोस् (Simplify):
    \( \dfrac{2}{x^2 + 3x + 2} + \dfrac{5x}{x^2 - x - 6} - \dfrac{x + 2}{x^2 - 2x - 3} \) [3U]
 
   The simplification is
\( \dfrac{2}{x^2 + 3x + 2} + \dfrac{5x}{x^2 - x - 6} - \dfrac{x + 2}{x^2 - 2x - 3} \)
or\( \dfrac{2}{(x + 2)(x + 1)} + \dfrac{5x}{(x - 3)(x + 2)} - \dfrac{x + 2}{(x - 3)(x + 1)} \)
or \( \dfrac{2(x - 3) + 5x(x + 1) - (x + 2)(x + 2)}{(x + 2)(x + 1)(x - 3)} \)
or \( \dfrac{2x - 6 + 5x^2 + 5x - (x^2 + 4x + 4)}{(x + 2)(x + 1)(x - 3)} \)
or \( \dfrac{5x^2 + 7x - 6 - x^2 - 4x - 4}{(x + 2)(x + 1)(x - 3)} \)
or \( \dfrac{(5x^2 - x^2) + (7x - 4x) + (- 6 - 4)}{(x + 2)(x + 1)(x - 3)} \)
or \( \dfrac{4x^2 + 3x - 10}{(x + 2)(x + 1)(x - 3)} \)
or \( \dfrac{(4x - 5) \cdot (x + 2)}{(x + 2) \cdot (x + 1) \cdot (x - 3)} \)
or \( \dfrac{4x - 5}{(x + 1)(x - 3)} \)
25. सरल गर्नुहोस् (Simplify):
    \( \dfrac{2(a + 4)}{a + 3} \cdot \dfrac{a^2 + 2a - 24}{a^2 - a - 12} \) [2U]
 
   The simplification is
\( \dfrac{2(a + 4)}{a + 3} \cdot \dfrac{a^2 + 2a - 24}{a^2 - a - 12} \)
or \( \dfrac{2(a + 4)}{a + 3} \cdot \dfrac{(a + 6) \cdot (a - 4)}{(a - 4) \cdot (a + 3)} \)
or \( \dfrac{2(a + 4)(a + 6)}{(a + 3)^2} \)
26. सरल गर्नुहोस् (Simplify):
    \( \dfrac{1}{x + 2} + \dfrac{2}{2(x - 2)} - \dfrac{4}{x^2 - 4} \) [2U]
 
   The simplification is
\( \dfrac{1}{x + 2} + \dfrac{2}{2(x - 2)} - \dfrac{4}{x^2 - 4} \)
or\( \dfrac{1}{x + 2} + \dfrac{2}{2(x - 2)} - \dfrac{4}{(x + 2)(x - 2)} \)
or\( \dfrac{1 \cdot (x - 2) + 1 \cdot (x + 2) - 4}{(x + 2)(x - 2)} \)
or \( \dfrac{x - \cancel{2} + x + \cancel{2} - 4}{x^2 - 4} \)
or \( \dfrac{2x - 4}{x^2 - 4} \)
or \( \dfrac{2(x - 2)}{(x + 2)(x - 2)} \)
or \( \dfrac{2 \cdot (x - 2)}{(x + 2) \cdot (x - 2)} \)
or \( \dfrac{2}{x + 2} \)
27. सरल गर्नुहोस् (Simplify):
    \( \dfrac{1}{a - 3} - \dfrac{1}{2(a + 3)} + \dfrac{a}{a^2 - 9} \) [2U]
 
   The simplification is
\( \dfrac{1}{a - 3} - \dfrac{1}{2(a + 3)} + \dfrac{a}{a^2 - 9} \)
or\( \dfrac{1}{a - 3} - \dfrac{1}{2(a + 3)} + \dfrac{a}{(a - 3)(a + 3)} \)
or \( \dfrac{1 \cdot 2(a + 3) - 1 \cdot (a - 3) + a \cdot 2}{2(a - 3)(a + 3)} \)
or \( \dfrac{2a + 6 - a + 3 + 2a}{2(a^2 - 9)} \)
or \( \dfrac{(2a - a + 2a) + (6 + 3)}{2(a^2 - 9)} \)
or \( \dfrac{3a + 9}{2(a^2 - 9)} \)
or \( \dfrac{3(a + 3)}{2(a - 3)(a + 3)} \)
or \( \dfrac{3 \cdot (a + 3)}{2(a - 3) \cdot (a + 3)} \)
or \( \dfrac{3}{2(a - 3)} \)
28. सरल गर्नुहोस् (Simplify): \( \dfrac{2x + 6}{x^2 - 9} - \dfrac{4x}{2x^2 - 6x} \) [2U]
 
