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Question 11 Mark

Do the following pair of linear equations have no solution? Justify your answer:
2x + 4y = 3, 12y + 6x = 6

Answer

The system of linear equations has no solution if $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
2x + 4y = 3 and 6x + 12y = 6
Here, $\frac{\text{a}_1}{\text{a}_2}=\frac{2}{6}=\frac{1}{3}$, $\frac{\text{b}_1}{\text{b}_2}=\frac{4}{12}=\frac{1}{3}$, $\frac{\text{c}_1}{\text{c}_1}=\frac{3}{6}=\frac{1}{2}$
$\therefore\ \frac{2}{6}=\frac{4}{12}\neq\frac{3}{6}$
So, the given system of linear equations has no solution.

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Question 21 Mark

Are the following pair of linear equations consistent? Justify your answer:
-3x - 4y = 12 and 4y + 3x = 12.

Answer

For consistent system of linear equations $\text{a},\text{ b}\neq0$
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$ (infinitely many solutions)
$\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$ (unique solurion)
-3x - 4y = 12 and 4y + 3x = 12
Here, $\frac{\text{a}_1}{\text{a}_2}=\frac{-3}{3}=-1$, $\frac{\text{b}_1}{\text{b}_2}=\frac{-4}{4}=-1$ and $\frac{\text{c}_1}{\text{c}_2}=\frac{12}{12}=1$
$\therefore\ \frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
So, the given pair of linear equations is inconsistent and has no solution.

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Question 31 Mark

Do the following pair of linear equations have no solution? Justify your answer:x = 2y, y = 2x

Answer

The system of linear equations has no solution if $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
x - 2y = 0 and 2x - y = 0
Here, $\frac{\text{a}_1}{\text{a}_1}=\frac{1}{2}$
and $\frac{\text{b}_1}{\text{b}_2}=\frac{-2}{-1}=2$
So, the given system of linear equations does not satisfy $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$.

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Question 41 Mark

Are the following pair of linear equations consistent? Justify your answer:
$\frac{3}{5}\text{x}-\text{y}=\frac{1}{2}$ and $\frac{1}{5}\text{x}-3\text{y}=\frac{1}{6}$

Answer

For consistent system of linear equations $\text{a},\text{ b}\neq0$
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$ (infinitely many solutions)
$\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$ (unique solurion)
$\frac{3}{5}\text{x}-\text{y}=\frac{1}{2}$ and $\frac{1}{5}\text{x}-3\text{y}=\frac{1}{6}$
Here, $\frac{\text{a}_1}{\text{a}_2}=\frac{\frac{3}{5}}{\frac{1}{5}}=3$, $\frac{\text{b}_1}{\text{b}_2}=\frac{-1}{-3}=\frac{1}{3}$ and $\frac{\text{c}_1}{\text{c}_2}=\frac{\frac{1}{2}}{\frac{1}{6}}=\frac{6}{2}=3$
$\therefore\ \frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$
So, the givne pair of linear equations is consistent and has unique solution.

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Question 51 Mark

Are the following pair of linear equations consistent? Justify your answer:
2ax + by = a and 4ax + 2by - 2a = 0; a, b ≠ 0

Answer

For consistent system of linear equations a, $\text{b}\neq0$
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$ (infinitely many solutions)
$\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$ (unique solurion)
2ax + by = a and 4ax + 2by – 2a = 0
Here, $\frac{\text{a}_1}{\text{a}_2}=\frac{2\text{a}}{4\text{a}}=\frac{1}{2}$, $\frac{\text{b}_1}{\text{b}_2}=\frac{\text{b}}{2\text{b}}=\frac{1}{2}$ and $\frac{\text{c}_1}{\text{c}_2}=\frac{\text{a}}{2\text{a}}=\frac{1}{2}$
$\therefore\ \frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
So, the given pair of linear equations si consistent and has infinitely many solutions.

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Question 61 Mark

Do the following pair of linear equations have no solution? Justify your answer:$3\text{x}+\text{y}-3=0$, $2\text{x}+\frac{2}{3}\text{y}=2$

Answer

The system of linear equations has no solution if $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
$3\text{x}+\text{y}-3=0$ and $2\text{x}+\frac{2}{3}\text{y}=2$
Here, $\frac{\text{a}_1}{\text{a}_2}=\frac{3}{2}$, $\frac{\text{b}_1}{\text{b}_2}=\frac{1}{\frac{2}{3}}$, $\frac{\text{c}_1}{\text{c}_2}=\frac{-3}{-2}=\frac{3}{2}$
So, the given system of linear equations does not satisfy $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$.

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Question 71 Mark

Do the following equations represent a pair of coincident lines? Justify your answer:
$3\text{x}+\frac{1}{7}\text{y}=3$ and $7\text{x}+3\text{y}=7$

Answer

Condition for coincident lines $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}\ .....(\text{i})$
$3\text{x}+\frac{1}{7}\text{y}=3$ and $7\text{x}+3\text{y}=7$
Here, $\frac{\text{a}_1}{\text{a}_2}=\frac{3}{7}$, $\frac{\text{b}_1}{\text{b}_2}=\frac{\frac{1}{7}}{3}=\frac{1}{21}$ and $\frac{\text{c}_1}{\text{c}_2}=\frac{3}{7}$
So, the given system of linear equation does not satisfy condition (i).

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Question 81 Mark

Do the following equations represent a pair of coincident lines? Justify your answer:
$\frac{\text{x}}{2}+\text{y}+\frac{2}{5}=0$ and $4\text{x}+8\text{y}+\frac{5}{16}=0$

Answer

Condition for coincident lines $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}\ .....(\text{i})$
$\frac{\text{x}}{2}+\text{y}+\frac{2}{5}=0$ and $4\text{x}+8\text{y}+\frac{5}{16}=0$
Here, $\frac{\text{a}_1}{\text{a}_2}=\frac{\frac{1}{2}}{8}=\frac{1}{8}$, $\frac{\text{b}_1}{\text{b}_2}=\frac{1}{8}$ and $\frac{\text{c}_1}{\text{c}_2}=\frac{\frac{2}{5}}{\frac{5}{16}}=\frac{32}{25}$
$\therefore\ \frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
So, the given system of linear equations does not satisfy condition (i).

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Question 91 Mark

For all real values of c, the pair of equations x - 2y = 8 and 5x - 10y = c have a unique solution. Justify whether it is true or false.

Answer

(False) System of linear equations are
x - 2y = 8 .....(i)
5x - 10y = c .....(ii)
$\therefore\ \frac{\text{a}_1}{\text{a}_2}=\frac{1}{5}$, $\frac{\text{b}_1}{\text{b}_2}=\frac{-2}{-10}=\frac{1}{5}$ and $\frac{\text{c}_1}{\text{c}_2}=\frac{-8}{\text{-c}}=\frac{8}{\text{c}}$
As $\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$ so system of linear equations can never have unique solution.
Hence, the given statement is fasle.

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Question 101 Mark

Do the following equations represent a pair of coincident lines? Justify your answer:
-2x - 3y = 1 and 6y + 4x = -2

Answer

Condition for coincident lines $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}\ .....(\text{i})$
-2x - 3y = 1 and 6y + 4x = -2
Here, $\frac{\text{a}_1}{\text{a}_2}=\frac{-2}{4}=\frac{-1}{2}$, $\frac{\text{b}_1}{\text{b}_2}=\frac{-3}{6}=\frac{-1}{2}$ and $\frac{\text{c}_1}{\text{c}_2}=\frac{1}{-2}$
So, the given system of linear equations does not satisfy given condition (i).

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