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

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EXERCISE 3.3 1–10 (part 3 of 5)

  1. Exercise 1

    For which value(s) of λ\displaystyle \lambda, do the pair of linear equations λx+y=λ2 and x+λy=1 have \lambda x+y=\lambda^2 \text { and } x+\lambda y=1 \quad \text { have }
    (i)
    no solution?
    (ii)
    infinitely many solutions?
    (iii)
    a unique solution?

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    NCERT’s answer
    (i)
    \(\displaystyle \lambda=-1\)
    (ii)
    \(\displaystyle \lambda=1\)
    (iii)
    All real values of \(\displaystyle \lambda\) except \(\displaystyle \pm 1\).
    \[a_1=\lambda,\ b_1=1,\ c_1=\lambda^2; \qquad a_2=1,\ b_2=\lambda,\ c_2=1 \]
    \[\frac{a_1}{a_2}=\lambda, \quad \frac{b_1}{b_2}=\frac{1}{\lambda}, \quad \frac{c_1}{c_2}=\lambda^2 \]
    (i)
    No solution needs \(\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2} \).
    \[\lambda=\frac{1}{\lambda} \implies \lambda^2=1 \implies \lambda=\pm1 \]
    \[\lambda=1:\ \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=1 \quad\text{(rejected)} \]
    \[\lambda=-1:\ \frac{a_1}{a_2}=\frac{b_1}{b_2}=-1\neq\frac{c_1}{c_2}=1 \]
    (ii)
    Infinitely many needs all three ratios equal, true only at \(\displaystyle \lambda=1\).
    (iii)
    Unique solution needs \(\displaystyle \frac{a_1}{a_2}\neq\frac{b_1}{b_2} \):
    \[\lambda\neq\frac{1}{\lambda} \implies \lambda\neq\pm1 \]
    Answer: (i) \(\displaystyle \lambda=-1\) (ii) \(\displaystyle \lambda=1\) (iii) \(\displaystyle \lambda\neq\pm1\)

  2. Exercise 2

    For which value(s) of k\displaystyle k will the pair of equations kx+3y=k−312x+ky=k\begin{aligned} & k x+3 y=k-3 \\ & 12 x+k y=k \end{aligned} have no solution?

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    NCERT’s answer
    \(\displaystyle k=-6\)
    No solution needs \(\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2} \). \[\frac{k}{12}=\frac{3}{k} \implies k^2=36 \implies k=\pm6 \] \[k=6:\ \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=\frac12 \quad\text{(rejected)} \] \[k=-6:\ \frac{a_1}{a_2}=\frac{b_1}{b_2}=-\frac12, \quad \frac{c_1}{c_2}=\frac{k-3}{k}=\frac{-9}{-6}=\frac32 \] Ratios differ, so \(\displaystyle k=-6\) gives no solution. Answer: \(\displaystyle k=-6\)

  3. Exercise 3

    For which values of a\displaystyle a and b\displaystyle b, will the following pair of linear equations have infinitely many solutions? x+2y=1(a−b)x+(a+b)y=a+b−2\begin{aligned} & x+2 y=1 \\ & (a-b) x+(a+b) y=a+b-2 \end{aligned}

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    NCERT’s answer
    \(\displaystyle a=3, b=1\)
    Infinitely many solutions need \[\frac{1}{a-b}=\frac{2}{a+b}=\frac{1}{a+b-2} \] First and second: \[a+b=2(a-b) \implies a=3b \] First and third: \[a+b-2=a-b \implies 2b=2 \implies b=1 \] So \(\displaystyle a=3b=3\). Check: \[a-b=2,\ a+b=4,\ a+b-2=2 \implies \frac12=\frac24=\frac12 \] Answer: \(\displaystyle a=3,\ b=1\)

  4. Exercise 4

    Find the value(s) of p\displaystyle p in (i) to (iv) and p\displaystyle p and q\displaystyle q in (v) for the following pair of equations:
    (i)
    3x−y−5=0\displaystyle 3 x-y-5=0 and 6x−2y−p=0\displaystyle 6 x-2 y-p=0, if the lines represented by these equations are parallel.
    (ii)
    −x+py=1\displaystyle -x+p y=1 and px−y=1\displaystyle p x-y=1, if the pair of equations has no solution.
    (iii)
    −3x+5y=7\displaystyle -3 x+5 y=7 and 2px−3y=1\displaystyle 2 p x-3 y=1, if the lines represented by these equations are intersecting at a unique point.
    (iv)
    2x+3y−5=0\displaystyle 2 x+3 y-5=0 and px−6y−8=0\displaystyle p x-6 y-8=0, if the pair of equations has a unique solution.
    (v)
    2x+3y=7\displaystyle 2 x+3 y=7 and 2px+py=28−qy\displaystyle 2 p x+p y=28-q y, if the pair of equations have infinitely many solutions.