   The simplification is
\( \dfrac{2x + 6}{x^2 - 9} - \dfrac{4x}{2x^2 - 6x} \)
or \( \dfrac{2(x + 3)}{(x+3)(x-3)} - \dfrac{4x}{2x(x - 3)} \)
or \( \dfrac{2}{(x-3)} - \dfrac{2}{(x - 3)} \)
or \( 0 \)

Unit 10: समीकरण र लेखाचित्र (Equation and Graph)

For Q.No.7 (a) in BLE Examination

1. कस्ता समीकरणहरूलाई युगपतरैखीय समीकरण भनिन्छ ?
   What type of equations are called simultaneous equations? [1K]
 
Two or more linear equations involving the same variables and that are solved together to find a common solution, is called Simultaneous linear equations
2. कस्ता समीकरणहरूलाई वर्ग समीकरण भनिन्छ ?
   What type of equations are called quadratic equations? [1K]
 
An equation in which the highest power (degree) of the variable is two is called a quadratic equation. The standard form is \( ax^2 + bx + c = 0 \), where \( a \neq 0 \).
3. हल गर्नुहोस् (Solve): \( 2 - x = 17 - 4x \) [1A]
 
The solution is
\( 2 - x = 17 - 4x \)
or \( -x + 4x = 17 - 2 \)
or \( 3x = 15 \)
or \( x = \dfrac{15}{3} \)
or \( x = 5 \)
4. हल गर्नुहोस् (Solve): \( \dfrac{7x + 3}{4} = 6 \) [1A]
 
The solution is
\( \dfrac{7x + 3}{4} = 6 \)
or \( 7x + 3 = 6 \cdot 4 \)
or \( 7x + 3 = 24 \)
or \( 7x = 24 - 3 \)
or \( 7x = 21 \)
or \( x = \dfrac{21}{7} \)
or \( x = 3 \)
5. x को मान कति हुँदा समीकरण \( 8x + 1 = 57 \) सत्य हुन्छ ?
   For what value of x the equation \( 8x + 1 = 57 \) becomes true? [1A]
 
The solution is
\( 8x + 1 = 57 \)
or \( 8x = 57 - 1 \)
or \( 8x = 56 \)
or \( x = \dfrac{56}{8} \)
or \( x = 7 \) For what value of \(x=7\) the equation \( 8x + 1 = 57 \) becomes true.
6. यदि \( 5x + 8 = 13 \), x को मान पत्ता लगाउनुहोस्।
   If \( 5x + 8 = 13 \), find the value of x. [1A]
 
The solution is
\( 5x + 8 = 13 \)
or \( 5x = 13 - 8 \)
or \( 5x = 5 \)
or \( x = \dfrac{5}{5} \)
or \( x = 1 \)
7. वर्ग समीकरण \( x^2 = 16 \) मा x का मानहरू के के हुन् ?
   What are the values of x in the quadratic equation \( x^2 = 16 \)? [1A]
 
The solution is
\( x^2 = 16 \)
or \( x = \pm \sqrt{16} \)
or \( x = \pm 4 \)

8. x को मान पत्ता लगाउनुहोस् (Find the value of x): \( \dfrac{x}{3} = \dfrac{3}{x} \) [1A]
 
The solution is
\( \dfrac{x}{3} = \dfrac{3}{x} \)
or \( x \cdot x = 3 \cdot 3 \)
or \( x^2 = 9 \)
or \( x = \pm \sqrt{9} \)
or \( x = \pm 3 \)

9. हल गर्नुहोस् (Solve): \( x^2 - 4 = 21 \) [1A]
 
The solution is
\( x^2 - 4 = 21 \)
or \( x^2 = 21 + 4 \)
or \( x^2 = 25 \)
or \( x = \pm \sqrt{25} \)
or \( x = \pm 5 \)

10. दिइएको समीकरण हल गर्नुहोस् (Solve the given equation): \( x^2 + 2x = 0 \) [1A]
 
