### Mathematics Assignments | PDF | 8 to 12

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### Maths Formulas Chapter-03 | Pair of Linear Equations in Two Variables

Pair of Linear Equations in Two Variables
Basic points and complete explanations of all methods of solving pair of linear equations in two variables, for class X, chapter 3 Mathematics.

General pair of linear equations in two variables

ax1 + by1 + c1 = 0  ….. (i)

ax2 + by2 + c2 = 0  ….. (ii)

There are four methods of solving pair of linear equations in two variables

1) Graphical Method :

In this method the point of intersection of two lines on the graph is the solution of two equations.

2) Substitution Method

3) Elimination Method

4) Cross Multiplication Method (Deleted)

## Types of graphs, nature of solutions, consistency/ inconsistency all are discussed in the following table.

 Conditions Nature of Solution Type of Graph Consistent /Inconsistent $\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}$ Unique Solution Two intersecting lines Consistent $\large \begin{matrix}\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}\end{matrix}$ Unique Solution Two intersecting lines Consistent $\large \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}$ Infinitely many solutions Coincident lines Consistent $\large \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$ No Solution Two Parallel Lines Inconsistent

Explanation of Algebraic Methods of solving Pair of Linear Equations in two Variables

Substitution Method of solving

Pair of linear equations in two variables

Let Given Pair of linear equations in two variables
x + y = 3………………(1)
2x + 3y = 10…………..(2)
From equation (1)
x = 3 – y………….(3)
Putting eqn. (3) in eqn. (2) we get
2(3 - y) + 3y = 10
6 - 2y + 3y = 10
6 + y = 10
y = 10 - 6
y = 4
Putting y = 4 in eqn. (3) we get
x = 3 - y
x = 3 - 4
x = -1
x = -1 and y = 4 is the required solution.
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Elimination Method of solving

Pair of linear equations in two variables
x + y = 3…………(1)
2x + 3y = 10…………..(2)

Eqn. (1) x 2 – eqn. (2) x 1 we get
y = 4
Putting y = 4 in equation (1)
x + 4 = 3
x = 3 - 4 = -1
x = -1 and y = 4 is the required solution.

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Cross Multiplication Method of solving(Deleted)

Pair of linear equations in two variables

There are two cases to explain this method

Case I :  Let general pair of linear equations in two variables :
$a_{1}x+b_{1}y=c_{1}$$a_{2}x+b_{2}y=c_{2}$
If constant terms on the right hand side then place the coefficients as shown below

$\large \frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{y}{c_{1}a_{2}-c_{2}a_{1}}=\frac{-1}{a_{1}b_{2}-a_{2}b_{1}}$

Case II :  If constant terms are on right hand side then we must follow the following steps
General pair of linear equations in two variables
$a_{1}x+b_{1}y+c_{1}=0$$a_{2}x+b_{2}y+c_{2}=0$

$\large \frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{y}{c_{1}a_{2}-c_{2}a_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}$

Explanation of cross multiplication method by taking an example
Let given a pair of equations in two variables
x + y = 3…………(1)
2x + 3y = 10…………..(2)
Exercise 3.7
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.

Sol. :         Let age of Ani = x
Let age of Biju = y
According to the question
x - y = 3 .....................(1)
Age of Dharam = 2x
Age of cathy = y/2
A.T.Q                 2x - y/2 = 30
Multiplying this equation by 2 we get
4x - y = 60 ....................(2)
Solving equation (1) and eqn(2) by elimination method we get
x = 19 years, y=16 years.

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 capital?
Sol. :           Let first friend have = Rs x
Second friend have = Rs y
A.T.Q.
x + 100 = 2(y - 100)
x + 100 = 2y - 200
x - 2y = - 200 - 100
x - 2y = - 300 ...............(1)
Again ATQ
6(x - 10) = y + 10
6x - 60 = y + 10
6x - y = 10 + 60
6x - y = 70 ......................(2)
Solving equation (1) and (2) by elimination method we get
x = 40 and y = 170

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.

