U will get 25 marks ....Plzzz solve it it's urgent ....plz solve it by option...
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Answered by
4
The correct option is (d)
----------------------------
Now, see....
choose the equation which have to simplify easily.
Here, x+ 2y+3z= 13.
Put the value of;
x= -10
y= 10
z= 1
You, will get = to RHS.
_____________________
Hope you get it✌✌
#Amrit⭐
----------------------------
Now, see....
choose the equation which have to simplify easily.
Here, x+ 2y+3z= 13.
Put the value of;
x= -10
y= 10
z= 1
You, will get = to RHS.
_____________________
Hope you get it✌✌
#Amrit⭐
Answered by
16
Here is your answer :
=> 3x - 4y + 70z = 0 ------- ( 1 )
=> 2x + 3y - 10z = 0 -------- ( 2 )
=> x + 2y + 3z = 13 --------- ( 3 )
From ( 1 ) and ( 2 ),we get ,
=> 3x - 4y + 70z = 2x + 3y - 10z
=> 3x - 2x - 4y - 3y + 70z + 10z = 0
=> x - 7y + 80z = 0
•°• x = 7y - 80z. -------- ( 4 )
Substitute the value of ( 4 ) in ( 2 ),
=> 2x + 3y - 10z = 0
=> 2( 7y - 80z ) + 3y - 10z = 0
=> 14y - 160z + 3y - 10z = 0
=> 14y + 3y - 160z - 10z = 0
=> 17y - 170z = 0
Talking 17 as common,
=> 17 ( y - 10z) = 0
=> ( y - 10z ) = 0 ÷ 17
•°• ( y - 10z ) = 0 --------------- ( 5 )
Substitute the value of ( 4 ) in ( 3 ),
=> x + 2y + 3z = 13
=> 7y - 80z + 2y + 3z = 13
=> 7y + 2y - 80z + 3z = 13
=> 9y - 77z = 13 --------------- ( 6 )
By multiplying to 9 in ( 5 ),
=> 9( y - 10z ) = 0 × 9
=> 9y - 90z = 0. ---------- ( 7 )
Subtracting ( 6 ) from ( 7 ),
=> 9y - 90z - ( 9y - 77z ) = 0 - 13
=> 9y - 90z - 9y + 77z = - 13
=> - 13z = -13
=> z = ( -13 ) ÷ ( -13 )
•°• z = 1
Substitute the value of z in ( 5 ),
=> y - 10z = 0
=> y - 10 × 1 = 0
=> y - 10 = 0
•°• y = 10
Now, substitute the value of y and z in ( 1 ),
=> 3x - 4y + 70z = 0
=> 3x - 4( 10 ) + 70 ( 1 ) = 0
=> 3x - 40 + 70 = 0
=> 3x + 30 = 0
=> 3x = -30
=> x = ( -30 ) ÷ 3
•°• x = ( -10 ).
Hence,
( x = -10 ) , ( y = 10 ) , ( z = 1 ).
Ans : ( d ) ( -10 , 10 , 1 )
Verification :
( i )
=> 3x - 4y + 70z = 0
=> 3( -10 ) - 4( 10 ) + 70( 1 ) = 0
=> -30 - 40 + 70 = 0
=> -70 + 70 = 0
•°• 0 = 0
( ii )
=> 2x + 3y - 10z = 0
=> 2( -10 ) + 3 ( 10 ) - 10( 1 ) = 0
=> -20 + 30 -10 = 0
=> -30 + 30 = 0
•°• 0 = 0
( iii )
=> x + 2y + 3z = 13
=> ( -10 ) + 2 ( 10 ) + 3 ( 1 ) = 13
=> -10 + 20 + 3 = 13
=> 23 - 10 = 13
•°• 13 = 13
Verified
Ans : ( d ) ( -10 , 10 , 1 )
Hope it helps !!
=> 3x - 4y + 70z = 0 ------- ( 1 )
=> 2x + 3y - 10z = 0 -------- ( 2 )
=> x + 2y + 3z = 13 --------- ( 3 )
From ( 1 ) and ( 2 ),we get ,
=> 3x - 4y + 70z = 2x + 3y - 10z
=> 3x - 2x - 4y - 3y + 70z + 10z = 0
=> x - 7y + 80z = 0
•°• x = 7y - 80z. -------- ( 4 )
Substitute the value of ( 4 ) in ( 2 ),
=> 2x + 3y - 10z = 0
=> 2( 7y - 80z ) + 3y - 10z = 0
=> 14y - 160z + 3y - 10z = 0
=> 14y + 3y - 160z - 10z = 0
=> 17y - 170z = 0
Talking 17 as common,
=> 17 ( y - 10z) = 0
=> ( y - 10z ) = 0 ÷ 17
•°• ( y - 10z ) = 0 --------------- ( 5 )
Substitute the value of ( 4 ) in ( 3 ),
=> x + 2y + 3z = 13
=> 7y - 80z + 2y + 3z = 13
=> 7y + 2y - 80z + 3z = 13
=> 9y - 77z = 13 --------------- ( 6 )
By multiplying to 9 in ( 5 ),
=> 9( y - 10z ) = 0 × 9
=> 9y - 90z = 0. ---------- ( 7 )
Subtracting ( 6 ) from ( 7 ),
=> 9y - 90z - ( 9y - 77z ) = 0 - 13
=> 9y - 90z - 9y + 77z = - 13
=> - 13z = -13
=> z = ( -13 ) ÷ ( -13 )
•°• z = 1
Substitute the value of z in ( 5 ),
=> y - 10z = 0
=> y - 10 × 1 = 0
=> y - 10 = 0
•°• y = 10
Now, substitute the value of y and z in ( 1 ),
=> 3x - 4y + 70z = 0
=> 3x - 4( 10 ) + 70 ( 1 ) = 0
=> 3x - 40 + 70 = 0
=> 3x + 30 = 0
=> 3x = -30
=> x = ( -30 ) ÷ 3
•°• x = ( -10 ).
Hence,
( x = -10 ) , ( y = 10 ) , ( z = 1 ).
Ans : ( d ) ( -10 , 10 , 1 )
Verification :
( i )
=> 3x - 4y + 70z = 0
=> 3( -10 ) - 4( 10 ) + 70( 1 ) = 0
=> -30 - 40 + 70 = 0
=> -70 + 70 = 0
•°• 0 = 0
( ii )
=> 2x + 3y - 10z = 0
=> 2( -10 ) + 3 ( 10 ) - 10( 1 ) = 0
=> -20 + 30 -10 = 0
=> -30 + 30 = 0
•°• 0 = 0
( iii )
=> x + 2y + 3z = 13
=> ( -10 ) + 2 ( 10 ) + 3 ( 1 ) = 13
=> -10 + 20 + 3 = 13
=> 23 - 10 = 13
•°• 13 = 13
Verified
Ans : ( d ) ( -10 , 10 , 1 )
Hope it helps !!
27jenny:
by in the exam i have objective que nd i have no time to do ful process that's why i want by option
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