Question

How do you solve the system of equations $- 5x - y = 8$ and $5x + y = - 1$?

Answer 1

We use the substitution method to solve two linear equations given in the question. We find the value of x from the first equation in terms of y and substitute in the second equation which becomes an equation in y entirely. Solve for the value of y and substitute back the value of y to obtain the value of x.

Complete step-by-step answer:
We have two linear equations $- 5x - y = 8$ and $5x + y = - 1$
Let us solve the first equation to obtain the value of x in terms of y.
We have $- 5x - y = 8$
Shift the value of y to the right hand side of the equation.
$\Rightarrow - 5x = 8 + y$
Multiply both sides by -5
$\Rightarrow x = \dfrac{{ - 8 - y}}{5}$ … (1)
Now we substitute the value of $x = \dfrac{{ - 8 - y}}{5}$from equation (1) in the second linear equation.
Substitute $x = \dfrac{{ - 8 - y}}{5}$in $5x + y = - 1$
$\Rightarrow 5\left( {\dfrac{{ - 8 - y}}{5}} \right) + y = - 1$
Cancel same factors from numerator and denominator on left hand side of the equation
$\Rightarrow - 8 - y + y = - 1$
Add or subtract same variables on left hand side of the equation’
$\Rightarrow - 8 = - 1$
Cancel same factors from both sides of the equation i.e. -1
$\Rightarrow 8 = 1$
This is not possible as $8 \ne 1$

$\therefore$Solution of the system of linear equations does not exist.

Note:
Students many times make the mistake of not changing the sign when shifting the values from one side of the equation to another side, always keep in mind the sign changes from positive to negative and vice versa when we shift a number from one side to another side of the equation.
Alternate method:
$- 5x - y = 8$ and $5x + y = - 1$
Multiply first equation by -1
$5x + y = - 8$ and $5x + y = - 1$
Since left hand side of both equations is same, we can say right hand side of the equation must be same
$\Rightarrow - 8 = - 1$
Cancel same factors from both sides of the equation i.e. -1
$\Rightarrow 8 = 1$
This is not possible as $8 \ne 1$
$\therefore$Solution of the system of linear equations does not exist.