5ax+6by=28; 3ax+4by=18,Solve the given pair of linear equations by elimination method.
Answers
Answered by
64
Answer :
Given equations are
5ax + 6by = 28 ...(i)
3ax + 4by = 18 ...(ii)
Now, multiplying (i) by 3 and (ii) by 5, we get
15ax + 18by = 84
15ax + 20by = 90
On subtraction, we get
2by = 6
or, y = 3/b
Now, putting y = 3/b in (i), we get
5ax + 6b (3/b) = 28
or, 5ax + 18 = 28
or, 5ax = 28 - 18
or, 5ax = 10
or, x = 2/a
Therefore, the required solution be
x = 2/a and y = 3/b
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Answered by
22
Solution :
5ax + 6by = 28 ---( 1 )
3ax + 4by = 18 --- ( 2 )
[ 3 × ( 1 ) - 5 × ( 2 ) ] we get
15ax + 18by = 84 --- ( 3 )
15ax + 20by = 90---( 4 )
_______________
•••••••• - 2by = - 6
=> y = ( -6 )/( -2b )
=> y = 3/b
Put y = 3/b in equation ( 1 ), we get
5ax + 6b ( 3/b ) = 28
=> 5ax + 18 = 28
=>5ax = 10
=> x = 10/5a
=> x = 2/a
Therefore ,
( x , y ) = ( 2/a , 3/b)
•••••
5ax + 6by = 28 ---( 1 )
3ax + 4by = 18 --- ( 2 )
[ 3 × ( 1 ) - 5 × ( 2 ) ] we get
15ax + 18by = 84 --- ( 3 )
15ax + 20by = 90---( 4 )
_______________
•••••••• - 2by = - 6
=> y = ( -6 )/( -2b )
=> y = 3/b
Put y = 3/b in equation ( 1 ), we get
5ax + 6b ( 3/b ) = 28
=> 5ax + 18 = 28
=>5ax = 10
=> x = 10/5a
=> x = 2/a
Therefore ,
( x , y ) = ( 2/a , 3/b)
•••••
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