Ex.3.2 Q4 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 are consistent / inconsistent? If consistent, obtain the Solution graphically:

(i) \(x + y = 5, \;2x + 2y = 10\)

(ii) \(x-y = 8, \;3x-3y = 16\)

(iii) \(2x + y-6 = 0, \;4x-2y-4 = 0\)

(iv) \(2x-2y-2 = 0, \;4x-4y-5 = 0\)

 Video Solution
Pair Of Linear Equations In Two Variables
Ex 3.2 | Question 4

Text Solution

What is Unknown?

Whether the linear equations are consistent or inconsistent and graphical solution, if consistent.

Reasoning:

Consistent means pair of linear equations have one solution or infinitely many solutions.

\[\begin{align}{a_1}x + {b_1}y + {c_1} &= 0\\{a_2}x + {b_2}y + {c_2} &= 0\end{align}\]

\[\begin{align}\frac{{{a_1}}}{{{a_2}}} &\ne \frac{{{b_1}}}{{{b_2}}}\,\,\,\,\,\,\,\,\,\,\,\;\; \quad \;\left( {{\bf{Intersecting}}{\rm{\; }}{\bf{Lines}}/{\bf{One}}\;{\bf{Solution}}} \right)\\
\,\frac{{{a_1}}}{{{a_2}}} &= \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}} \quad \,\;\left( {{\bf{Coincident}}{\rm{ \;}}{\bf{Lines}}/{\bf{Infinitely}}\;{\bf{many}}\;{\bf{Solutions}}} \right)\end{align}\]

Inconsistent means, the lines may be parallel and do not have any Solution)

\[\begin{align}\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \,\,\frac{{{c_1}}}{{{c_2}}}\,\,\,\,\,\,\,\,\,\left( {{\bf{Parallel}}{\rm{\; }}{\bf{lines}}/{\bf{No}}\;{\bf{Solution}}} \right)\end{align}\]

 (i) What is Known?

\(\begin{align}x + y-5 &= 0\\2x + 2y-10 &= 0\end{align}\)

Steps:

\[\begin{align}\frac{{{a_1}}}{{{a_2}}} &= \frac{1}{2}\\\frac{{{b_1}}}{{{b_2}}} &= \frac{1}{2}\\\frac{{{c_1}}}{{{c_2}}} &= \frac{{ - 5}}{{ - 10}} \\&= \frac{1}{2}\end{align}\]

From above

\[\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \,\,\frac{{{c_1}}}{{{c_2}}}\]

Therefore, lines are coincident and have infinitely many solutions.

Hence, they are consistent.

\[\begin{align}x + y-5 &= 0\\y &= - x + 5\\y& = 5-x\end{align}\]

 

\(x\)

\(1\)

\(2\)

\(y = 5-x\)

\(4\)

\(3\)

\[\begin{align}2x + 2y-10 &= 0\\2y &= 10-2x\\y &= 5-x\end{align}\]

\(x\) \(3\) \(4\)
\(y = 5-x\) \(2\) \(1\)

All the points on coincident line are solutions for the given pair of equations.

(ii) What is Known?

\(\begin{align}x-y-8 &= 0\\3x-3y-16 &= 0\end{align}\)

Steps:

\[\begin{align}\frac{{{a_1}}}{{{a_2}}} &= \frac{1}{3}\\\frac{{{b_1}}}{{{b_2}}} &= \frac{{ - 1}}{{ - 3}}\\&= \frac{1}{3}\\\frac{{{c_1}}}{{{c_2}}} &= \frac{{ - 8}}{{ - 16}} \\&= \frac{1}{2}\end{align}\]

From above

\[\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \,\,\frac{{{c_1}}}{{{c_2}}}\,\]

Therefore, lines are parallel and have no solution,

Hence, the pair of equations are inconsistent.

\[\begin{align}x - y - 8 &= 0\\y &= x - 8\end{align}\]

\(x\) \(8\) \(6\)
\(y{\rm{ }} = {\rm{ }}x{\rm{ }}-{\rm{ }}8\) \(0\) \( - 2\)

\[\begin{align}3x - 3y - 16 &= 0\\3y &= 3x - 16\\y &= \frac{{3x - 16}}{3}\end{align}\]

\(x\) \(2\) \(4\)
\(y = \frac{{3x - 16}}{3}\) \( - 3.3\) \( - 1.3\)

(iii) What is Known?

\(\begin{align}2x + y-6 &= 0\\4x-2y-4 &= 0\end{align}\)

Steps:

\[\begin{align}\frac{{{a_1}}}{{{a_2}}} &= \frac{2}{4}\\& = \frac{1}{2}\\\frac{{{b_1}}}{{{b_2}}} &= \frac{1}{{ - 2}} \\&= - \frac{1}{2}\\\frac{{{c_1}}}{{{c_2}}} &= \frac{{ - 6}}{{ - 4}} \\&= \frac{3}{2}\end{align}\]

From above:

\[\frac{{{a_1}}}{{{a_2}}} \ne \frac{{{b_1}}}{{{b_2}}}\,\]

Therefore, lines are intersecting and have one solution.

Hence, they are consistent.

\[\begin{align}2x + y-6 &= 0\\y &= 6-2x\end{align}\]

\(x\) \(0\) \(2\)
\(y = 6-2x\) \(6\) \(2\)

\[\begin{align}4x - 2y - 4 &= 0\\2y &= 4x - 4\\y& = 2x - 2\end{align}\]

\(x\) \(2\) \(3\)
\(y = 2x - 2\) \(2\) \(4\)

\(x=2\) and \(y=2\) are solutions for the given pair of equations.

(iv) What is Known?

\(\begin{align}2x-2y-2 &= 0\\4x-4y-5& = 0 \\\end{align} \)

Steps:

\[\begin{align}\frac{{{a_1}}}{{{a_2}}} &= \frac{2}{4} \\&= \frac{1}{2}\\\frac{{{b_1}}}{{{b_2}}} &= \frac{{ - 2}}{{ - 4}}\\&= \frac{1}{2}\\\frac{{{c_1}}}{{{c_2}}} &= \frac{{ - 2}}{{ - 5}}\\& = \frac{2}{5}\end{align}\]

From above:

\[\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \,\,\frac{{{c_1}}}{{{c_2}}}\,\]

Therefore, lines are parallel and have no solution,

Hence, the pair of equations are inconsistent.

\[\begin{align}2x - 2y - 2 &= 0\\2y &= 2x - 2\\y &= x - 1\end{align}\]

\(x\) \(1\) \(3\)
\(y = x-1\) \(0\) \(2\)

\[\begin{align}4x - 4y - 5 &= 0\\4y &= 4x - 5\\y& = \frac{{4x - 5}}{4}\end{align}\]

\(x\) \(4\) \(3\)
\(y = \frac{{4x - 5}}{4}\) \(2.8\) \(1.8\)