The solution is
\( x^2 + 2x = 0 \)
or \( x(x + 2) = 0 \)
Either \( x = 0 \) or \( x + 2 = 0 \).
Therefore, the values of \( x \) are
\( x=0 \) and \( x=-2 \)
11. हल गर्नुहोस् (Solve): \( \dfrac{1}{x} = \dfrac{x}{16} \) [1A]
 
The solution is
\( \dfrac{1}{x} = \dfrac{x}{16} \)
or \( x \cdot x = 1 \cdot 16 \)
or \( x^2 = 16 \)
or \( x = \pm \sqrt{16} \)
or \( x = \pm 4 \)
   
12. हल गर्नुहोस् (Solve): \( 3x^2 - 4x = 0 \) [1A]
 
The solution is
\( 3x^2 - 4x = 0 \)
or \( x(3x - 4) = 0 \)
Either \( x = 0 \) or \( 3x - 4 = 0 \).
Therefore, the values of \( x \) are
\(x= 0 \) and \( x=\dfrac{4}{3} \).
   
13. x को मान २ र ३ हुने वर्ग समीकरण पत्ता लगाउनुहोस्।
    Find the quadratic equation in which values of x are 2 and 3. [1U]
 
The quadratic equation is
\( x^2 - (sum)x + (product) = 0 \)
or \( x^2 - (5)x + 6 = 0 \)
or \( x^2 - 5x + 6 = 0 \)
   
14. मूलहरू १ र २ हुने वर्ग समीकरण लेख्नुहोस्।
    Write the quadratic equations whose roots are 1 and 2. [1A]
 
The quadratic equation is
\( x^2 - (sum)x + (product) = 0 \)
or \( x^2 - (3)x + 2 = 0 \)
or \( x^2 - 3x + 2 = 0 \)
   
15. एउटा वर्ग समीकरणका जम्मा कतिओटा मूलहरू हुन्छन् ?
    How many roots are there in a quadratic equation? [1A]
 
There are two roots in a quadratic equation.

For Q.No.7(b) in BLE Examination

1. हल गर्नुहोस् (Solve): \( \dfrac{x + 1}{2} = \dfrac{7x - 3}{3x} \) [2A]
 
The solution is
\( \dfrac{x + 1}{2} = \dfrac{7x - 3}{3x} \)
or \( 3x(x + 1) = 2(7x - 3) \)
or \( 3x^2 + 3x = 14x - 6 \)
or \( 3x^2 + 3x - 14x + 6 = 0 \)
or \( 3x^2 - 11x + 6 = 0 \)
or \( 3x^2 - (9x + 2x) + 6 = 0 \)
or \( 3x(x - 3) - 2(x - 3) = 0 \)
or \( (x - 3)(3x - 2) = 0 \)
Therefore, the values of \( x \) are
or\(x= 3 \) and \(x= \dfrac{2}{3} \).
2. हल गर्नुहोस् (Solve): \( 8x^2 - 32 = 0 \) [2A]
 
The solution is
\( 8x^2 - 32 = 0 \)
or \( 8x^2 = 32 \)
or \( x^2 = \dfrac{32}{8} \)
or \( x^2 = 4 \)
or \( x = \pm \sqrt{4} \)
Therefore, the values of \( x \) are
or \( x = 2 \) and \( x = -2 \).
   
3. हल गर्नुहोस् (Solve): \( 3x - 9x^2 = 0 \) [2A]
 
The solution is
\( 3x - 9x^2 = 0 \)
or \( 3x(1 - 3x) = 0 \)
Therefore, the values of \( x \) are
or \( x = 0 \) and \( x = \dfrac{1}{3} \).
   
4. हल गर्नुहोस् (Solve): \( \dfrac{x^2}{4} - 25 = 0 \) [2A]
 
The solution is
\( \dfrac{x^2}{4} - 25 = 0 \)
or \( \dfrac{x^2}{4} = 25 \)
or \( x^2 = 25 \times 4 \)
or \( x^2 = 100 \)
or \( x = \pm \sqrt{100} \)
Therefore, the values of \( x \) are
or \( x = 10 \) and \( x = -10 \).
   
