SolveItNCERT · CBSE Boards

NCERT Exemplar · Class 9 Mathematics Linear Equations in Two Variables

42 questions · 42 still being checked

EXERCISE 4.4 1–6 (part 5 of 5)

  1. Exercise 1

    Show that the points A(1,2),B(−1,−16)\displaystyle \mathrm{A}(1,2), \mathrm{B}(-1,-16) and C(0,−7)\displaystyle \mathrm{C}(0,-7) lie on the graph of the linear equation y=9x−7\displaystyle y=9 x-7.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[y = 9x - 7 \] \[x=1 \implies y = 9(1)-7 = 2 \quad \text{(matches A)} \] \[x=-1 \implies y = 9(-1)-7 = -16 \quad \text{(matches B)} \] \[x=0 \implies y = 9(0)-7 = -7 \quad \text{(matches C)} \] Each pair satisfies the equation.Answer: A, B and C all lie on \(\displaystyle y = 9x - 7\).

  2. Exercise 2

    The following observed values of x\displaystyle x and y\displaystyle y are thought to satisfy a linear equation. Write the linear equation :
    x\displaystyle x6\displaystyle 6-6\displaystyle 6
    y\displaystyle y-2\displaystyle 26\displaystyle 6
    Draw the graph using the values of x,y\displaystyle x, y as given in the above table. At what points the graph of the linear equation (i) cuts the x\displaystyle x-axis (ii) cuts the y\displaystyle y-axis

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    The graph cuts the $\displaystyle x$-axis at $\displaystyle (3,0)$ and the $\displaystyle y$-axis at $\displaystyle (0,2)$.
    NCERT_Solution_Class9_Maths_Exemplar_Ch4_Ex4-4_Q2 The line through the plotted points \(\displaystyle (6,-2)\) and \(\displaystyle (-6,6)\), read off the graph, meets \[\text{the } x\text{-axis at } (3,0), \qquad \text{the } y\text{-axis at } (0,2) \] Intercept form of the line: \[\frac{x}{3}+\frac{y}{2}=1 \implies 2x+3y=6 \] Checking on the table's own points: \[(6,-2):\ 2(6)+3(-2)=6, \qquad (-6,6):\ 2(-6)+3(6)=6 \] Answer: \(\displaystyle 2x+3y=6\); (i) it cuts the \(\displaystyle x\)-axis at \(\displaystyle (3,0)\) and (ii) the \(\displaystyle y\)-axis at \(\displaystyle (0,2)\).

  3. Exercise 3

    Draw the graph of the linear equation 3x+4y=6\displaystyle 3 x+4 y=6. At what points, the graph cuts the x\displaystyle x-axis and the y\displaystyle y-axis.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    The graph cuts the $\displaystyle x$-axis at $\displaystyle (2,0)$ and the $\displaystyle y$-axis $\displaystyle \left(0, \frac{3}{2}\right)$.
    \[3x+4y=6 \] \[y=0 \implies 3x=6 \implies x=2 \] \[x=0 \implies 4y=6 \implies y=\frac{3}{2} \] NCERT_Solution_Class9_Maths_Exemplar_Ch4_Ex4-4_Q3 Answer: The graph cuts the \(\displaystyle x\)-axis at \(\displaystyle (2,0)\) and the \(\displaystyle y\)-axis at \(\displaystyle \left(0,\frac{3}{2}\right)\).

  4. Exercise 4

    The linear equation that converts Fahrenheit (F) to Celsius (C) is given by the relation C=5 F−1609\mathrm{C}=\frac{5 \mathrm{~F}-160}{9} (i) If the temperature is 86\displaystyle 86°F, what is the temperature in Celsius? (ii) If the temperature is 35\displaystyle 35°C, what is the temperature in Fahrenheit? (iii) If the temperature is 0\displaystyle 0°C what is the temperature in Fahrenheit and if the temperature is 0\displaystyle 0°F, what is the temperature in Celsius? (iv) What is the numerical value of the temperature which is same in both the scales?

