(a – b)x + (a + b) y = a 2 – 2ab – b 2 (a + b)(x + y) = a 2 + b 2

 10th cbse mathematics  chapter 3 miscellaneous / optional exercise 

solve

(iv)

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

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

 

rearrange the second equation 

making the RHS zero for both equations

(a – b)x + (a + b) y -[ (a^2) – 2ab – (b^2)] = 0

(a + b)x +(a + b)y -[ (a^2) + (b^2 )] =0

 

using method of cross multiplication

 

                 x                                           y                     1

(a+b)            {-[ (a^2)–2ab–(b^2)]}        (a-b)                 (a+b)

(a+b)             { -[ (a^2) + (b^2 )]}         (a+b)               (a+b)

 

 

x /{(a+b){-[ (a^2) + (b^2 )]} -(a+b){-[ (a^2)–2ab–(b^2)]}}

=

y/ {{-[ (a^2)–2ab–(b^2)]}(a+b) - { -[ (a^2) + (b^2 )]}(a-b)} 

=

1 /{(a-b)(a+b) - (a+b)(a+b)}

simplifying

x / {(a+b)[-2(b^2 ) -2ab]} taking (a+b) out in the Dr.

=

y/ {(a^2) [-a-b+a-b] +2ab(a+b) +(b^2)[a+b+a-b]} grouping (a^2) and (b^2)

= 

1/ {(a+b)[a-b-a-b]} taking (a+b) out

 

OR

x /{(a+b)(-2b)(b+a)}

=

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

=

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

 

OR

x /{(a+b)(-2b)(b+a)}

=

y / {[(a^2)(-2b) + 2(a^2)b + 2a(b^2) + (b^2)(2a)] } opening middle term

=

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

 

OR

x /{(a+b)(-2b)(b+a)}

=

y / {[2a(b^2) + (b^2)(2a)] } cancelling off

=

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

 

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

 

using the first and third

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

 

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

x= (a+b) 

using the second and the third

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

 

y =  {{4a(b^2) } / {(a+b)(-2b)}

y =(-2ab) / (a+b) 

(v)

152x – 378y = – 74

–378x + 152y = – 604

here note that the coefficients of x and y are interchanged

namely 152 and (-378)

Therefore we add and subtract the two equations to get another two equations which will usually be easier to solve than the two given equations

  152x – 378y = – 74       --------------------(1)

–378x + 152y = – 604   ---------------------(2)

---------------------------------------adding

-226x -226y = (-678) 

divide each term with (-226)

we get

x + y = 3 -------------------------(3)

  152x – 378y = – 74       --------------------(1)

–378x + 152y = – 604   ---------------------(2)

---------------------------------------subtracting

530x -530y = 530

dividing each term with 530

x - y = 1 ----------------------------(4)

now we concentrate on the new equations (3) and (4)

x + y = 3 -------------------------(3)

x -  y = 1 ----------------------------(4)

-------------------adding

2x   = 4

x =2

 

substitute x=2 in  x + y = 3

2+y = 3 

y = 3-2

y =1

 

 

 

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ncert cbse 10th mathematics chapter 3 optional exercise 3.7 

 The ages of two friends Ani and Biju differ by 3 years. Ani’s father  is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Ani’s father differ by 30 years. Find the ages of Ani and Biju

solution

 

2. One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? 

solution

3. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train. 

solution  

 

4. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

solution  

5. In a ∆ ABC, ∠ C = 3 ∠ B = 2 (∠ A + ∠ B). Find the three angles.

solution

 

Solve the following pair of linear equations:

 px + qy = p – q 

 qx – py = p + q

solution

 

(ii) ax + by = c
     bx + ay = 1 + c

solution  

 

(iii) 

(x/a) -(y/b) = 0

ax +by = (a^2)  + (b^2)

solution  

 

(iv)

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

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

solution

 

(v)

152x – 378y = – 74

–378x + 152y = – 604

solution   

 

  ncert cbse 10th  mathematics chapter 2 optional exercise
 If the zeroes of the polynomial (x^3) – 3(x^2) + x + 1 are a – b, a, a + b, find a and b.
solution
 
2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively
solution



4. If two zeroes of the polynomial (x^4) – 6(x^3) – 26(x^2) + 138x – 35 are 
 [2 ±sqrt(3) ] , find other zeroes

 solution
 
5. If the polynomial (x^4) – 6(x^3) + 16(x^2) – 25x + 10 is divided by another polynomial (x^2) – 2x + k, the remainder comes out to be x + a, find k and a
solution  
 
exercise 2.3
 

3. obtain all other zeroes of 3(x^4)+6(x^3)-2(x^2)-10x-5 if two of its zeroes are sqrt(5/3) and [sqrt(5/3)]
solution
 
4. On dividing (x^3) – 3(x^2) + x + 2 by a polynomial g(x), the quotient and remainder were x – 2
and –2x + 4, respectively. Find g(x).   
solution
 
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