5. हल गर्नुहोस् (Solve): \( x^2 - x - 2 = 0 \) [2A]
 
The solution is
\( x^2 - x - 2 = 0 \)
or \( x^2 - (2x - x) - 2 = 0 \)
or \( x^2 - 2x + x - 2 = 0 \)
or \( x(x - 2) + 1(x - 2) = 0 \)
or \( (x - 2)(x + 1) = 0 \)
Therefore, the values of \( x \) are
or \( x = 2 \) and \( x = -1 \).
   
6. हल गर्नुहोस् (Solve): \( x^2 - x - 6 = 0 \) [2A]
 
The solution is
\( x^2 - x - 6 = 0 \)
or \( x^2 - (3x - 2x) - 6 = 0 \)
or \( x^2 - 3x + 2x - 6 = 0 \)
or \( x(x - 3) + 2(x - 3) = 0 \)
or \( (x - 3)(x + 2) = 0 \)
Therefore, the values of \( x \) are
or \( x = 3 \) and \( x = -2 \).
7. समीकरण \( x + y = 4 \) लाई लेखाचित्रमा देखाउनुहोस्।
   Show the equation \( x + y = 4 \) in a graph. [2A]
 
Given lines are
\( x + y = 4 \)
The table value of the line is
\( x \)	0	4	2
\( y \)	4	0	2
The line is shown in the graph.
8. दिइएको समीकरणलाई लेखाचित्रमा देखाउनुहोस्।
   (Show the given equation in a graph): \( x = y + 4 \) [2A]
 
  
    
  Given line is
  \( x = y + 4 \) वा \( y = x - 4 \)
  The table value of the line is                                        
\( x \)	0	4	2
\( y \)	-4	0	-2
The line is shown in the graph.   
  
   
 
 
9. दिइएको समीकरणलाई लेखाचित्रमा देखाउनुहोस्।
   (Show the given equation in a graph): \( y = 4x + 8 \) [2A]
 
  
    
  Given line is
  \( y = 4x + 8 \)
  The table value of the line is                                        
\( x \)	0	-2	-1
\( y \)	8	0	4
The line is shown in the graph.   
  
 
 
 
10. दिइएको समीकरणलाई लेखाचित्रमा देखाउनुहोस्।
    (Show the given equation in a graph): \( 3x + 4y = 12 \) [2A]
 
  
    
  Given line is
\( 3x + 4y = 12 \)
\( y = \dfrac{12 - 3x}{4} \)
  The table value of the line is                                        
\( x \)	0	4	-4
\( y \)	3	0	6
The line is shown in the graph.   
 
 
 
11. हल गर्नुहोस् (Solve): \( x(x + 1) = 4 + x \) [2A]
 
The solution is
\( x(x + 1) = 4 + x \)
or \( x^2 + x = 4 + x \)
or \( x^2 + x - x - 4 = 0 \)
or \( x^2 - 4 = 0 \)
or \( x^2 = 4 \)
or \( x = \pm \sqrt{4} \)
Therefore, the values of \( x \) are
or \( x = 2 \) and \( x = -2 \).
   
12. हल गर्नुहोस् (Solve): \( (x - 7)^2 - 64 = 0 \) [2HA]
 
The solution is
\( (x - 7)^2 - 64 = 0 \)
or \( (x - 7)^2 = 64 \)
or \( x - 7 = \pm \sqrt{64} \)
or \( x - 7 = \pm 8 \)
Taking the positive sign (+)
\( x - 7 = 8 \Rightarrow x = 15 \)
Taking the negative sign (-)
\( x - 7 = -8 \Rightarrow x = -1 \)
Therefore, the values of \( x \) are
or \( x = 15 \) and \( x = -1 \).
   
13. हल गर्नुहोस् (Solve): \( 7x^2 + 13x - 2 = 0 \) [2HA]
 
The solution is
\( 7x^2 + 13x - 2 = 0 \)
or \( 7x^2 + (14x - x) - 2 = 0 \)
or \( 7x^2 + 14x - x - 2 = 0 \)
or \( 7x(x + 2) - 1(x + 2) = 0 \)
or \( (x + 2)(7x - 1) = 0 \)
Either \( x + 2 = 0 \) or \( 7x - 1 = 0 \)
Therefore, the values of \( x \) are
or \( x = -2 \) and \( x = \dfrac{1}{7} \).
   