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    NCERT’s answer
    (i)
    All real values of \(\displaystyle p\) except 10.
    (ii)
    \(\displaystyle p=1\)
    (iii)
    All real values of \(\displaystyle p\) except \(\displaystyle \frac{9}{10}\).
    (iv)
    All real values of \(\displaystyle p\) except \(\displaystyle -4\).
    (v)
    \(\displaystyle p=4, q=8\)
    (i)
    \(\displaystyle 3x-y-5=0,\ 6x-2y-p=0\):
    \[\frac{a_1}{a_2}=\frac36=\frac12=\frac{-1}{-2}=\frac{b_1}{b_2} \]
    Parallel needs \(\displaystyle \frac{c_1}{c_2}\neq\frac12 \):
    \[\frac{-5}{-p}\neq\frac12 \implies p\neq10 \]
    (ii)
    \(\displaystyle -x+py=1,\ px-y=1\). No solution needs \(\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2} \):
    \[\frac{-1}{p}=-p \implies p^2=1 \implies p=\pm1 \]
    \[p=1:\ \frac{a_1}{a_2}=\frac{b_1}{b_2}=-1\neq\frac{c_1}{c_2}=1 \]
    \[p=-1:\ \text{all three ratios}=1 \quad\text{(rejected)} \]
    (iii)
    \(\displaystyle -3x+5y=7,\ 2px-3y=1\). Unique intersection needs \(\displaystyle a_1b_2-a_2b_1\neq0\):
    \[(-3)(-3)-(2p)(5)=9-10p\neq0 \implies p\neq\frac{9}{10} \]
    (iv)
    \(\displaystyle 2x+3y-5=0,\ px-6y-8=0\). Unique solution needs \(\displaystyle a_1b_2-a_2b_1\neq0\):
    \[2(-6)-p(3)=-12-3p\neq0 \implies p\neq-4 \]
    (v)
    \(\displaystyle 2x+3y=7,\ 2px+(p+q)y=28\). Infinitely many needs
    \[\frac{2}{2p}=\frac{3}{p+q}=\frac{7}{28} \]
    \[\frac1p=\frac14 \implies p=4, \qquad \frac{3}{p+q}=\frac14 \implies p+q=12 \implies q=8 \]
    Answer: (i) \(\displaystyle p\neq10\) (ii) \(\displaystyle p=1\) (iii) \(\displaystyle p\neq\frac{9}{10}\) (iv) \(\displaystyle p\neq-4\) (v) \(\displaystyle p=4,\ q=8\)

  5. Exercise 5

    Two straight paths are represented by the equations x−3y=2\displaystyle x-3 y=2 and −2x+6y=5\displaystyle -2 x+6 y=5. Check whether the paths cross each other or not.

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    NCERT’s answer
    Do not cross each other.
    \[\frac{a_1}{a_2}=\frac{1}{-2}, \quad \frac{b_1}{b_2}=\frac{-3}{6}=\frac{1}{-2}, \quad \frac{c_1}{c_2}=\frac{2}{5} \] \[\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2} \] The lines are parallel and distinct, so they never meet. Answer: The paths do not cross; they run parallel.

  6. Exercise 6

    Write a pair of linear equations which has the unique solution x=−1,y=3\displaystyle x=-1, y=3. How many such pairs can you write?

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    NCERT’s answer
    \(\displaystyle x-y=-4\) \(\displaystyle 2x+3y=7\); infinitely many pairs.
    Any two independent lines through \(\displaystyle (-1,3)\) work, so infinitely many pairs exist. One choice: \[x+y=2 \quad(-1+3=2), \qquad x-y=-4 \quad(-1-3=-4) \] \[\frac{a_1}{a_2}=\frac{1}{1}\neq\frac{b_1}{b_2}=\frac{1}{-1} \] so the solution is unique; adding the two equations gives \(\displaystyle 2x=-2\), i.e. \(\displaystyle x=-1,\ y=3\). Answer: e.g. \(\displaystyle x+y=2,\ x-y=-4\); infinitely many such pairs are possible.

