Show that $x = 2, y = 1$ is not a solution of the system of simultaneous linear equations.
$2 x + 7 y = 11$
$x - 3 y = 5$

Open in App

Solution

The given system of equations is
$2 x + 7 y = 11$ .(i)
$x - 3 y = 5$ .(ii)

Putting $x = 2, y = 1$ in equation (i), we have
LHS $= 2 \times 2 + 7 \times 1 = 11 =$ RHsSo, $x = 2$ and $y = 1$ satisfy equation (i)

Putting $x = 2, y = 1$ in equation (i), we have
LHS $= 2 \times 1 - 3 \times 1 = - 1 \neq$ RHS

So, $x = 2$ and $y = 1$ do not satisfy equation (ii)

Hence, $x = 2, y = 1$ is not a solution of the given system of equations.

Suggest Corrections

Similar questions

x = 2, y = 1

3 x - 2 y = 4

2 x + y = 5

x + y + z = 2

2 x + 3 y + 2 z = 5

2 x + 3 y + (a^{2} - 1) z = a + 1

x + y + z = 2

2 x + 3 y + 2 z = 5

2 x + 3 y + (a^{2} - 1) z = a + 1

x + 2 y + z = 8

2 x + 3 y - z = 11

3 x - y - 2 z = 5

x = 2, y = - 1,

7 x + 3 y = 11

Join BYJU'S Learning Program