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Q7.    Solve the following pair of linear equations:

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

Given Equations,

\\152x - 378y = -74............(1)\\ -378x + 152y = -604............(2)

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,

-226x-226y=-678

\Rightarrow x+y=3...........(3)

Subtracting (2) from (1) we get,

530x-530y=530

\Rightarrow x-y=1...........(4)

Now, Adding (3) and (4) we get,

2x=4

\Rightarrow x=2

Putting this value in (3) 

 2+y=3

\Rightarrow y=1

Hence,

x=2\:and\:y=1.

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Pankaj Sanodiya

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,

\\(a-b)x + (a+b)y = a^2 -2ab - b^2..........(1)

And

\\ (a+b)(x+y) = a^2 +b^2\\\Rightarrow (a+b)x+(a+b)y=a^2+b^2...........(2)

Now, Subtracting (1) from (2), we get

(a+b)x-(a-b)x=a^2+b^2-a^2+2ab+b^2

\Rightarrow(a+b-a+b)x=2b^2+2ab

\Rightarrow 2bx=2b(b+2a)

\Rightarrow x=(a+b)

Substituting this in (1), we get,

(a-b)(a+b)+(a+b)y=a^2-2ab-b^2

\Rightarrow a^2-b^2+(a+b)y=a^2-2ab-b^2

\Rightarrow (a+b)y=-2ab

\Rightarrow y=\frac{-2ab}{a+b}.

Hence,

x=(a+b),\:and\:y=\frac{-2ab}{a+b}

 

 

 

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Pankaj Sanodiya

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Q7.    Solve the following pair of linear equations:

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

Given equation,

\\\frac{x}{a} - \frac{y}{b} = 0\\\Rightarrow bx-ay=0...........(1)\\ ax + by = a^2 +b^2...............(2)

Now By Cross multiplication method,

\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}

\frac{x}{(-a)(-a^2-b^2)-(b)(0)}=\frac{y}{(0)(a)-(b)(-a^2-b^2)}=\frac{1}{(b)(b)-(-a)(a)}

\frac{x}{a(a^2+b^2)}=\frac{y}{b(a^2+b^2)}=\frac{1}{a^2+b^2}

x=a,\:and\:y=b

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Pankaj Sanodiya

Q7.    Solve the following pair of linear equations:

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

Given Two equations, 

\\ax + by = c.........(1)\\ bx +ay = 1 + c.........(2)

Now By Cross multiplication method,

\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}

\frac{x}{(b)(-1-c)-(-c)(a)}=\frac{y}{(-c)(b)-(a)(-1-c)}=\frac{1}{(a)(a)-(b)(b)}

\frac{x}{-b-bc+ac}=\frac{y}{-cb+a+ac}=\frac{1}{a^2-b^2}

x=\frac{-b-bc+ac}{a^2-b^2},\:and\:y=\frac{a-bc+ac}{a^2-b^2}

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Pankaj Sanodiya

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Q7.    Solve the following pair of linear equations:

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

Given Equations,

\\px + qy = p - q.........(1)\\ qx - py = p + q.........(2)

Now By Cross multiplication method,

\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}

\frac{x}{(q)(-p-q)-(q-p)(-p)}=\frac{y}{(q-p)(q)-(p)(-p-q)}=\frac{1}{(p)(-p)-(q)(q)}

\frac{x}{-p^2-q^2}=\frac{y}{p^2+q^2}=\frac{1}{-p^2-q^2}

x=1,\:and\:y=-1

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Pankaj Sanodiya

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,

4y+20-4x=180

\Rightarrow 4y-4x=160

\Rightarrow y-x=40............(1)

Also,

3y-5-7x+5=180

\Rightarrow 3y-7x=180........(2)

Multiplying (1) by 3 we get,

\Rightarrow 3y-3x=120........(3)

Now,

Subtracting, (2) from (3) we get,

4x=-60

\Rightarrow x=-15

Substituting this value in (1) we get,

y-(-15)=40

\Rightarrow y=40-15

\Rightarrow y=25

Hence four angles of a quadrilateral are :

\angle A =4y+20=4(25)+20=100+20=120^0

\angle B =3y-5=3(25)-5=75-5=70^0

\angle C =-4x=-4(-15)=60^0

\angle D =-7x+5=-7(-15)+5=105+5=110^0

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Pankaj Sanodiya

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

Given two equations,

5x - y =5.........(1)

And

3x - y = 3........(2)

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:

Graph

As we can see from the graph, the three points of the triangle are, (0,-3),(0,-5) and (1,0).

 

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Pankaj Sanodiya

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

Given,

\angle C = 3 \angle B = 2(\anlge A + \angle B)

\Rightarrow 3 \angle B = 2(\anlge A + \angle B)

\Rightarrow \angle B = 2 \angle A

\Rightarrow 2 \angle A -\angle B = 0..........(1)

 Also, As we know that the sum of angles of a triangle is 180, so

\angle A +\angle B+ \angle C=180

\angle A +\angle B+ 3\angle B=180^0

\angle A + 4\angle B=180^0..........(2)

Now From (1) we have 

\angle B = 2 \angle A.......(3)

Putting this value in (2) we have 

\angle A + 4(2\angle A)=180^0.

\Rightarrow 9\angle A=180^0.

\Rightarrow \angle A=20^0.

Putting this in (3)

\angle B = 2 (20)=40^0

And 

\angle C = 3 \angle B =3(40)=120^0

Hence three angles of triangles 20^0,40^0\:and\:120^0.

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Pankaj Sanodiya

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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,

speed=\frac{distance }{time}

\Rightarrow v=\frac{d}{t}


\Rightarrow d=vt..........(1)
Now, According to the question,
 (v+10)=\frac{d}{t-2}

\Rightarrow (v+10){t-2}=d

\Rightarrow vt +10t-2v-20=d

Now, Using equation (1), we have 

\Rightarrow -2v+10t=20............(2)
Also,

(v-10)=\frac{d}{t+3}

\Rightarrow (v-10)({t+3})=d

\Rightarrow vt+3v-10t-30=d
 \Rightarrow3v-10t=30..........(3)

Adding equations (2) and (3), we obtain:
v=50.
Substituting the value of x in equation (2), we obtain:
(-2)(50)+10t=20

\Rightarrow -100+10t=20

\Rightarrow 10t=120

\Rightarrow t=12
Putting this value in (1) we get,

d=vt=(50)(12)=600

Hence the distance covered by train is 600km.

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Pankaj Sanodiya

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
                                                           =xy

Now, According to the question,

\\xy = (x - 1) (y + 3) \\\Rightarrow xy= xy - y + 3x - 3 \\\Rightarrow 3x - y - 3 = 0 \\ \Rightarrow 3x - y = 3 ...... ... (1)

 Also,

   \\xy=(x+2)(y-3)\\\Rightarrow xy = xy + 2y - 3x - 6 \\ \Rightarrow 3x - 2y = -6 ... (2)
Subtracting equation (2) from (1), we get:
y=9
Substituting the value of y in equation (1), we obtain:
\\3x - 9 = 3 \\\Rightarrow 3x = 9 + 3 = 12 \\\Rightarrow x = 4
Hence,
The number of rows is 4 and Number of students in a row is 9.

Total number of students in a class

:xy=(4)(9)=36

Hence there is 36 number of students in the class.

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Pankaj Sanodiya

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