Question

Without actually solving the simultaneous equations given below, decide whether the simultaneous equations have a unique solution, no solution, or infinitely many solutions.
$\dfrac{{x - 2y}}{3} = 1$; $2x - 4y = \dfrac{9}{2}$
(a) No solution
(b) Infinitely many solutions
(c) Unique solution
(d) Data insufficient

Answer 1

Here, we need to check whether the simultaneous equations have a unique solution, no solution, or infinitely many solutions. We will convert the given equations to the standard form of a linear equation in two variables. Then, we will find the ratios of the coefficients of the variables and the constant. Finally, we will compare the ratios to check whether the equations have a unique solution, no solution, or infinitely many solutions.

Formula used: We will use the formula of the linear equations in two variables ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$ have no solution if $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$.

Complete step by step solution:
The given equations are linear equations in two variables.
The standard form of a linear equation in two variables is given by $ax + by + c = 0$, where $a$ and $b$ are not equal to 0.
We will rewrite the equations in the standard form of a linear equation in two variables.
Multiplying both sides of the equation $\dfrac{{x - 2y}}{3} = 1$ by 3, we get
$\begin{array}{l} \Rightarrow \dfrac{{x - 2y}}{3} \times 3 = 1 \times 3\\\ \Rightarrow x - 2y = 3\end{array}$
Subtracting 3 from both sides, we get
$\begin{array}{l} \Rightarrow x - 2y - 3 = 3 - 3\\\ \Rightarrow x - 2y - 3 = 0\end{array}$
Thus, we have rewritten $\dfrac{{x - 2y}}{3} = 1$ as $x - 2y - 3 = 0$.
Comparing $x - 2y - 3 = 0$ to the standard form ${a_1}x + {b_1}y + {c_1} = 0$, we get
${a_1} = 1$, ${b_1} = - 2$, and ${c_1} = - 3$
Now, we will rewrite the equation $2x - 4y = \dfrac{9}{2}$.
Multiplying both sides of the equation $2x - 4y = \dfrac{9}{2}$ by 2, we get
$\Rightarrow 2\left( {2x - 4y} \right) = 2 \times \dfrac{9}{2}$
Multiplying using the distributive law of multiplication, we get
$\Rightarrow 4x - 8y = 9$
Subtracting 9 from both sides, we get
$\begin{array}{l} \Rightarrow 4x - 8y - 9 = 9 - 9\\\ \Rightarrow 4x - 8y - 9 = 0\end{array}$
Thus, we have rewritten $2x - 4y = \dfrac{9}{2}$ as $4x - 8y - 9 = 0$.
Comparing $4x - 8y - 9 = 0$ to the standard form ${a_2}x + {b_2}y + {c_2} = 0$, we get
${a_2} = 4$, ${b_2} = - 8$, and ${c_2} = - 9$
Now, we will find the ratios of the coefficients of $x$, $y$, and the constant.
Dividing ${a_1} = 1$ by ${a_2} = 4$, we get
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{1}{4}$
Dividing ${b_1} = - 2$ by ${b_2} = - 8$, we get
$\dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{ - 2}}{{ - 8}} = \dfrac{1}{4}$
Dividing ${c_1} = - 3$ by ${c_2} = - 9$, we get
$\dfrac{{{c_1}}}{{{c_2}}} = \dfrac{{ - 3}}{{ - 9}} = \dfrac{1}{3}$
The linear equations in two variables ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$ have unique solution if $\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}$.
The linear equations in two variables ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$ have infinitely many solutions if $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$.
The linear equations in two variables ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$ have no solution if $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$.
Since $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{1}{4}$, $\dfrac{{{b_1}}}{{{b_2}}} = \dfrac{1}{4}$, and $\dfrac{{{c_1}}}{{{c_2}}} = \dfrac{1}{3}$, we get
$\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$
Therefore, the simultaneous equations $\dfrac{{x - 2y}}{3} = 1$ and $2x - 4y = \dfrac{9}{2}$ have no solution.

Thus, the correct option is option (a).

Note:
We used the term linear equation in two variables to express the statements in mathematical form. An equation is said to be a linear equation in two variables only if the equation has two distinct variables and the coefficient of both variables should not be zero. If the coefficient of any of the two variables will be zero then the equation becomes a linear equation of one variable.
For example, The equation $px + qy + r = s$ is a linear equation of 2 variables. Here, $x$ and $y$ are variables whereas $p$ and $q$ are their respective coefficients. Also $r$ and $s$are constants. If either $p$ or $q$ is 0 then the equation becomes a linear equation of one variable. In addition to this, the equation must follow the condition that is $x \ne y$ .