Ex.3.5 Q1 Pair of Linear Equations in Two Variables Solution - NCERT Maths Class 10

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Question

Which of the following pairs of linear equations has unique solution, no solution or infinitely many solutions? In case there is a unique solution, find it by using cross multiplication method.

(i) \(\begin{align}\,&x-3y-3  =  0 \\&3x-9y-2  =  0\end{align}\)

(ii) \(\begin{align}&2x + y  = 5\\&3x + 2y = 8\end{align}\)

(iii) \(\begin{align}3x-5y &= 20\\6x-10y &= 40\end{align}\)

(iv) \(\begin{align}x-3y-7 = 0\\3x-3y-15 = 0\end{align}\)

 Video Solution
Pair Of Linear Equations In Two Variables
Ex 3.5 | Question 1

Text Solution

Steps:

\(\begin{align}\rm{(i)} \quad x-3y-3 &= 0\\3x-9y-2 &= 0\end{align}\)

\[\begin{align}& \frac{{{a}_{1}}}{{{a}_{2}}}=\frac{1}{3},\quad \frac{{{b}_{1}}~}{{{b}_{2}}}=\frac{-3}{-9}~=~\frac{1}{3},\quad\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{-3}{-2}=~\frac{3}{2} \\ & \frac{{{a}_{1}}}{{{a}_{2}}~}=\frac{{{b}_{1}}}{{{b}_{2}}~}\ne \frac{{{c}_{1}}}{~{{c}_{2}}} \\\end{align}\]

Therefore, the given sets of lines are parallel to each other and will not intersect each other thus, there will be no solution for these equations.

Steps:

\(\begin{align}(\rm{ii}) \quad 2x + y &= 5\\3x + 2y &= 8\end{align}\)

\[\begin{align}& 2x+y-5=0 \\ & 3x+2y-8=0 \\ & \frac{{{a}_{1}}~}{{{a}_{2}}}=\frac{2}{3},\quad\frac{{{b}_{1}}~}{{{b}_{2}}}=\frac{1}{2~}\,,\quad\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{-5}{-8}=\frac{5}{8} \\ & \frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}} \\\end{align}\]

Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.

By cross-multiplication method,

\[\begin{align} \frac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}&=\frac{y}{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}=\frac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \\  \frac{x}{-8+10}&=\frac{y}{-15+16}=\frac{1}{4-3} \\  \frac{x}{2}=\frac{y}{1}&=1 \\  \frac{x}{2}\,\,&=\,\,1\;\;\text{ and }\;\;\frac{y}{1}\,\,=\,\,1 \\\end{align}\]

\(\therefore\) \(x = 2\) and  \(y = 1\)

Steps:

\(\begin{align}( \rm{iii})\quad 3x-5y &= 20\\6x-10y &= 40 \end{align}\)

\[\begin{align}3x-5y - 20& = 0\\
6x-10y - 40& = 0\\\frac{{{a_1}}}{{{a_2}}}& = \,\frac{{3}}{6} = \frac{1}{2},\frac{{{b_1}}}{{{b_2}}} = \frac{{5}}{{10}} = \frac{1}{2},\frac{{{c_1}}}{{{c_2}}} = \frac{{ - 20}}{{ - 40}} = \frac{1}{2}\\\frac{{{a_1}}}{{{a_2}}} &= \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}\end{align}\]

Therefore, the given sets of lines will be overlapping each other i.e., the lines will be coincident to each other and thus, there are infinite solutions possible for these equations.

Steps:

\(\begin{align}(\rm{iv})\quad x-3y-7 &= 0\\3x - 3y - 15 &= 0\end{align}\)

\[\begin{align} \frac{{{a}_{1}}}{{{a}_{2}}}&=\frac{1}{3},\quad\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{-3}{-3}~=1,\quad \frac{{{c}_{1}}}{{{c}_{2}}}=\frac{-7}{-15}=\frac{7}{15} \\  \frac{{{a}_{1}}}{{{a}_{2}}}\,\,&\ne \,\frac{{{b}_{l}}}{{{b}_{2}}}\end{align}\]

Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.

By cross-multiplication method,

\[\begin{align}\frac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}&=\frac{y}{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}=\frac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}} \\\frac{x}{45-21}&=\frac{y}{-21-(-15)}=\frac{1}{-3-(-9)} \\\frac{x}{24}&=\frac{y}{-6}=\frac{1}{6} \\\frac{x}{24}&=\frac{1}{6}\,\quad\text{and}\,\quad\frac{y}{-6}=\frac{1}{6} \\  x&=4\,\quad\text{and}\,\quad y=-1\end{align}\]

\(\therefore\)  \(x=4,\) and \(y=-1\) 

  
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