14. x को मान कति हुँदा, \( x^2 - 5x + 6 \) को मान शून्य हुन्छ ?
    For what value of x, the value of \( x^2 - 5x + 6 \) is zero? [2HA]
 
The solution is
\( x^2 - 5x + 6 = 0 \)
or \( x^2 - (3x + 2x) + 6 = 0 \)
or \( x^2 - 3x - 2x + 6 = 0 \)
or \( x(x - 3) - 2(x - 3) = 0 \)
or \( (x - 3)(x - 2) = 0 \)
Either \( x - 3 = 0 \) or \( x - 2 = 0 \)
Therefore, the values of \( x \) are
or \( x = 3 \) and \( x = 2 \).
   
15. x को मान कति हुँदा, \( 2x^2 - x - 6 \) को मान शून्य हुन्छ ?
    For what value of x, the value of \( 2x^2 - x - 6 \) is zero? [2HA]
 
The solution is
\( 2x^2 - x - 6 = 0 \)
or \( 2x^2 - (4x - 3x) - 6 = 0 \)
or \( 2x^2 - 4x + 3x - 6 = 0 \)
or \( 2x(x - 2) + 3(x - 2) = 0 \)
or \( (x - 2)(2x + 3) = 0 \)
Either \( x - 2 = 0 \) or \( 2x + 3 = 0 \)
Therefore, the values of \( x \) are
or \( x = 2 \) and \( x = -\dfrac{3}{2} \).

For Q.No.8 (b) in BLE Examination

1. हल गर्नुहोस् (Solve): \( x^2 + \dfrac{1}{16} = \dfrac{x}{2} \) [2A]
 
The solution is
\( x^2 + \dfrac{1}{16} = \dfrac{x}{2} \)
or \( 16x^2 + 16 \cdot \dfrac{1}{16} = 16 \cdot \dfrac{x}{2} \)
or \( 16x^2 + 1 = 8x \)
or \( 16x^2 - 8x + 1 = 0 \)
or\( (4x)^2 - 2(4x)(1) + (1)^2 = 0 \)
or \( (4x - 1)^2 = 0 \)
or \( 4x - 1 = 0 \)
or \( 4x = 1 \)
or \( x = \dfrac{1}{4} \)
2. हल गर्नुहोस् (Solve): \( \dfrac{3 - 4x}{5} - \dfrac{4 + 5x}{9} + \dfrac{7x + 11}{15} = 0 \) [2A]
 
The solution is
\( \dfrac{3 - 4x}{5} - \dfrac{4 + 5x}{9} + \dfrac{7x + 11}{15} = 0 \)
or \( 9(3 - 4x) - 5(4 + 5x) + 3(7x + 11) = 0 \)
or \( 27 - 36x - 20 - 25x + 21x + 33 = 0 \)
or \( (-36x - 25x + 21x) + (27 - 20 + 33) = 0 \)
or \( (-61x + 21x) + (40) = 0 \)
or \( -40x + 40 = 0 \)
or \( 40 = 40x \)
or \( x = 1 \).
   
3. हल गर्नुहोस् (Solve): \( (x + 1)(x + 2) = x(x + 7) - 6 \) [2A]
 
The solution is
\( (x + 1)(x + 2) = x(x + 7) - 6 \)
or \( x^2 + 2x + x + 2 = x^2 + 7x - 6 \)
or \( x^2 + 3x + 2 = x^2 + 7x - 6 \)
or \( 3x + 2 = 7x - 6 \)
or \( 2 + 6 = 7x - 3x \)
or \( 8 = 4x \)
or \( x = 2 \)
4. हल गर्नुहोस् (Solve): \( \dfrac{6x - 3}{2x + 7} = \dfrac{3x - 2}{x + 5} \) [2A]
 
The solution is
\( \dfrac{6x - 3}{2x + 7} = \dfrac{3x - 2}{x + 5} \)
or \( (6x - 3)(x + 5) = (3x - 2)(2x + 7) \)
or \( 6x^2 + 30x - 3x - 15 = 6x^2 + 21x - 4x - 14 \)
or \( 6x^2 + 27x - 15 = 6x^2 + 17x - 14 \)
or \( 27x - 17x = -14 + 15 \)
or \( 10x = 1 \)
or \( x = \dfrac{1}{10} \).
      
