SolveItClass 10 · NCERT

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

12 questions · 12 still being checked

EXERCISE 3.2 1–3 (part 2 of 3)

  1. Exercise 1

    Solve the following pair of linear equations by the substitution method.
    (i)
    x+y=14\displaystyle x+y=14 x−y=4x-y=4
    (ii)
    s−t=3\displaystyle s-t=3 s3+t2=6\frac{s}{3}+\frac{t}{2}=6
    (iii)
    3x−y=3\displaystyle 3 x-y=3 9x−3y=99 x-3 y=9
    (iv)
    0.2x+0.3y=1.30.4x+0.5y=2.3\begin{aligned} & 0.2 x+0.3 y=1.3 \\ & 0.4 x+0.5 y=2.3 \end{aligned}
    (v)
    2x+3y=0\displaystyle \sqrt{2} x+\sqrt{3} y=0 3x−8y=0\sqrt{3} x-\sqrt{8} y=0
    (vi)
    3x2−5y3=−2x3+y2=136\begin{aligned} & \frac{3 x}{2}-\frac{5 y}{3}=-2 \\ & \frac{x}{3}+\frac{y}{2}=\frac{13}{6} \end{aligned}
    NCERT’s answer
    (i)
    $\displaystyle x=9, y=5$ (ii) $\displaystyle s=9, t=6$ (iii) $\displaystyle y=3 x-3$, where $\displaystyle x$ can take any real value, i.e., infinitely many solutions. (iv) $\displaystyle x=2, y=3$ (v) $\displaystyle x=0, y=0$ (vi) $\displaystyle x=2, y=3$

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  2. Exercise 2

    Solve 2x+3y=11\displaystyle 2 x+3 y=11 and 2x−4y=−24\displaystyle 2 x-4 y=-24 and hence find the value of ' m\displaystyle m ' for which y=mx+3\displaystyle y=m x+3.
    NCERT’s answer
    $\displaystyle x=-2, y=5 ; m=-1$

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  3. Exercise 3

    Form the pair of linear equations for the following problems and find their solution by substitution method.
    (i)
    The difference between two numbers is 26\displaystyle 26 and one number is three times the other. Find them.
    (ii)
    The larger of two supplementary angles exceeds the smaller by 18\displaystyle 18 degrees. Find them.
    (iii)
    The coach of a cricket team buys 7\displaystyle 7 bats and 6\displaystyle 6 balls for ₹ 3800. Later, she buys 3\displaystyle 3 bats and 5\displaystyle 5 balls for ₹ 1750\displaystyle 1750 . Find the cost of each bat and each ball.
    (iv)
    The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10\displaystyle 10 km, the charge paid is ₹ 105\displaystyle 105 and for a journey of 15\displaystyle 15 km, the charge paid is ₹ 155\displaystyle 155 . What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25\displaystyle 25 km?
    (v)
    A fraction becomes 911\displaystyle \frac{9}{11}, if 2\displaystyle 2 is added to both the numerator and the denominator. If, 3\displaystyle 3 is added to both the numerator and the denominator it becomes 56\displaystyle \frac{5}{6}. Find the fraction.
    (vi)
    Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob's age was seven times that of his son. What are their present ages?
    NCERT’s answer
    (i)
    $\displaystyle x-y=26, x=3 y$, where $\displaystyle x$ and $\displaystyle y$ are two numbers $\displaystyle (x>y) ; x=39, y=13$. (ii) $\displaystyle x-y=18, x+y=180$, where $\displaystyle x$ and $\displaystyle y$ are the measures of the two angles in degrees; $\displaystyle x=99, y=81$. (iii) $\displaystyle 7 x+6 y=3800,3 x+5 y=1750$, where $\displaystyle x$ and $\displaystyle y$ are the costs (in ₹) of one bat and one ball respectively; $\displaystyle x=500, y=50$. (iv) $\displaystyle x+10 y=105, x+15 y=155$, where $\displaystyle x$ is the fixed charge (in ₹) and $\displaystyle y$ is the charge (in ₹ per km); $\displaystyle x=5, y=10$; ₹ $\displaystyle 255$ . (v) $\displaystyle 11 x-9 y+4=0,6 x-5 y+3=0$, where $\displaystyle x$ and $\displaystyle y$ are numerator and denominator of the fraction; $\displaystyle \frac{7}{9}(x=7, y=9)$. (vi) $\displaystyle x-3 y-10=0, x-7 y+30=0$, where $\displaystyle x$ and $\displaystyle y$ are the ages in years of Jacob and his son; $\displaystyle x=40, y=10$.

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