Sol. :            Let speed of train = x km/h
Time taken by train = y Km/h
Distance = Speed x Time
= xy
Case 1
New speed  = (x + 10) km/h
New time taken = y - 2) hours
Distance = Speed x Time
= (x + 10) x (y - 2)
= xy - 2x + 10y - 20
A.T.O.
xy - 2x + 10y - 20 =  xy
-2x + 10y = 20 .......................(1)
Case II
New speed = (x - 10) km/h
New time taken  = (y + 3) hours
Distance = (x - 10) x (y + 3)
= xy + 3x - 10y - 30
A.T.Q.
xy + 3x - 10y - 30 = xy
3x - 10y - 30 = 0
3x - 10y = 30 .....................(2)
By elimination method solve equation (1) and (2) we get
x = 50, y = 12
Distance = xy = 50 x 12 = 600km

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

Sol.                   Let number of rows = x
Number of students in each row = y
Total students in the class = xy
A.T.Q.
(x + 3)(y - 1) = xy
xy - x + 3y - 3 = xy
- x + 3y - 3 = 0
- x + 3y = 3 .................(1)
Again ATQ
(x - 3)(y + 2) = xy
xy + 2x - 3y - 6 = xy
2x - 3y = 6 ....................(2)
Solving equations (1) and (2) by elimination method we get
x = 9, y = 4
Number of students in the class = 9 x 4 = 36

Q7. Solve the following equations
i) Solve  the following By cross multiplication method
px + qy = p – q
qx - py  + = pq

Solution
$\large \frac{x}{q(p+q)+p(p-q)}=\frac{y}{q(p-q)-p(p+q)}=\frac{-1}{-p^{2}-q^{2}}$

$\large \frac{x}{pq+q^{2}+p^{2}-pq}=\frac{y}{pq-q^{2}-p^{2}-pq}=\frac{-1}{-\left (p^{2}+q^{2} \right )}$

$\large \frac{x}{p^{2}+q^{2}}=\frac{y}{-\left (p^{2}+q^{2} \right )}=\frac{1}{\left (p^{2}+q^{2} \right )}$

$\large x=1, y=-1$

ii) Solve the following equations By Cross multiplication method
ax + by = c
bx + ay = 1+c

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

$\large x=\frac{-\left ( b+bc-ac \right )}{a^{2}-b^{2}},\; \; y=\frac{-\left ( bc-a-ac \right )}{a^{2}-b^{2}}$

$\large x=\frac{ac-b-bc}{a^{2}-b^{2}},\; \; y=\frac{a+ac-bc}{a^{2}-b^{2}}$

iii) Solve : $\frac{x}{a}-\frac{y}{b}=0 ........(1)$$ax+by=a^{2}+b^{2} .......(2)$
From equation (1)
$\frac{x}{a}=\frac{y}{b}\Rightarrow x=\frac{a}{b}y .........(3)$
Putting equation (3) in equation (2) we get
$a\times \frac{ay}{b}+by=a^{2}+b^{2}$

$\frac{a^{2}y+b^{2}y}{b}=a^{2}+b^{2}$

$\frac{\left (a^{2}+b^{2} \right )y}{b}=a^{2}+b^{2}\: \: \Rightarrow y=b$
Putting y = b in equation (3) we get
$x=\frac{a}{b}\times b\Rightarrow x=a$

Q7 iv)            (a - b)x + (a + b)y = a2 - 2ab - b2  …..(1)
(a + b)(x + y) = a2 + b2
(a + b)x + (a + b)y = a2 + b2 …….(2)

Eqn. (1) – Eqn.(2)
2bx = - 2b(a – b)    ⇒   x  = a – b

Putting x = a - b in equation  (2)

(a + b)(a - b) + (a + b)y = a2 + b2

a2 - b2  + (a + b)y = a2 + b2

$(a+b)y=b^{2}+b^{2}$

$\Rightarrow\: \: \: y = \frac{2b^{2}}{a+b}$

Q7(v)       152x – 378y = -74 …….(1)

-378x + 152y = - 604 ……(2)

Dividing by -226 we get

x + y = 3 …………..(3)

Subtracting eqn.(1) and eqn.(2) we get

Dividing by 530 we get

x – y =1 ……………(4)

By elimination method solving eqn. (3) and (4) we get

x = 2, y = 1
Q 8)

ABCD is a cyclic

Sol. Since opposite angles of a cyclic quadrilateral are supplementary.
Therefore
∠A + ∠C = 180o
4y + 20 - 4x = 180
- 4x + 4y = 180 - 20
- 4x + 4y = 160
Dividing by 4 we get
-x + y = 40..........(1)
∠B + ∠D = 180o
3y - 5 - 7x + 5 = 180
-7x + 3y = 180  ..................(2)
Solving these equations by elimination method:-
Eqn.(1) x 3 - eqn. (2) x 1 we get
x = -60/4 ⇒ x = -15
Putting x = -15 in eqn. (1) we get
-(-15) + y = 40
y = 40 - 15
y = 25
x = -15, y = 25
Now we find the angles of quadrilateral
∠A = 4y + 20 = 4 x 25 + 20 = 120o
∠B = 3y - 5    = 3 x 25 - 5    = 70o
∠C = - 4x       = -4 x -15       = 60o
∠D = -7x + 5 = -7 x - 15 + 5 =  105 + 5 = 110o