  7. Exercise 7

    If 2x+y=23\displaystyle 2 x+y=23 and 4x−y=19\displaystyle 4 x-y=19, find the values of 5y−2x\displaystyle 5 y-2 x and yx−2\displaystyle \frac{y}{x}-2.

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    NCERT’s answer
    \(\displaystyle 31, \frac{-5}{7}\)
    \[2x+y=23, \qquad 4x-y=19 \] \[\text{Adding: } 6x=42 \implies x=7 \] \[y=23-2x=23-14=9 \] \[5y-2x=5(9)-2(7)=45-14=31 \] \[\frac{y}{x}-2=\frac{9}{7}-2=-\frac{5}{7} \] Answer: \(\displaystyle 5y-2x=31,\quad \dfrac{y}{x}-2=-\dfrac{5}{7}\)

  8. Exercise 8

    Find the values of x\displaystyle x and y\displaystyle y in the following rectangle [see Fig. 3.2\displaystyle 3.2]. NCERT_Question_Class10_Maths_Exemplar_Ch3_Ex3-3_Q8

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    NCERT’s answer
    \(\displaystyle x=1, y=4\)
    Opposite sides of a rectangle are equal.\[x+3y=13, \qquad 3x+y=7 \] \[y=7-3x \] \[x+3(7-3x)=13 \implies x+21-9x=13 \implies -8x=-8 \implies x=1 \] \[y=7-3(1)=4 \] Answer: \(\displaystyle x=1,\ y=4\)

  9. Exercise 9

    Solve the following pairs of equations:
    (i)
    x+y=3.30.63x−2y=−1,3x−2y≠0\begin{aligned} & x+y=3.3 \\ & \frac{0.6}{3 x-2 y}=-1, \quad 3 x-2 y \neq 0 \end{aligned}
    (ii)
    x3+y4=45x6−y8=4\begin{aligned} & \frac{x}{3}+\frac{y}{4}=4 \\ & \frac{5 x}{6}-\frac{y}{8}=4 \end{aligned}
    (iii)
    4x+6y=156x−8y=14,y≠0\begin{aligned} & 4 x+\frac{6}{y}=15 \\ & 6 x-\frac{8}{y}=14, y \neq 0 \end{aligned}
    (iv)
    12x−1y=−11x+12y=8,x,y≠0\begin{aligned} & \frac{1}{2 x}-\frac{1}{y}=-1 \\ & \frac{1}{x}+\frac{1}{2 y}=8, \quad x, y \neq 0 \end{aligned}
    (v)
    43x+67y=−2467x+43y=24\begin{aligned} & 43 x+67 y=-24 \\ & 67 x+43 y=24 \end{aligned}
    (vi)
    xa+yb=a+bxa2+yb2=2,a,b≠0\begin{aligned} & \frac{x}{a}+\frac{y}{b}=a+b \\ & \frac{x}{a^2}+\frac{y}{b^2}=2, \quad a, b \neq 0 \end{aligned}
    (vii)
    2xyx+y=32xy2x−y=−310,x+y≠0,2x−y≠0\begin{aligned} & \frac{2 x y}{x+y}=\frac{3}{2} \\ & \frac{x y}{2 x-y}=\frac{-3}{10}, \quad x+y \neq 0,2 x-y \neq 0 \end{aligned}