5. हल गर्नुहोस् (Solve): \( 2x + y = 6 \), \( 3x + y = 7 \) [2A]
 
The solution is
Given equations are
\( 2x + y = 6 \) (i)
\( 3x + y = 7 \) (ii)
Subtracting equation (i) from (ii), we get
\( 3x + y=7 \)
\( _{(-)}2x +_{(-)} y = _{(-)}6 \)

---
\( x = 1 \)
Now, substituting the value of \( x \) in equation (i), we get
\( 2x + y = 6 \)
or \( 2(1) + y = 6 \)
or \( 2 + y = 6 \)
or \( y = 6 - 2 \)
or \( y = 4 \)
Therefore, the values of \( x \) and \( y \) are
\( x = 1 \) and \( y = 4 \)
   
6. हल गर्नुहोस् (Solve): \( \dfrac{x}{2} + \dfrac{2}{x} = 2 \) [2A]
 
The solution is
\( \dfrac{x}{2} + \dfrac{2}{x} = 2 \)
or \( \dfrac{x \cdot x + 2 \cdot 2}{2x} = 2 \)
or \( \dfrac{x^2 + 4}{2x} = 2 \)
or \( x^2 + 4 = 2 \cdot 2x \)
or \( x^2 + 4 = 4x \)
or \( x^2 - 4x + 4 = 0 \)
or \( (x - 2)^2 = 0 \)
or \( x - 2 = 0 \)
Therefore, the value of \( x \) is
or \( x = 2 \).
7. हल गर्नुहोस् (Solve): \( \dfrac{x - 4}{4} + \dfrac{6}{x + 1} = \dfrac{5}{4} \) [2A]
 
The solution is
\( \dfrac{x - 4}{4} + \dfrac{6}{x + 1} = \dfrac{5}{4} \)
Isolating the fraction with x in the denominator:
or \( \dfrac{6}{x + 1} = \dfrac{5}{4} - \dfrac{x - 4}{4} \)
or \( \dfrac{6}{x + 1} = \dfrac{5 - (x - 4)}{4} \)
or \( \dfrac{6}{x + 1} = \dfrac{5 - x + 4}{4} \)
or \( \dfrac{6}{x + 1} = \dfrac{9 - x}{4} \)
or \( 6 \cdot 4 = (9 - x)(x + 1) \)
or \( 24 = 9x + 9 - x^2 - x \)
or \( 24 = -x^2 + 8x + 9 \)
or \( x^2 - 8x + 24 - 9 = 0 \)
or \( x^2 - 8x + 15 = 0 \)
or \( x^2 - (5x + 3x) + 15 = 0 \)
or \( x^2 - 5x - 3x + 15 = 0 \)
or \( x(x - 5) - 3(x - 5) = 0 \)
or \( (x - 5)(x - 3) = 0 \)
Therefore, the values of \( x \) are
or \( x = 5 \) and \( x = 3 \).
8. लेखाचित्रद्वारा हल गर्नुहोस् (Solve graphically):
   \( 2x - y = 5 \) and \( x - y = 1 \) [2A]
 
  
    
  Given lines are:
\( 2x - y = 5 \)
\( y = 2x - 5 \)) (i)
                               
\( x \)	4	2	3
\( y \)	3	-1	1
Next
\( x - y = 1 \)
\( y = x - 1 \) (ii)
                               
\( x \)	0	1	3
\( y \)	-1	0	2
The solution is \( x = 4 \) and \( y = 3 \)    
  
 
 
 
9. लेखाचित्र विधिबाट हल गर्नुहोस् (Solve by graphical method):
   \( x + y = 6 \), \( 2x - y = 9 \) [2A]
Given lines are:
\( x + y = 6 \)
\( y = -x + 6 \) (i)

\( x \)	0	3	6
\( y \)	6	3	0
Next
\( 2x - y = 9 \)
\( y = 2x - 9 \) (ii)

\( x \)	3	4	5
\( y \)	-3	-1	1
The solution is \( x = 5 \) and \( y = 1 \)
10. लेखाचित्र विधिबाट हल गर्नुहोस् (Solve by graphical method):
    \( 5x - 3y = 5 \), \( -3x + 2y = -2 \) [2A]
Given lines are:
\( 5x - 3y = 5 \)
\( y = \frac{5x-5}{3} \) (i)

\( x \)	1	4	-5
\( y \)	0	5	-10
Next
\( -3x + 2y = -2 \)
\( y = \frac{3x-2}{2}\) (ii)

