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NCERT Exemplar · Class 10 Mathematics Pair of Linear Equations in Two Variables

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EXERCISE 3.2 1–6 (part 2 of 5)

  1. Exercise 1

    Do the following pair of linear equations have no solution? Justify your answer.
    (i)
    2x+4y=312y+6x=6\begin{aligned} & 2 x+4 y=3 \\ & 12 y+6 x=6 \end{aligned}
    (ii)
    x=2yy=2x\begin{aligned} & x=2 y \\ & y=2 x \end{aligned}
    (iii)
    3x+y−3=02x+23y=2\begin{aligned} & 3 x+y-3=0 \\ & 2 x+\frac{2}{3} y=2 \end{aligned}

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    NCERT’s answer
    (i)
    Yes
    (ii)
    No
    (iii)
    No
    \[(i)\quad 2x+4y-3=0,\qquad 6x+12y-6=0 \] \[\frac{a_1}{a_2}=\frac{2}{6}=\frac13,\quad \frac{b_1}{b_2}=\frac{4}{12}=\frac13,\quad \frac{c_1}{c_2}=\frac{-3}{-6}=\frac12 \] \[\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\ \Rightarrow\ \text{parallel lines, no solution} \] \[(ii)\quad x-2y=0,\qquad 2x-y=0 \] \[\frac{a_1}{a_2}=\frac12,\quad \frac{b_1}{b_2}=\frac{-2}{-1}=2 \] \[\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\ \Rightarrow\ \text{intersecting lines, unique solution} \] \[(iii)\quad 3x+y-3=0,\qquad 2x+\frac23y-2=0\ \xrightarrow{\times 3}\ 6x+2y-6=0 \] \[\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=\frac12\ \Rightarrow\ \text{coincident lines, infinitely many solutions} \] Answer: (i) Yes (ii) No (iii) No

  2. Exercise 2

    Do the following equations represent a pair of coincident lines? Justify your answer.
    (i)
    3x+17y=37x+3y=7\begin{aligned} & 3 x+\frac{1}{7} y=3 \\ & 7 x+3 y=7 \end{aligned}
    (ii)
    −2x−3y=16y+4x=−2\begin{aligned} & -2 x-3 y=1 \\ & 6 y+4 x=-2 \end{aligned}
    (iii)
    x2+y+25=04x+8y+516=0\begin{aligned} & \frac{x}{2}+y+\frac{2}{5}=0 \\ & 4 x+8 y+\frac{5}{16}=0 \end{aligned}

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    NCERT’s answer
    (i)
    No
    (ii)
    Yes
    (iii)
    No
    \[(i)\quad 3x+\frac{1}{7}y-3=0,\qquad 7x+3y-7=0 \] \[\frac{a_1}{a_2}=\frac{3}{7},\quad \frac{b_1}{b_2}=\frac{1/7}{3}=\frac{1}{21} \] \[\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\ \Rightarrow\ \text{intersecting lines, not coincident} \] \[(ii)\quad -2x-3y-1=0,\qquad 4x+6y+2=0 \] \[\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=-\frac12\ \Rightarrow\ \text{coincident lines} \] \[(iii)\quad \frac{x}{2}+y+\frac{2}{5}=0,\qquad 4x+8y+\frac{5}{16}=0 \] \[\frac{a_1}{a_2}=\frac18=\frac{b_1}{b_2},\quad \frac{c_1}{c_2}=\frac{2/5}{5/16}=\frac{32}{25} \] \[\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\ \Rightarrow\ \text{parallel lines, not coincident} \] Answer: (i) No (ii) Yes (iii) No

