Question

How do you solve the system of equations \[ - 5x - y = 8\] and $5x + y = - 1$.

Answer 1

Hint: We have given two linear equations in two variables. To solve the system of the linear equation, we can use the following method:
Graphical method
Substitution method
Elimination method
Cross multiplication method

Complete step-by-step solution:
Step 1: We have given two equations of lines \[ - 5x - y = 8\] and $5x + y = - 1$. For the first line, the coefficient of $x$ is ${a_1} = - 5$ , the coefficient of $y$ is ${b_1} = - 1$, and the constant term is${c_1} = 8$.
For the second line, the coefficient of $x$ is ${a_2} = 5$ , the coefficient of $y$ is ${b_2} = 1$, and the constant term is${c_2} = - 1$.
Step 2: Now we determine the ratios of the coefficient of $x$ of both lines.
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{ - 5}}{5} = - 1$ .
The ratios of the coefficient of $y$ of both lines is $\dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{ - 1}}{1} = - 1$.
The ratios of the constant term is $\dfrac{{{c_1}}}{{{c_2}}} = \dfrac{8}{{ - 1}} = - 8$
Step 3: Now for the given system of equation, $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$ so, given lines are parallel lines.
For parallel lines, no solution exists.

Note: Before solving the system of linear equations in two variables, first check where the lines are coincident, parallel, or inclined to each other.
 For two lines ${a_1}x + {b_1}y = {c_1}$ and ${a_2}x + {b_2}y = {c_2}$
If $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$ then, both lines are a coincidence line, which means both equations belong to the same line. For this situation, there is an infinite solution, which is that all the points lie on the line.
If $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$ then, both lines are a parallel line, which means both equations belong to the same line. For this situation, there exist NO solution as parallel lines never intersect each other.
If $\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}$ then, there exists a unique solution for the system of equation.