Question

How do you solve the system of equations $y - x = 5$ and $3y = 3x + 15$?

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 solution:
First, we have to move $x$ to the right side of the equation, $y - x = 5$. Thus, adding $x$ to both sides of the equation.
$y = 5 + x$
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 $x = 1$, $y = 5 + 1 = 6$
When $x = 2$, $y = 5 + 2 = 7$
Now we have to make a table of these values of $x$ and $y$.

$x$	1	2
$y$	6	7

Now we have to plot the points $A\left( {1,6} \right)$ and $B\left( {2,7} \right)$ 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, dividing both sides of the equation by $3$.
$y = \dfrac{{3x + 15}}{3}$
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 $x = 0$, $y = \dfrac{{3 \times 0 + 15}}{3} = 5$
When $x = - 1$, $y = \dfrac{{3 \times - 1 + 15}}{3} = 4$
Now we have to make a table of these values of $x$ and $y$.

$x$	$0$	$ - 1$
$y$	$5$	$4$

Now we have to plot the points $C\left( {0,5} \right)$ and $D\left( { - 1,4} \right)$ on the graph paper and draw a line passing through $C$ and $D$.

We find that $C$ and $D$ both lie on the graph paper of $y - x = 5$. Thus, the graphs of the two equations are coincident. Consequently, every solution of one equation is a solution of the other.
Final solution: Hence, the given system of equations has infinitely many solutions.

Note:
We can directly check whether the system of equations is consistent with infinitely many solutions 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 consistent with infinitely many solutions, if
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$……(i)
Step by step solution:
First, we have to compare $y - x = 5$ and $3y = 3x + 15$ with ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$.
${a_1} = - 1,{b_1} = 1,{c_1} = - 5$ and ${a_2} = - 3,{b_2} = 3,{c_2} = - 15$
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.
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{ - 1}}{{ - 3}} = \dfrac{1}{3}$, $\dfrac{{{b_1}}}{{{b_2}}} = \dfrac{1}{3}$ and $\dfrac{{{c_1}}}{{{c_2}}} = \dfrac{{ - 5}}{{ - 15}} = \dfrac{1}{3}$
Therefore, $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$.
Final solution: Hence, the given system of equations has infinitely many solutions.