  3. Exercise 3

    Are the following pair of linear equations consistent? Justify your answer.
    (i)
    −3x−4y=124y+3x=12\begin{aligned} & -3 x-4 y=12 \\ & 4 y+3 x=12 \end{aligned}
    (ii)
    35x−y=1215x−3y=16\begin{aligned} & \frac{3}{5} x-y=\frac{1}{2} \\ & \frac{1}{5} x-3 y=\frac{1}{6} \end{aligned}
    (iii)
    2ax+by=a4ax+2by−2a=0;a,b≠0\begin{aligned} & 2 a x+b y=a \\ & 4 a x+2 b y-2 a=0 ; a, b \neq 0 \end{aligned}
    (iv)
    x+3y=112(2x+6y)=22\begin{aligned} & x+3 y=11 \\ & 2(2 x+6 y)=22 \end{aligned}

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    NCERT’s answer
    (i)
    No
    (ii)
    Yes
    (iii)
    Yes
    (iv)
    No
    \[(i)\quad 3x+4y+12=0,\qquad 3x+4y-12=0 \] \[\frac{a_1}{a_2}=\frac{b_1}{b_2}=1\neq\frac{c_1}{c_2}=-1\ \Rightarrow\ \text{parallel, inconsistent} \] \[(ii)\quad \frac35x-y-\frac12=0,\qquad \frac15x-3y-\frac16=0 \] \[\frac{a_1}{a_2}=3\neq\frac{b_1}{b_2}=\frac13\ \Rightarrow\ \text{intersecting, consistent} \] \[(iii)\quad 2ax+by-a=0,\qquad 4ax+2by-2a=0\quad (a,b\neq0) \] \[\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=\frac12\ \Rightarrow\ \text{coincident, consistent} \] \[(iv)\quad x+3y-11=0,\qquad 4x+12y-22=0 \] \[\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac14\neq\frac{c_1}{c_2}=\frac12\ \Rightarrow\ \text{parallel, inconsistent} \] Answer: (i) No (ii) Yes (iii) Yes (iv) No

  4. Exercise 4

    For the pair of equations λx+3y=−72x+6y=14\begin{aligned} & \lambda x+3 y=-7 \\ & 2 x+6 y=14 \end{aligned} to have infinitely many solutions, the value of λ\displaystyle \lambda should be 1. Is the statement true? Give reasons.

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    NCERT’s answer
    No
    False. \[\lambda x+3y+7=0,\qquad 2x+6y-14=0 \] \[\frac{b_1}{b_2}=\frac{3}{6}=\frac12,\qquad \frac{c_1}{c_2}=\frac{7}{-14}=-\frac12 \] \[\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\quad\text{for every }\lambda \] Infinitely many solutions needs all three ratios equal; since \(\displaystyle b_1/b_2\neq c_1/c_2\) always, no \(\displaystyle \lambda\) gives that -- at \(\displaystyle \lambda=1\) the lines are merely parallel, so there is no solution at all.

  5. Exercise 5

    For all real values of c\displaystyle c, the pair of equations x−2y=85x−10y=c\begin{aligned} & x-2 y=8 \\ & 5 x-10 y=c \end{aligned} have a unique solution. Justify whether it is true or false.

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    NCERT’s answer
    False
    False. \[x-2y-8=0,\qquad 5x-10y-c=0 \] \[\frac{a_1}{a_2}=\frac{1}{5}=\frac{b_1}{b_2}=\frac{-2}{-10}\quad\text{for every }c \] Since \(\displaystyle a_1/a_2=b_1/b_2\) always, the lines are never intersecting: coincident (infinitely many solutions) when \(\displaystyle c=40\), parallel (no solution) otherwise -- a unique solution never occurs.

  6. Exercise 6

    The line represented by x=7\displaystyle x=7 is parallel to the x\displaystyle x-axis. Justify whether the statement is true or not.

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    NCERT’s answer
    Not true
    False. \[x=7 \] is the set of points \(\displaystyle (7,y)\) for every \(\displaystyle y\) -- a line perpendicular to the \(\displaystyle x\)-axis, meeting it at \(\displaystyle (7,0)\), and parallel instead to the \(\displaystyle y\)-axis. NCERT_Solution_Class10_Maths_Exemplar_Ch3_Ex3-2_Q6