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
\[a_{1}x+b_{1}y=c_{1}\]\[a_{2}x+b_{2}y=c_{2}\]
There are four methods of solving pair of linear equations in two variables
In this method the point of intersection of two lines on the graph is the solution of two equations.
4) Cross Multiplication Method
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

\[\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}\]

Unique Solution

Two intersecting lines

Consistent

\[\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}= \frac{c_{1}}{c_{2}}\]

Infinitely many solutions

Coincident lines

Consistent

\[\frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}\]

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
Putting eqn. (3) in eqn. (2) we get
Putting y = 4 in eqn. (3) we get
x = 1 and y = 4 is the required solution.
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Elimination Method of solving
Pair of linear equations in two variables
Eqn. (1) x 2 – eqn. (2) x 1 we get
Putting y = 4 in equation (1)
x = 1 and y = 4 is the required solution.
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Cross Multiplication Method of solving
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
\[\frac{x}{b_{1}c_{2}b_{2}c_{1}}=\frac{y}{c_{1}a_{2}c_{2}a_{1}}=\frac{1}{a_{1}b_{2}a_{2}b_{1}}\]
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
\[a_{1}x+b_{1}y+c_{1}=0\]\[a_{2}x+b_{2}y+c_{2}=0\]
\[\frac{x}{b_{1}c_{2}b_{2}c_{1}}=\frac{y}{c_{1}a_{2}c_{2}a_{1}}=\frac{1}{a_{1}b_{2}a_{2}b_{1}}\]
Explanation of cross multiplication method by taking an example
Let given a pair of equations in two variables
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
By cross multiplication method
\[\frac{x}{q(p+q)+p(pq)}=\frac{y}{q(pq)p(p+q)}=\frac{1}{p^{2}q^{2}}\]\[\frac{x}{pq+q^{2}+p^{2}pq}=\frac{y}{pqq^{2}p^{2}pq}=\frac{1}{\left (p^{2}+q^{2} \right )}\]\[\frac{x}{p^{2}+q^{2}}=\frac{y}{\left (p^{2}+q^{2} \right )}=\frac{1}{\left (p^{2}+q^{2} \right )}\]\[x=1, y=1\]
ii) Solve :
ax + by = c
bx + ay = 1+c
By Cross multiplication method
\[\frac{x}{b+bcac}=\frac{y}{bcaac}=\frac{1}{a^{2}b^{2}}\]\[x=\frac{\left ( b+bcac \right )}{a^{2}b^{2}},\; \; y=\frac{\left ( bcaac \right )}{a^{2}b^{2}}\]\[x=\frac{acbbc}{a^{2}b^{2}},\; \; y=\frac{a+acbc}{a^{2}b^{2}}\]
iii) Solve : \[\frac{x}{a}\frac{y}{b}=0 ........(1)\]\[ax+by=a^{2}+b^{2} .......(2)\]\[From \; \; equation\; \; (1),\; \; \frac{x}{a}=\frac{y}{b}\]\[x=\frac{a}{b}y .........(3)\]Putting equation (3) in equation (2) we get\[a\times \frac{ay}{b}+by=a^{2}+b^{2}\]\[\frac{a^{2}y+b^{2}y}{b}=a^{2}+b^{2}\] \[\frac{\left (a^{2}+b^{2} \right )y}{b}=a^{2}+b^{2}\]\[y=b\]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 = a^{2}  2ab  b^{2} …..(1)
(a + b)(x +
y) = a^{2} + b^{2}
^{}
(a + b)x + (a
+ b)y = a^{2} + b^{2} …….(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 = a^{2} + b^{2}
^{}
a^{2}  b^{2} + (a + b)y = a^{2} + b^{2}
^{}
\[(a+b)y=b^{2}+b^{2}\]\[y =
\frac{2b^{2}}{a+b}\]
Q7(iv) 152x – 378y = 74 …….(1)
378x +
152y =  604 ……(2)
Adding
eqn.(1) and eqn.(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
quadrilateral,
find the measure of angles of the cyclic quadrilateral ?
Sol. Since opposite angles of a cyclic quadrilateral are supplementary.
Therefore
∠A + ∠C = 180^{o}
4y + 20  4x = 180
 4x + 4y = 180  20
 4x + 4y = 160
Dividing by 4 we get
x + y = 40..........(1)
∠B + ∠D = 180^{o}
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 = 120^{o}
∠B = 3y  5 = 3 x 25  5 = 70^{o}
∠C =  4x = 4 x 15 = 60^{o}
∠D = 7x + 5 = 7 x  15 + 5 = 105 + 5 = 110^{o}
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