Question

explain substitution....... ❤ ❤ ❤ ⤵ ⤵ ⤵ ⤵ ⤵ ⤵ ⤵ ⤵

Answer 1

Step 1 : Find out the value of one of the unknown quantities in one of the given equations, in terms of the other unknown quantities.

Step 2 : The value from step 1, substitute in the second equation for that unknown quantity, we get an equation withone unknown quantity. Now if we slove it, we will get the value of one unknown quantity.

Step 3 : Substitute the value of the unknown quantity from step 2 in step 1 equation, we will get the value of other unknown quantity.

Example :

4x + y + 1 = 0

4x + y + 1 = 03x+2y-3=0

4x + y + 1 = 03x+2y-3=0Solution :

4x + y + 1 = 03x+2y-3=0Solution :4x + y + 1 = 0.......... (i)

4x + y + 1 = 03x+2y-3=0Solution :4x + y + 1 = 0.......... (i) 3x+2y-3=0...................... (ii)

4x + y + 1 = 03x+2y-3=0Solution :4x + y + 1 = 0.......... (i) 3x+2y-3=0...................... (ii) From (1), y=-1 - 4r ........ (iii)

4x + y + 1 = 03x+2y-3=0Solution :4x + y + 1 = 0.......... (i) 3x+2y-3=0...................... (ii) From (1), y=-1 - 4r ........ (iii)Putting y=-4x-1, in (ii) we get

4x + y + 1 = 03x+2y-3=0Solution :4x + y + 1 = 0.......... (i) 3x+2y-3=0...................... (ii) From (1), y=-1 - 4r ........ (iii)Putting y=-4x-1, in (ii) we get3x+2 (-4x - 1)-3=0.

4x + y + 1 = 03x+2y-3=0Solution :4x + y + 1 = 0.......... (i) 3x+2y-3=0...................... (ii) From (1), y=-1 - 4r ........ (iii)Putting y=-4x-1, in (ii) we get3x+2 (-4x - 1)-3=0.⇒ 3x-8x-2-3=0

4x + y + 1 = 03x+2y-3=0Solution :4x + y + 1 = 0.......... (i) 3x+2y-3=0...................... (ii) From (1), y=-1 - 4r ........ (iii)Putting y=-4x-1, in (ii) we get3x+2 (-4x - 1)-3=0.⇒ 3x-8x-2-3=0⇒=> -5x-5=0

4x + y + 1 = 03x+2y-3=0Solution :4x + y + 1 = 0.......... (i) 3x+2y-3=0...................... (ii) From (1), y=-1 - 4r ........ (iii)Putting y=-4x-1, in (ii) we get3x+2 (-4x - 1)-3=0.⇒ 3x-8x-2-3=0⇒=> -5x-5=0⇒ -5x = 5

4x + y + 1 = 03x+2y-3=0Solution :4x + y + 1 = 0.......... (i) 3x+2y-3=0...................... (ii) From (1), y=-1 - 4r ........ (iii)Putting y=-4x-1, in (ii) we get3x+2 (-4x - 1)-3=0.⇒ 3x-8x-2-3=0⇒=> -5x-5=0⇒ -5x = 5⇒x = -1

4x + y + 1 = 03x+2y-3=0Solution :4x + y + 1 = 0.......... (i) 3x+2y-3=0...................... (ii) From (1), y=-1 - 4r ........ (iii)Putting y=-4x-1, in (ii) we get3x+2 (-4x - 1)-3=0.⇒ 3x-8x-2-3=0⇒=> -5x-5=0⇒ -5x = 5⇒x = -1Putting x=-1 in (iii) we get

4x + y + 1 = 03x+2y-3=0Solution :4x + y + 1 = 0.......... (i) 3x+2y-3=0...................... (ii) From (1), y=-1 - 4r ........ (iii)Putting y=-4x-1, in (ii) we get3x+2 (-4x - 1)-3=0.⇒ 3x-8x-2-3=0⇒=> -5x-5=0⇒ -5x = 5⇒x = -1Putting x=-1 in (iii) we gety = - 1 - 4(-1)

4x + y + 1 = 03x+2y-3=0Solution :4x + y + 1 = 0.......... (i) 3x+2y-3=0...................... (ii) From (1), y=-1 - 4r ........ (iii)Putting y=-4x-1, in (ii) we get3x+2 (-4x - 1)-3=0.⇒ 3x-8x-2-3=0⇒=> -5x-5=0⇒ -5x = 5⇒x = -1Putting x=-1 in (iii) we gety = - 1 - 4(-1) = - 1 + 4

4x + y + 1 = 03x+2y-3=0Solution :4x + y + 1 = 0.......... (i) 3x+2y-3=0...................... (ii) From (1), y=-1 - 4r ........ (iii)Putting y=-4x-1, in (ii) we get3x+2 (-4x - 1)-3=0.⇒ 3x-8x-2-3=0⇒=> -5x-5=0⇒ -5x = 5⇒x = -1Putting x=-1 in (iii) we gety = - 1 - 4(-1) = - 1 + 4= 3

∴ Required root (-1,3)