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (i)
    $\displaystyle 30$°C (ii) $\displaystyle 95$°F (iii) $\displaystyle 32^{\circ} \mathrm{F}, \frac{-160}{9}^{\circ} \mathrm{C}$ (iv) -$\displaystyle 40$
    \[C = \frac{5F-160}{9} \]
    (i)
    \[C = \frac{5(86)-160}{9} = \frac{270}{9} = 30 \]
    (ii)
    \[35 = \frac{5F-160}{9} \implies 5F = 315+160 = 475 \implies F = 95 \]
    (iii)
    \[C=0 \implies 0=\frac{5F-160}{9} \implies F=32 \]
    \[F=0 \implies C=\frac{5(0)-160}{9} = -\frac{160}{9} \]
    (iv)
    \[t = \frac{5t-160}{9} \implies 9t = 5t-160 \implies t = -40 \]
    Answer: (i) \(\displaystyle 30^{\circ}\text{C}\) (ii) \(\displaystyle 95^{\circ}\text{F}\) (iii) \(\displaystyle 32^{\circ}\text{F}\) and \(\displaystyle -\frac{160}{9}^{\circ}\text{C}\) (iv) \(\displaystyle -40^{\circ}\), same on both scales.

  5. Exercise 5

    If the temperature of a liquid can be measured in Kelvin units as x∘K\displaystyle x^{\circ} \mathrm{K} or in Fahrenheit units as y∘F\displaystyle y^{\circ} \mathrm{F}, the relation between the two systems of measurement of temperature is given by the linear equation y=95(x−273)+32y=\frac{9}{5}(x-273)+32 (i) Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is 313\displaystyle 313°K. (ii) If the temperature is 158\displaystyle 158°F, then find the temperature in Kelvin.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    (i)
    $\displaystyle 104$°F (ii) $\displaystyle 343$°K
    \[y = \frac{9}{5}(x-273)+32 \]
    (i)
    \[y = \frac{9}{5}(313-273)+32 = \frac{9}{5}(40)+32 = 72+32 = 104 \]
    (ii)
    \[158 = \frac{9}{5}(x-273)+32 \]
    \[126 = \frac{9}{5}(x-273) \implies x-273 = 70 \implies x = 343 \]
    Answer: (i) \(\displaystyle 104^{\circ}\text{F}\) (ii) \(\displaystyle 343\ \text{K}\).

  6. Exercise 6

    The force exerted to pull a cart is directly proportional to the acceleration produced in the body. Express the statement as a linear equation of two variables and draw the graph of the same by taking the constant mass equal to 6\displaystyle 6 kg. Read from the graph, the force required when the acceleration produced is (i) 5 m/sec2\displaystyle 5 \mathrm{~m} / \mathrm{sec}^{2}, (ii) 6 m/sec2\displaystyle 6 \mathrm{~m} / \mathrm{sec}^{2}.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    $\displaystyle y=m x$, where $\displaystyle y$ denotes the force, $\displaystyle x$ denotes the acceleration and $\displaystyle m$ denotes the constant mass. (i) $\displaystyle 30$ Newton (ii) $\displaystyle 36$ Newton
    \[F \propto a \implies F = ma \quad \text{(Newton's second law)} \]
    \[m = 6\ \text{kg} \implies F = 6a \]
    NCERT_Solution_Class9_Maths_Exemplar_Ch4_Ex4-4_Q6
    (i)
    \[a=5 \implies F = 6(5) = 30\ \text{N} \]
    (ii)
    \[a=6 \implies F = 6(6) = 36\ \text{N} \]
    Answer: (i) \(\displaystyle F=30\ \text{N}\) at \(\displaystyle a=5\ \text{m/s}^2\); (ii) \(\displaystyle F=36\ \text{N}\) at \(\displaystyle a=6\ \text{m/s}^2\), from \(\displaystyle F=6a\).