\( x \)	0	2	4
\( y \)	-1	2	5
The solution is \( x = 4 \) and \( y = 5 \)
11. लेखाचित्र विधिबाट हल गर्नुहोस् (Solve by graphical method):
    \( x + y = 5 \), \( x - y = 3 \) [2A]
Given lines are:
\( x + y = 5 \)
\( y = -x + 5 \) (i)

\( x \)	0	2	5
\( y \)	5	3	0
Next
\( x - y = 3 \)
\( y = x - 3 \) (ii)

\( x \)	0	3	4
\( y \)	-3	0	1
The solution is \( x = 4 \) and \( y = 1 \)
12. लेखाचित्र विधिबाट हल गर्नुहोस् (Solve by graphical method):
    \( x + y = 6 \), \( y = x - 4 \) [2A]
Given lines are:
\( x + y = 6 \)
\( y = -x + 6 \) (i)

\( x \)	0	3	6
\( y \)	6	3	0
Next
\( y = x - 4 \)
\( y = x - 4 \) (ii)

\( x \)	0	4	5
\( y \)	-4	0	1
The solution is \( x = 5 \) and \( y = 1 \)
13. लेखाचित्र विधिबाट हल गर्नुहोस् (Solve by graphical method):
    \( 2x - 1 = y \), \( 3x - 2y = 0 \) [2A]
Given lines are:
\( 2x - 1 = y \)
\( y = 2x - 1 \) (i)

\( x \)	0	1	2
\( y \)	-1	1	3
Next
\( 3x - 2y = 0 \)
\( y = \frac{3}{2}x \) (ii)

\( x \)	0	2	-2
\( y \)	0	3	-3
The solution is \( x = 2 \) and \( y = 3 \)
14. लेखाचित्र विधिबाट हल गर्नुहोस् (Solve by using graph):
    \( 5x + 7y = 1 \) and \( x + 4y = -5 \) [2A]
Given lines are:
\( 5x + 7y = 1 \)
\( y = \frac{1-5x}{7} \) (i)

\( x \)	3	-4
\( y \)	-2	3
Next
\( x + 4y = -5 \)
\( x = -5-4y \) (ii)

\( x \)	-5	-1	3
\( y \)	0	-1	-2
The solution is \( x = 3 \) and \( y = -2 \)
15. लेखाचित्र विधिबाट हल गर्नुहोस् (Solve by graphical method):
    \( 2x + y = 2 \), \( x - y = -5 \) [2A]
Given lines are:
\( 2x + y = 2 \)
\( y = -2x + 2 \) (i)

\( x \)	0	1	-1
\( y \)	2	0	4
Next
\( x - y = -5 \)
\( y = x + 5 \) (ii)

\( x \)	-2	0	2
\( y \)	3	5	7
The solution is \( x = -1 \) and \( y = 4 \)
16. बाबुको उमेर छोराको उमेरको ३ गुणा छ। तिनीहरूको उमेरको योगफल ४० वर्ष भए तिनीहरूको उमेर पत्ता लगाउनुहोस्।
    Father's age is three times the son's age. If the sum of their ages is 40 years, find their ages. [2A]
 
The solution is
Let the son's age be \( x \) years.
The father's age is \( y \) years.
Then
First Condition
Father's age is three times the son's age.
\(y=3x\)(i)
Second Condition
Sum of their ages is 40 years
\(x+y=40\)(ii)
Solving (i) and (ii), we get
\(x+y=40\)
or\(x+3x=40\)
or \( 4x = 40 \)
or \( x = 10 \)
\(y=3x = 3 \times 10 = 30 \)
Therefore,
The son's age \( x =10\) years.
The father's age is \( y=30 \) years.
17. ६ ओटा कलम र ३ ओटा पेन्सिलको संयुक्त मूल्य रु. ६० छ। ५ ओटा कलम र २ ओटा पेन्सिलको संयुक्त मूल्य रु. ४८ छ। प्रत्येक कलम र प्रत्येक पेन्सिलको मूल्य कति पर्छ ?
    The combined price of 6 pens and 3 pencils is Rs. 60. The combined price of 5 pens and 2 pencils is Rs. 48. What is the price of each pen and a pencil? [2HA]
 
