Question

How do you solve the system of equations $2x + y = 6$ and $ - 2x - y = 6$?

Answer 1

Hint: First we have to make the first linear equation in Slope-intercept form and then calculate the value of $y$ for any two arbitrary values of $x$. Next make a table of these values of $x$ and $y$. Next plot the obtained points on the graph paper and draw a line passing through these points. Now repeat the process with the second equation and determine the solution of the given system of equations using the graph obtained.

Formula used:
Slope Intercept of a line:
The equation of a line with slope $m$ and making an intercept $c$ on $y$-axis is $y = mx + c$.

Complete step by step answer:
First, we have to move $2x$ to the right side of the equation, $2x + y = 6$. Thus, subtracting $2x$ from both sides of the equation.
$y = 6 - 2x$
Now, we have to calculate the value of $y$ for any two arbitrary values of $x$. Thus, finding the value of $y$ when $x = 1$ and $x = 2$.
When $x = 1$, $y = 6 - 2 = 4$
When $x = 2$, $y = 6 - 4 = 2$
Now we have to make a table of these values of $x$ and $y$.

$x$	$1$	$2$
$y$	$4$	$2$

Now we have to plot the points $A\left( {1,4} \right)$ and $B\left( {2,2} \right)$ on the graph paper and draw a line passing through $A$ and $B$.

Now we have to make the second equation in Slope-intercept form. Thus, multiply both sides of the equation by $ - 1$.
$2x + y = - 6$
Subtract $2x$ from both sides of the equation, we get
$y = - 6 - 2x$
Now, we have to calculate the value of $y$ for any two arbitrary values of $x$. Thus, finding the value of $y$ when $x = 0$ and $x = - 1$.
When $x = 0$, $y = - 6$
When $x = - 1$, $y = - 6 + 2 = - 4$
Now we have to make a table of these values of $x$ and $y$.

$x$	$0$	$ - 1$
$y$	$ - 6$	$ - 4$

Now we have to plot the points $C\left( {0, - 6} \right)$ and $D\left( { - 1, - 4} \right)$ on the graph paper and draw a line passing through $C$ and $D$.

We find the lines represented by equations $2x + y = 6$ and $ - 2x - y = 6$ are parallel. So, the two lines have no common point.
Hence, the given system of equations has no solution.

Note: We can directly check whether the system of equations is inconsistent or not using below property:
The system of equations ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$ is inconsistent, if
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$……(i)
Step by step solution:
First, we have to compare $2x + y = 6$ and $ - 2x - y = 6$ with ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$.
${a_1} = 2,{b_1} = 1,{c_1} = - 6$ and ${a_2} = - 2,{b_2} = - 1,{c_2} = - 6$
Now we have to find $\dfrac{{{a_1}}}{{{a_2}}},\dfrac{{{b_1}}}{{{b_2}}},\dfrac{{{c_1}}}{{{c_2}}}$ and check whether it satisfy (i) or not.
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{2}{{ - 2}} = - 1$, $\dfrac{{{b_1}}}{{{b_2}}} = \dfrac{1}{{ - 1}} = - 1$ and $\dfrac{{{c_1}}}{{{c_2}}} = \dfrac{{ - 6}}{{ - 6}} = 1$
Therefore, $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$.
Hence, the given system of equations has no solution.