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Pair of Linear Equations in Two Variables Test - 43

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Pair of Linear Equations in Two Variables Test - 43
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  • Question 1
    1 / -0
    On comparing the ratios $$\dfrac {a_1}{a_2}, \dfrac {b_1}{b_2}$$ and $$\dfrac {c_1}{c_2}$$, find out whether the following pairs of linear equations are consistent, or inconsistent.
    $$5x - 3y = 11; - 10x + 6y =  22$$
    Solution
     The given linear equation are
    $$\Rightarrow 5x-3y-11=0....eq1$$
    $$\Rightarrow  a_{1}=5,b_{1}=-3,c_{1}=-11$$
    $$\Rightarrow -10x+6y-22=0...eq2$$
    $$\Rightarrow a_{2}=-10,b_{2}=6,c_{2}=-22$$
    $$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{5}{-10} \Rightarrow- \frac{1}{2}$$
    $$\Rightarrow \frac{b_{1}}{b_{2}}=\frac{-3}{6} \Rightarrow -\frac{1}{2}$$
    $$\Rightarrow \frac{c_{1}}{c_{2}}=\frac{-11}{-22} \Rightarrow \frac {1}{2}$$
    comparing
    $$\Rightarrow  \frac{a_{1}}{a_{2}},\frac{b_{1}}{b_{2}},\frac{c_{1}}{c_{2}}$$
    $$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$$
    Hence, the following pairs of linear equations are  inconsistent.




  • Question 2
    1 / -0
    Draw the graphs of the equations $$x - y + 1 = 0$$ and $$3x + 2y - 12 = 0$$. Determine the co-ordinates of the vertices of the triangle formed by these lines and the $$x$$-axis, and shade the triangular region.
    Solution
    For equation $$x – y + 1 = 0$$ or $$x=y-1$$, we have following points which lie on the line

    Plug $$y =0,1$$ we get

    $$x=0-1=-1$$
    $$x=1-1=0$$

    Therefore, the required points are $$(-1,0), (0,1)$$.
     
    For equation $$3x + 2y – 12 = 0$$ or $$y=\frac { 12-3x }{ 2 }$$, we have following points which lie on the line

    Plug $$x= 0 ,4$$ we get

    $$y=\frac { 12-0 }{ 2 }=6$$
    $$y=\frac { 12-12 }{ 2 }=0$$

    Therefore, the required points are $$(0,6), (4,0)$$.


    We can see from the graphs that points of intersection of the lines with the x–axis are $$(–1, 0), (2, 3)$$ and $$(4, 0)$$.


  • Question 3
    1 / -0
    Aftab tells his daughter, "Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be." (Isn't this interesting?) What are their present ages? Solve graphically introducing necessary varaibles.
    Solution
    Let the present age of Aftab's daughter $$= x$$ years. and the present age of Aftab $$= y$$ years, $$(y > x) $$
    According to the given conditions, we have
    Seven years ago,
    $$(y-7)=7\times (x-7)$$
    $$\Rightarrow y - 7 = 7x - 49$$
    $$\Rightarrow 7x - y - 42 = 0$$ ...(i)
    Three years later,
    $$(y+ 3) = 3 \times (x + 3)$$
    $$\Rightarrow y+ 3 = 3x + 9$$
    $$\Rightarrow 3x- y + 6 = 0$$ ...(ii)
    Thus, the algebraic relations are
    $$7x - y - 42 = 0, 3x - y + 6 = 0$$.
    Now, we represent the problem graphically as below: $$7x - y - 42 = 0$$ ...(i)
    $$3x - y + 6 = 0$$ ...(ii)
    From the graph, we find that
    $$x = 12$$
    and $$y = 42$$
    Thus, the present age of Aftab's daugther $$= 12$$ years and the present age of Aftab $$= 42$$ years.

  • Question 4
    1 / -0
    Determine whether the following pairs of linear equations are consistent/inconsistent. 
    $$x-y=8, 3x-3y= 16$$
    If consistent, obtain the solution graphically.
    Solution

  • Question 5
    1 / -0
    $$10$$ students of class $$X$$ took part in a Mathematics quiz. If the number of girls is $$4$$ more than the number of boys, find the number of boys and girls who took part in the quiz by graphical method.
    Solution
    Let number of boys be $$x$$.
    Number of girls be $$y$$.

    Given that total number of student  is $$10$$ so that
    $$x + y= 10$$

    Subtract $$y$$ both side we get

    $$x= 10 – y$$

    Plug $$y = 0 , 5, 10$$ we get

    $$x= 10 – 0= 10$$
    $$x= 10 – 5= 5$$
    $$x= 10 – 10= 0$$

    Therefore, the required points are $$(0,10),(5,5),(10,0)$$.

    Given that If the number of girls is $$4$$ more than the number of boys
    So that

    $$y=  x + 4$$

    Plug $$x = -4, 0, 4$$, and we get

    $$y= - 4 + 4= 0$$
    $$y=  0 + 4= 4$$
    $$y=  4 + 4= 8$$

    Therefore, the required points are $$(-4,0),(0,4),(4,8)$$.

    The graph is as shown above:

    Since the point of intersection in the above graph is $$(3,7)$$, 

    Hence, the number of boys are $$3$$ and the number of girls are $$7$$.

