Question

show that x=2 and y=1 is not a solution of the system of simultaneous linear equation 2x+7y=11 and x-3y =5 .​

Answer 1

2x + 7y = 2 ( 2 ) + 7 ( 1 )

= 4 + 7

= 11

x – 3y = 2 – 3 ( 1 )

= 2 – 3

= – 1

≠ 5

Point ( 2 , 1 ) satisfies the first equation but not the second one hence it is not a solution of the above given equations

Answer 2

Answer

The given system of equations is

The given system of equations is2x+7y=11 .(i)

The given system of equations is2x+7y=11 .(i)x−3y=5 .(ii)

The given system of equations is2x+7y=11 .(i)x−3y=5 .(ii)Putting x=2,y=1 in equation (i), we have

The given system of equations is2x+7y=11 .(i)x−3y=5 .(ii)Putting x=2,y=1 in equation (i), we haveLHS=2×2+7×1=11=RHsSo, x=2 and y=1 satisfy equation (i)

The given system of equations is2x+7y=11 .(i)x−3y=5 .(ii)Putting x=2,y=1 in equation (i), we haveLHS=2×2+7×1=11=RHsSo, x=2 and y=1 satisfy equation (i)Putting x=2,y=1 in equation (i), we have

The given system of equations is2x+7y=11 .(i)x−3y=5 .(ii)Putting x=2,y=1 in equation (i), we haveLHS=2×2+7×1=11=RHsSo, x=2 and y=1 satisfy equation (i)Putting x=2,y=1 in equation (i), we haveLHS =2×1−3×1=−1= RHS

The given system of equations is2x+7y=11 .(i)x−3y=5 .(ii)Putting x=2,y=1 in equation (i), we haveLHS=2×2+7×1=11=RHsSo, x=2 and y=1 satisfy equation (i)Putting x=2,y=1 in equation (i), we haveLHS =2×1−3×1=−1= RHSSo, x=2 and y=1 do not satisfy equation (ii)

The given system of equations is2x+7y=11 .(i)x−3y=5 .(ii)Putting x=2,y=1 in equation (i), we haveLHS=2×2+7×1=11=RHsSo, x=2 and y=1 satisfy equation (i)Putting x=2,y=1 in equation (i), we haveLHS =2×1−3×1=−1= RHSSo, x=2 and y=1 do not satisfy equation (ii)Hence, x=2,y=1 is not a solution of the given system of equations

Step-by-step explanation:

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