The solution is
Let the price of a pen be \( x \) (Rs.)
Let the price of a pencil be \( y \) (Rs.)
Then
First Condition (6 pens and 3 pencils cost Rs. 60):
\( 6x + 3y = 60 \)
or\( 2x + y = 20 \)(i)
Second Condition (5 pens and 2 pencils cost Rs. 48)
\( 5x + 2y = 48 \)(ii)
To solve (i) and (ii), we multiply (i) by (2), and subtract from (ii), then we get
\(5x + 2y = 48 \)
\( _{(-)}4x +_{(-)} 2y = _{(-)}40 \)

---
\( x = 8 \)
Substituting \( x=8 \) back into (i), we get
\( 2x + y = 20 \)
or\( 2(8) + y = 20 \)
or\(16 + y = 20 \)
or\(y = 4 \)
Therefore,
  The price of each pen \( x = \) Rs. 8.
The price of each pencil \( y = \) Rs. 4.
18. यदि दुईओटा सङ्ख्याहरूको योगफल २५ र फरक १५ छ भने ती सङ्ख्याहरू पत्ता लगाउनुहोस्।
    If the sum of two numbers is 25 and their difference is 15, find the numbers. [2HA]
 
The solution is
Let the two numbers be \( x \) and \( y \).
Then
First Condition (The sum of two numbers is 25):
\( x + y = 25 \)(i)
Second Condition (Their difference is 15):
\( x - y = 15 \)(ii)
To solve (i) and (ii), we add equation (i) and (ii):
\( x + y = 25 \)
\( x - y = 15 \)
  
---
\( 2x = 40 \)
or \( x = 20 \)
   Substituting \( x=20 \) back into (i), we get
\( x + y = 25 \)
or\( 20 + y = 25 \)
or\(y = 25 - 20 \)
  or\(y = 5 \)
Therefore,
  The first number \( x = 20 \).
The second number \( y = 5 \).
19. दुईओटा सङ्ख्याहरूको अन्तर २८ छ। यदि ठूलो सङ्ख्या सानो सङ्ख्याको ३ गुणा छ भने उक्त सङ्ख्याहरू पत्ता लगाउनुहोस्।
    The difference of two numbers is 28. If the larger number is 3 times the smaller, find the numbers. [2A]
 
The solution is
Let the larger number be \( x \).
Let the smaller number be \( y \).
Then
First Condition (The difference of two numbers is 28):
\( x - y = 28 \)(i)
Second Condition (The larger number is 3 times the smaller):
\( x = 3y \)(ii)
To solve (i) and (ii), we substitute (ii) into (i):
\( x - y = 28 \)
or \( 3y - y = 28 \)
or \( 2y = 28 \)
or \( y = 14 \)
   Substituting \( y=14 \) back into (ii), we get
\( x = 3y \)
or\( x = 3 \times 14 \)
or\( x = 42 \)
Therefore,
  The larger number \( x = 42 \).
The smaller number \( y = 14 \).
20. दुईओटा सङ्ख्याहरू ३:५ को अनुपातमा छन्। यदि तिनीहरूको योगफल ८० भए ती सङ्ख्याहरू पत्ता लगाउनुहोस्।
    Two numbers are in the ratio 3:5. If the sum is 80, find the numbers. [2A]
 
The solution is
Let the two numbers be \( 3x \) and \( 5x \).
Condition (The sum is 80):
\( 3x + 5x = 80 \)
or \( 8x = 80 \)
or \( x = 10 \)
Therefore,
  The first number \( 3x = 30 \).
The second number \( 5x = 50 \).
21. दुई सङ्ख्याहरूको योग १७ र अन्तर ३ भए ती सङ्ख्याहरू पत्ता लगाउनुहोस्।
    If the sum of two numbers is 17 and their difference is 3, find the numbers. [2A]
 
The solution is
Let the two numbers be \( x \) and \( y \).
Then
First Condition (The sum of two numbers is 17):
\( x + y = 17 \)(i)
Second Condition (Their difference is 3):
\( x - y = 3 \)(ii)
To solve (i) and (ii), we add equation (i) and (ii):
\( x + y = 17 \)
\( x - y = 3 \)
  
---
\( 2x = 20 \)
or \( x = 10 \)
   Substituting \( x=10 \) back into (i), we get
\( x + y = 17 \)
or\( 10 + y = 17 \)
or\(y = 17 - 10 \)
  or\(y = 7 \)
Therefore,
  The first number \( x = 10 \).
The second number \( y = 7 \).