  • Question 6
    1 / -0
    On comparing the ratios $$\dfrac {a_1}{a_2}, \dfrac {b_1}{b_2}$$ and $$\dfrac {c_1}{c_2}$$, find out whether the following pairs of linear equations are consistent, or inconsistent.
    $$2x - 3y = 8; 4x - 6y = 9$$
    Solution
    The given linear equation are
    $$\Rightarrow 2x-3y-8=0....eq1$$
    $$\Rightarrow  a_{1}=2,b_{1}=-3,c_{1}=-8$$
    $$\Rightarrow 4x-6y-9=0...eq2$$
    $$\Rightarrow a_{2}=4,b_{2}=-6,c_{2}=-9$$
    $$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{2}{4} \Rightarrow \frac{1}{2}$$
    $$\Rightarrow \frac{b_{1}}{b_{2}}=\frac{-3}{-6} \Rightarrow \frac{1}{2}$$
    $$\Rightarrow \frac{c_{1}}{c_{2}}=\frac{-8}{-9}$$
    comparing
    $$\Rightarrow  \frac{a_{1}}{a_{2}},\frac{b_{1}}{b_{2}},\frac{c_{1}}{c_{2}}$$
    $$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$$
    Hence, the following pairs of linear equations are  inconsistent.




  • Question 7
    1 / -0
    Determine whether the following pair of linear equations are consistent/inconsistent.
     $$2x - 2y - 2 = 0, 4x - 4y - 5 = 0$$
    Solution

  • Question 8
    1 / -0
    On comparing the ratios $$\dfrac {a_1}{a_2}, \dfrac {b_1}{b_2}$$ and $$\dfrac {c_1}{c_2}$$, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident.
    $$6x - 3y + 10 = 0; 2x - y + 9 = 0$$
    Solution
    The given linear equation are
    $$\Rightarrow 6x-3y+10=0....eq1$$
    $$\Rightarrow  a_{1}=6,b_{1}=-3,c_{1}=10$$
    $$\Rightarrow 2x-y+9=0...eq2$$
    $$\Rightarrow a_{2}=2,b_{2}=-1,c_{2}=9$$
    $$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{6}{3} \Rightarrow=\frac{3}{1}$$
    $$\Rightarrow \frac{b_{1}}{b_{2}}=\frac{-3}{-1}$$
    $$\Rightarrow \frac{c_{1}}{c_{2}}=\frac{10}{9}$$
    comparing
    $$\Rightarrow  \frac{a_{1}}{a_{2}},\frac{b_{1}}{b_{2}},\frac{c_{1}}{c_{2}}$$
    $$\Rightarrow \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq  \frac{c_{1}}{c_{2}} $$
    Hence,the line represented by eq1 and eq2 are parallel.




  • Question 9
    1 / -0
    On comparing the ratios $$\frac {a_1}{a_2}, \frac {b_1}{b_2}$$ and $$\frac {c_1}{c_2}$$, find out whether the lines representing the following pairs of linear equations intersect at a point, or are parallel or coincident.
    $$9x + 3y + 12 = 0; 18x + 6y + 24 = 0$$
    Solution
    The given linear equation are
    $$\Rightarrow 9x+3y+12=0....(1)$$
    $$\Rightarrow  a_{1}=9,b_{1}=3,c_{1}=12$$

    $$\Rightarrow 18x+6y+24=0...(2)$$
    $$\Rightarrow a_{2}=18,b_{2}=6,c_{2}=24$$

    $$\Rightarrow \displaystyle \frac{a_{1}}{a_{2}}=\frac{9}{18} = \frac{1}{2}$$

    $$\Rightarrow \displaystyle \frac{b_{1}}{b_{2}}=\frac{3}{6} =\frac{1}{2} $$

    $$\Rightarrow \displaystyle \frac{c_{1}}{c_{2}}=\frac{12}{24}=\frac{1}{2}$$

    comparing
    $$\ \displaystyle  \frac{a_{1}}{a_{2}},\frac{b_{1}}{b_{2}},\frac{c_{1}}{c_{2}}$$

    We have $$\ \displaystyle \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$$
    Hence, the lines represented by eq1 and eq2 are coincident
  • Question 10
    1 / -0
    The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs. 105 and for a journey of 15 km, the charge paid is Rs. 155. What are the fixed charges and the charge per kilometer? How much does a person have to pay for traveling a distance of 25 km?
    Solution
    Let fixed charge be Rs x and charge per km be Rs y.
     $$\Rightarrow x + 10y = 105 ...(1)$$
     $$\Rightarrow x + 15y = 155  ..(2)$$.

    Now From eq 2
    $$\Rightarrow 15y = 155-x$$

    $$\Rightarrow y = \dfrac{155-x}{15}...(3)$$

    Substituting y from eq3 in eq1

    $$\Rightarrow x+\dfrac{10\left ( 155-x \right )}{15} = 105$$

    $$\Rightarrow 15x+1550-10x =1575$$
    $$\Rightarrow 5x = 1575-1550$$
    $$\Rightarrow 5x = 25$$
    $$\Rightarrow x = 5$$

    Substituting x in eq2
    $$\Rightarrow 5+15y = 155$$
    $$\Rightarrow 15y = 155-5$$
    $$\Rightarrow 15y = 150$$
    $$\Rightarrow y = 10$$

    Hence $$x= 5$$    and  $$ y = 10$$
    Fixed charge is Rs.55 and the charge per kilometer is Rs.1010 
    For 2525 km person have to pay = $$5+10 \times 25= 255 Rs.$$
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