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Q7.    Solve the following pair of linear equations:

(v)    $\\152x - 378y = -74\\ -378x + 152y = -604$

Given Equations, As we can see by adding and subtracting both equations we can make our equations simple to solve.  So, Adding (1) and )2) we get, Subtracting (2) from (1) we get, Now, Adding (3) and (4) we get, Putting this value in (3)    Hence, .

Q7.    Solve the following pair of linear equations:

(iv)    $\\(a-b)x + (a+b)y = a^2 -2ab - b^2\\ (a+b)(x+y) = a^2 +b^2$

Given, And Now, Subtracting (1) from (2), we get Substituting this in (1), we get, . Hence,

Q7.    Solve the following pair of linear equations:

(iii)    $\\\frac{x}{a} - \frac{y}{b} = 0\\ ax + by = a^2 +b^2$

Given equation, Now By Cross multiplication method,

Q7.    Solve the following pair of linear equations:

(ii)    $\\ax + by = c\\ bx +ay = 1 + c$

Given Two equations,  Now By Cross multiplication method,

Q7.    Solve the following pair of linear equations:

(i)    $\\px + qy = p - q\\ qx - py = p + q$

Given Equations, Now By Cross multiplication method,

Q8.    ABCD is a cyclic quadrilateral (see Fig. 3.7). Find the angles of the cyclic quadrilateral.

As we know that in a quadrilateral the sum of opposite angles is 180 degree. So, From Here, Also, Multiplying (1) by 3 we get, Now, Subtracting, (2) from (3) we get, Substituting this value in (1) we get, Hence four angles of a quadrilateral are :

Q6.    Draw the graphs of the equations $5x - y =5$and $3x - 7 = 3$. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y axis.

Given two equations, And Points(x,y) which satisfies equation (1) are: X 0 1 5 Y -5 0 20 Points(x,y) which satisfies equation (1) are: X 0 1 2 Y -3 0 3   GRAPH: As we can see from the graph, the three points of the triangle are, (0,-3),(0,-5) and (1,0).

Q5.    In a $\Delta\textup{ABC}$, $\angle C = 3 \angle B = 2( \angle A + \angle B)$. Find the three angles.

Given,  Also, As we know that the sum of angles of a triangle is 180, so Now From (1) we have  Putting this value in (2) we have  Putting this in (3) And  Hence three angles of triangles

Q3.    A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Let the speed of the train be v km/h and the time taken by train to travel the given distance be t hours and the distance to travel be d km.  Now As we Know, Now, According to the question,   Now, Using equation (1), we have  Also,   Adding equations (2) and (3), we obtain: Substituting the value of x in equation (2), we obtain: Putting this value in (1) we get, Hence...

Q4.    The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Let the number of rows be x and number of students in a row be y. Total number of students in the class = Number of rows * Number of students in a row                                                             Now, According to the question,  Also,     Subtracting equation (2) from (1), we get: Substituting the value of y in equation (1), we obtain: Hence, The number of rows is 4...

Q2.    One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II]

[Hint :$x + 100 = 2(y - 100), y + 10 = 6(x - 10)$].

Let the amount of money the first person and the second person having is x and y respectively

Noe, According to the question.

$x + 100 = 2(y - 100)$

$\Rightarrow x - 2y =-300...........(1)$

Also

$y + 10 = 6(x - 10)$

$\Rightarrow y - 6x =-70..........(2)$

Multiplying (2) by 2 we get,

$2y - 12x =-140..........(3)$

Now adding (1) and (3), we get

$-11x=-140-300$

$\Rightarrow 11x=440$

$\Rightarrow x=40$

Putting this value in (1)

$40-2y=-300$

$\Rightarrow 2y=340$

$\Rightarrow y=170$

Thus two friends had 40 Rs and 170 Rs respectively.

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Q1.    The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.

Let the age of Ani be , age of Biju be , Case 1: when Ani is older than Biju age of Ani's father Dharam:    and age of his sister Cathy : Now According to the question, Also, Now subtracting (1)  from (2), we get, putting this in (1) Hence the age of Ani and Biju is 19 years and 16 years respectively. Case 2: And Now Adding (3) and (4), we get, putting it in (3) Hence the age of...

Q2.    Formulate the following problems as a pair of equations, and hence find their solutions:
(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

Let the speed of Ritu in still water be x and speed of current be y, Let's solve this problem by using relative motion concept, the relative speed when they are going in the same direction (downstream)= x +y  the relative speed when they are going in the opposite direction (upstream)= x - y Now, As we know, Relative distance = Relative speed * time . So, According to the question, And, Now,...

Q1.    Solve the following pairs of equations by reducing them to a pair of linear equations:

(viii)    $\\\frac{1}{3x + y} + \frac{1}{3x -y} = \frac{3}{4}\\ \frac{1}{2(3x+y)} - \frac{1}{2(3x -y)} = \frac{-1}{8}$

Given Equations, Let,  Now, our equation becomes And Now, Adding (1) and (2), we get Putting this value in (1) Now, And Now, Adding (3) and (4), we get Putting this value in (3), Hence,

Q1.    Solve the following pairs of equations by reducing them to a pair of linear equations:

(vii)    $\\\frac{10}{x + y} + \frac{2}{x - y}= 4\\ \frac{15}{x+y} - \frac{5}{x - y} = -2$

Given Equations, Let,  Now, our equation becomes And By Cross Multiplication method, Now, And, Adding (3) and (4) we get, Putting this value in (3) we get, And Hence,

Q1.    Solve the following pairs of equations by reducing them to a pair of linear equations:

(vi)    $\\6x + 3y = 6xy\\ 2x + 4y = 5 xy$

Given Equations, Let,  Now, our equation becomes And By Cross Multiplication method, And Hence,

Q1.    Solve the following pairs of equations by reducing them to a pair of linear equations:

(v)    $\\\frac{7x - 2y}{xy} = 5\\ \frac{8x + 7y}{xy} = 15$

Given Equations, Let,  Now, our equation becomes And By Cross Multiplication method, And Hence,

Q1.    Solve the following pairs of equations by reducing them to a pair of linear equations:

(iv)    $\\\frac{5}{x - 1} + \frac{1}{y -2} = 2\\ \frac{6}{x-1} - \frac{3}{y -2} =1$

Given Equations, Let,  Now, our equation becomes And Multiplying (1) by 3 we get Now, adding (2) and (3) we get Putting this in (2) Now, Hence,

Q1.    Solve the following pairs of equations by reducing them to a pair of linear equations:

(iii)    $\\\frac{4}{x} + 3y = 14\\ \frac{3}{x} - 4y = 23$

Given Equations, Let,  Now, our equation becomes And By Cross Multiplication method, And Hence,

Q1.    Solve the following pairs of equations by reducing them to a pair of linear equations:

(ii)    $\\ \frac{2}{\sqrt x} + \frac{3}{\sqrt y} = 2\\ \frac{4}{\sqrt x} - \frac{9}{\sqrt y} = -1$

Given Equations, Let,  Now, our equation becomes And By Cross Multiplication method, So, . And hence
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