x - 2y = 1
3x + 2y = 19
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Hello users .....
Given equations are :
x - 2y = 1 ......(1.)
3x + 2y = 19 ....(2.)
Solution:-
Method 1st
in equation (1.)
x - 2y = 1
=> 2y = x - 1 ......(3.)
in equation (2.)
3x + 2y = 19
=> 2y = 19 - 3x ....(4.)
Comparing (3.) and (4.).....
=> (x - 1 ) = ( 19 - 3x )
=> x + 3x = 19 + 1
=> 4x = 20
=> x = 20/4 = 5
Putting x in equation (1.)
we get
x - 2y = 1
=> 5 - 2y = 1
=> -2y = 1 - 5
=> -2y = - 4
=> 2y = 4
=> y = 4/2 = 2
Method 2nd
Adding equations (1.) and (2.)
We get,
x - 2y = 1
+ 3x +2y = 19
-----------------------
4x + 0 = 20
-----------------------
=> 4x = 20
=> x = 20 / 4 = 5
putting value of x in equation (1.)
We get
5 - 2y = 1
=> -2y = 1 - 5
=> -2y = - 4
=> 2y = 4
=> y = 4/2 = 2
Hence,
x= 5 and y = 2 Answer
# hope it helps
Given equations are :
x - 2y = 1 ......(1.)
3x + 2y = 19 ....(2.)
Solution:-
Method 1st
in equation (1.)
x - 2y = 1
=> 2y = x - 1 ......(3.)
in equation (2.)
3x + 2y = 19
=> 2y = 19 - 3x ....(4.)
Comparing (3.) and (4.).....
=> (x - 1 ) = ( 19 - 3x )
=> x + 3x = 19 + 1
=> 4x = 20
=> x = 20/4 = 5
Putting x in equation (1.)
we get
x - 2y = 1
=> 5 - 2y = 1
=> -2y = 1 - 5
=> -2y = - 4
=> 2y = 4
=> y = 4/2 = 2
Method 2nd
Adding equations (1.) and (2.)
We get,
x - 2y = 1
+ 3x +2y = 19
-----------------------
4x + 0 = 20
-----------------------
=> 4x = 20
=> x = 20 / 4 = 5
putting value of x in equation (1.)
We get
5 - 2y = 1
=> -2y = 1 - 5
=> -2y = - 4
=> 2y = 4
=> y = 4/2 = 2
Hence,
x= 5 and y = 2 Answer
# hope it helps
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