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    NCERT’s answer
    (i)
    \(\displaystyle x=1.2, y=2.1\)
    (ii)
    \(\displaystyle x=6, y=8\)
    (iii)
    \(\displaystyle x=3, y=2\)
    (iv)
    \(\displaystyle x=\frac{1}{6}, y=\frac{1}{4}\)
    (v)
    \(\displaystyle x=1, y=-1\)
    (vi)
    \(\displaystyle x=a^2, y=b^2\)
    (vii)
    \(\displaystyle x=\frac{1}{2}, y=\frac{-3}{2}\)
    (i)
    \[x+y=3.3 \]
    \[\frac{0.6}{3x-2y}=-1 \implies 3x-2y=-0.6 \]
    \[x=3.3-y \implies 3(3.3-y)-2y=-0.6 \implies 9.9-5y=-0.6 \]
    \[y=2.1,\quad x=1.2 \]
    (ii)
    \[\frac{x}{3}+\frac{y}{4}=4 \implies 4x+3y=48 \]
    \[\frac{5x}{6}-\frac{y}{8}=4 \implies 20x-3y=96 \]
    \[24x=144 \implies x=6,\quad y=8 \]
    (iii)
    Let \(\displaystyle u=\frac{1}{y}\).
    \[4x+6u=15,\qquad 6x-8u=14 \]
    \[16x+24u=60,\qquad 18x-24u=42 \]
    \[34x=102 \implies x=3 \implies u=\frac12 \implies y=2 \]
    (iv)
    Let \(\displaystyle u=\frac1x,\ v=\frac1y\).
    \[\frac{u}{2}-v=-1,\qquad u+\frac{v}{2}=8 \]
    \[u-2v=-2,\qquad 2u+v=16 \]
    \[u=6,\quad v=4 \implies x=\frac16,\ y=\frac14 \]
    (v)
    \[43x+67y=-24,\qquad 67x+43y=24 \]
    \[\text{Add: } 110(x+y)=0 \implies x+y=0 \]
    \[\text{Subtract: } 24(x-y)=48 \implies x-y=2 \]
    \[x=1,\quad y=-1 \]
    (vi)
    \[\frac{x}{a}+\frac{y}{b}=a+b \quad (1),\qquad \frac{x}{a^2}+\frac{y}{b^2}=2 \quad (2) \]
    \[\frac{1}{a}\times(1):\ \frac{x}{a^2}+\frac{y}{ab}=\frac{a+b}{a} \]
    \[\text{Subtract (2): } y\left(\frac{1}{ab}-\frac{1}{b^2}\right)=\frac{a+b}{a}-2 \implies y\cdot\frac{b-a}{ab^2}=\frac{b-a}{a} \implies y=b^2 \]
    \[\text{From (1): } \frac{x}{a}=a+b-b=a \implies x=a^2 \]
    (vii)
    Let \(\displaystyle u=\frac1x,\ v=\frac1y\).
    \[\frac{2xy}{x+y}=\frac32 \implies \frac{x+y}{xy}=\frac43 \implies u+v=\frac43 \]
    \[\frac{xy}{2x-y}=-\frac{3}{10} \implies \frac{2x-y}{xy}=-\frac{10}{3} \implies 2v-u=-\frac{10}{3} \]
    \[3v=-2 \implies v=-\frac23,\quad u=2 \]
    \[x=\frac1u=\frac12,\quad y=\frac1v=-\frac32 \]
    Answer: (i) \(\displaystyle x=1.2,\ y=2.1\) (ii) \(\displaystyle x=6,\ y=8\) (iii) \(\displaystyle x=3,\ y=2\) (iv) \(\displaystyle x=\frac16,\ y=\frac14\) (v) \(\displaystyle x=1,\ y=-1\) (vi) \(\displaystyle x=a^2,\ y=b^2\) (vii) \(\displaystyle x=\frac12,\ y=-\frac32\)

  10. Exercise 10

    Find the solution of the pair of equations x10+y5−1=0\displaystyle \frac{x}{10}+\frac{y}{5}-1=0 and x8+y6=15\displaystyle \frac{x}{8}+\frac{y}{6}=15. Hence, find λ\displaystyle \lambda, if y=λx+5\displaystyle y=\lambda x+5.

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    NCERT’s answer
    \(\displaystyle x=340, y=-165\); \(\displaystyle \lambda=-\frac{1}{2}\)
    \[\frac{x}{10}+\frac{y}{5}-1=0 \implies x+2y=10 \quad (1) \] \[\frac{x}{8}+\frac{y}{6}=15 \implies 3x+4y=360 \quad (2) \] \[(2)-2\times(1):\ x=340 \] \[(1):\ 340+2y=10 \implies y=-165 \] \[y=\lambda x+5 \implies -165=340\lambda+5 \implies \lambda=\frac{-170}{340}=-\frac12 \]Answer: \(\displaystyle x=340,\ y=-165,\ \lambda=-\frac12\)