Question

Given below are three equations. Two of them have infinite solutions and two have a unique solutions. state the two pairs.

$3x-2y=4,6x+2y=4,9x-6y=12$

Answer 1

Step 1: Applying and Analysing

The given system of equation is of the forms

$$\begin{gathered} 3x-2y-4=0\Rightarrow \Rightarrow 1 \\ 6x+2y-4=0\Rightarrow \Rightarrow 2 \\ 9x-6y-12=0\Rightarrow \Rightarrow 3 \end{gathered}$$

The given system of equation is of the form

$$\begin{gathered} Thisisoftheform{a}_{1}x+b_{1}y+c_1=0 \\ {a}_{2}x+b_{2}y+c_2=0 \end{gathered}$$

Step 2: Equating the coefficients

Pairing equation 1 & 2,

$$\begin{gathered} where,a_1=3,b_1=-2,c_1=-4 \\ {a}_2=6,b_2=2,c_2=-4 \\ wehave,\frac{a_1}{a_2}\ne \frac{b_1}{b_2} \\ \frac{3}{6}\ne \frac{-2}{2} \\ \frac{1}{2}\ne \frac{-1}{1} \\ Here,\frac{a_1}{a_2}\ne \frac{b_1}{b_2} \end{gathered}$$

Therefore, We can say these pairs of equations have a unique solution.

Step 3: Pairing equation 2 & 3

$$\begin{gathered} Here,a_1=6,b_1=2,c_1=-4 \\ {a}_2=9,b_2=-6,c_2=-12 \\ wehave, \\ \frac{a_1}{a_2}=\frac{6}{9}=\frac{2}{3} \\ \frac{b_1}{b_2}=\frac{2}{-6}=\frac{1}{-3} \\ clearly,\frac{a_1}{a_2}\ne \frac{b_1}{b_2} \end{gathered}$$

Here $$\begin{gathered} \frac{2}{3}\ne \frac{1}{-3} \\ wecansaythesetwopairsofequationshaveauniquepairsofequationshaveauniquesolution. \end{gathered}$$

Step 4: Pairing equation 1 & 2

$$\begin{gathered} a_1=3,b_1=-2,c_1=-4 \\ {a}_2=9,b_2=-6,c_2=-12 \\ wehave\frac{a_1}{a_2}=\frac{3}{9}=\frac{1}{3} \\ \frac{b_1}{b_2}=\frac{-2}{-6}=\frac{1}{3} \\ \frac{c_1}{c_2}=\frac{-4}{-12}=\frac{1}{3} \end{gathered}$$

Clearly, $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=\frac{1}{3}$

we can say the two pairs of equation have infinitely many solutions.

Step 4: Result

If pairing the given equations$3x-2y=4$ and $6x+2y=4$ . The system have unique solution. If pairing the given equation $6x+2y=4$and $9x-6y=12$the system have a unique solution. if pairing$3x-2y=4$ and $9x-6y=12$ the system have infinitely many solutions.