If $2x + py = - 5$ and $3x + 3y = - 6$ has a unique solution, then find $p$.
Answer
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Hint: In this problem, we need to find the value of $p$ such that the given system of two linear equations has a unique solution. For this, we must know the general form of a system of two linear equations in two variables. The given system has unique solution so we can say that the condition $\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}$ is satisfied.
Complete step-by-step solution:
The general form of system of two linear equations in two variables $x$ and $y$ is given by$
{a_1}x + {b_1}y = {c_1} \cdots \cdots \left( 1 \right) \\
{a_2}x + {b_2}y = {c_2} \cdots \cdots \left( 2 \right) \\
$
In this problem, we have the following system of two linear equations in two variables $x$ and $y$.$
2x + py = - 5 \cdots \cdots \left( 3 \right) \\
3x + 3y = - 6 \cdots \cdots \left( 4 \right) \\
$
Let us compare equation $\left( 3 \right)$ with the equation $\left( 1 \right)$. Therefore, we get ${a_1} = 2,\;{b_1} = p$ and ${c_1} = - 5$.
Let us compare equation $\left( 4 \right)$ with the equation $\left( 2 \right)$. Therefore, we get ${a_2} = 3,\;{b_2} = 3$ and ${c_2} = - 6$.
In this problem, it is also given that the system has a unique solution. So, we can write
$\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}} \Rightarrow \dfrac{2}{3} \ne \dfrac{p}{3} \Rightarrow p \ne 2$
Therefore, we can say that if $2x + py = - 5$ and $3x + 3y = - 6$ has a unique solution, then $p \ne 2$.
In other words, we can say that the given system has a unique solution only if $p \ne 2$. If $p = 2$ then the system has either infinite solutions or no solution.
Note:In this problem, we have two equations of straight line. It is given that the system has a unique solution. So, we can say that the given lines will intersect at only one point if $p \ne 2$. If the condition $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$ is satisfied then we can say that the given system of equations have infinitely many solutions. For example, let us take two equations of line as $x + y - 4 = 0$ and $2x + 2y - 8 = 0$. This system of equations has infinitely many solutions. In this case, we can say that these lines are coincident lines. If the condition $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$ is satisfied then we can say that the given system of equations has no solution. For example, let us take two equations of line as $x + y - 4 = 0$ and $x + y - 8 = 0$. This system of equations has no solution. In this case, we can say that these lines are parallel lines.
Complete step-by-step solution:
The general form of system of two linear equations in two variables $x$ and $y$ is given by$
{a_1}x + {b_1}y = {c_1} \cdots \cdots \left( 1 \right) \\
{a_2}x + {b_2}y = {c_2} \cdots \cdots \left( 2 \right) \\
$
In this problem, we have the following system of two linear equations in two variables $x$ and $y$.$
2x + py = - 5 \cdots \cdots \left( 3 \right) \\
3x + 3y = - 6 \cdots \cdots \left( 4 \right) \\
$
Let us compare equation $\left( 3 \right)$ with the equation $\left( 1 \right)$. Therefore, we get ${a_1} = 2,\;{b_1} = p$ and ${c_1} = - 5$.
Let us compare equation $\left( 4 \right)$ with the equation $\left( 2 \right)$. Therefore, we get ${a_2} = 3,\;{b_2} = 3$ and ${c_2} = - 6$.
In this problem, it is also given that the system has a unique solution. So, we can write
$\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}} \Rightarrow \dfrac{2}{3} \ne \dfrac{p}{3} \Rightarrow p \ne 2$
Therefore, we can say that if $2x + py = - 5$ and $3x + 3y = - 6$ has a unique solution, then $p \ne 2$.
In other words, we can say that the given system has a unique solution only if $p \ne 2$. If $p = 2$ then the system has either infinite solutions or no solution.
Note:In this problem, we have two equations of straight line. It is given that the system has a unique solution. So, we can say that the given lines will intersect at only one point if $p \ne 2$. If the condition $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$ is satisfied then we can say that the given system of equations have infinitely many solutions. For example, let us take two equations of line as $x + y - 4 = 0$ and $2x + 2y - 8 = 0$. This system of equations has infinitely many solutions. In this case, we can say that these lines are coincident lines. If the condition $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$ is satisfied then we can say that the given system of equations has no solution. For example, let us take two equations of line as $x + y - 4 = 0$ and $x + y - 8 = 0$. This system of equations has no solution. In this case, we can say that these lines are parallel lines.
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