3.4               Algebraic Solution:

Finding solution for consistent pair of Linear Equations.

The solution of a pair of linear equations is of the form ( x,y ) which satisfies both the equations simultaneously. Solution for a consistent pair of linear equations can be found out using

i ) Elimination method                                                  ii ) Substitution Method

iii ) Cross  –  multiplication method                          iv ) Graphical method

Substitution Method of finding solution of a pair of Linear Equations:

y  –  2 x   =   1       and     x  +  2 y   =   12

( i ) Express one variable in terms of the other using one of the equations. In this case,  y   =   2 x  +  1.

( ii ) Substitute for this variable ( y ) in the second equation to get a linear equation in one variable, x.                                      x  +  2   ×   ( 2 x  +  1 )   =   12

⇒     5 x  +  2   =   12

( iii ) Solve the linear equation in one variable to find the value of that variable.
5 x  +  2   =   12
x   =   2

( iv ) Substitute this value in one of the equations to get the value of the other variable.

y   =   2   ×   2  +  1                        ⇒      y   =   5

So, ( 2,  5 ) is the required solution of the pair of linear equations   y  –  2 x   =   1  and  x  +  2 y   =   12.

# EXERCISE  3.3

1. Solve the following pair of linear equations by the substitution method.

( i ) Given:  –  x  +  y   =   14  and  x   –   y   =   4

Solution :

⇒   x  +  y   =   14       … ( I )

x   –   y   =   4 .        .. ( II )

From equation ( I ), we get

x   =   14   –   y .          .. ( III )

Putting this value in equation ( ii ), we get

⇒  ( 14   –   y )   –   y   =   4

⇒  14   –   2 y   =   4

⇒  10   =   2 y

⇒  y   =   10 / 2

⇒  y   =   5 .             .. ( IV )

Putting this in equation ( III ), we get

x   =   9

Hence, x   =   9  and  y   =   5.

( iv )  Given:   0.2 x  +  0.3 y   =  1.3       and     0.4 x  +  0.5 y   =   2.3   are the two equations.

Solution :            From 1st equation, we get,

x   =   ( 1.3  –   0.3 y ) / 0.2 _________________( 1 )

Now, substitute the value of x in the given second equation to get,

0.4 ( 1.3  –  0.3 y ) / 0.2  +  0.5 y   =   2.3

⇒  2 ( 1.3  –  0.3 y )  +  0.5 y   =   2.3

⇒  2.6  –  0.6 y  +  0.5 y   =   2.3

⇒  2.6  –  0.1 y   =   2.3

⇒  0.1 y   =   0.3

⇒  y   =   3

Now, substitute the value of y in equation ( 1 ), we get,

x   =   ( 1.3  –  0.3 ( 3 ) ) / 0.2   =   ( 1.3  –  0.9 ) / 0.2   =   0.4 / 0.2   =   2

∴    x   =   2  and  y   =   3.

( v ) Given: √2 x  +  √3 y   =  0    and    √3 x  –  √8 y   =   0

are the two equations.

Solution :    From 1st equation, we get,

x   =    –  ( √3 / √2 ) y      ______     ( 1 )

Putting the value of x in the given second equation to get,

√3(   –  √3 / √2 ) y  –  √8 y   =   0 ⇒ (   –  3 / √2 ) y  –   √8 y   =   0

⇒  y   =   0

Now, substitute the value of y in equation ( 1 ), we get,

x   =   0

Therefore, x   =   0 and y   =   0.

( vi ) Given:( 3 x / 2 )  –  ( 5 y / 3 )   =  –  2   and    ( x / 3 )  +  ( y / 2 )   =   13 / 6 are the two equations.

Solution :From 1st equation, we get,

( 3 / 2 ) x   =     –  2  +  ( 5 y / 3 )

x   =   2(   –  6 + 5 y ) / 9   =   (   –  12 + 10y ) / 9 ………………………( 1 )

Putting the value of x in the given second equation to get,

( –  12 + 10 y ) / 9 ) / 3  +  y / 2   =   13 / 6

⇒  y / 2   =   13 / 6  – (–  12 + 10 y ) / 27  )  +  y / 2   =   13 / 6

Now, substitute the value of y in equation ( 1 ), we get,

( 3 x / 2 )  –  5( 3 ) / 3   =   –  2

⇒   ( 3 x / 2 )  –  5   =   –  2

x   =   2

Therefore,    x   =   2   and   y   =   3.

2.  Solve  2 x  +  3 y   =   11   and  2 x  –  4 y   =  –  24 and hence find the value of ‘m’ for                         which   y  =  mx  +  3.

Solution:

2 x  +  3 y   =   11 …….( i )
2 x   –   4 y   =     –  24……( ii )
solving these equations by substitution method,
Now, from ( ii ), we get,
2 x  –  4 y  =    –  24
2 x  =    –  24 + 4 y
x  =  (   –  24 + 4 y ) / 2
x  =    –  24 / 2 + 4 y / 2
x  =    –  12 + 2 y………..( iii )

Put this value of x in ( i ), we get,
2(   –  12 + 2 y )  +  3 y   =   11
–  24 + 4 y + 3 y  =  11          ⇒          7 y  =  11 + 24
7 y  =  35          ⇒           y  =  35 / 7
y  =  5
now put this value of y in ( iii )
=    –  12 + 2 y          =  – 12 + 2( 5 )
=    –  12 + 10
=    –  2

y  =  m x + 3                                      ( y  =  5   and   x  = –  2 )
5  =  m  (  –  2 ) + 3
5  =    –  2 m + 3       ⇒         5  –  3  =    –  2 m
2  =    –  2 m              ⇒         –  m   =   2 / 2
–  m   =   1                  ⇒       m   =  –  1

3.Form the pair of linear equations for the following problems and find their solution by substitution method.

( i )The difference between two numbers is 26 and one number is three times the other. Find them.

Solution:    Let the two numbers be and  y.

x  –  y  =   26               ………       ( i )

and   x  =  3 y

Substituting the value of x in equation      ( i )

3  y  –  y  =  26        ⇒        2 y  =  26          ⇒            y  =  13

x  =  3  ×  13  =  39

So,the numbers are 39 and 13.

( ii )The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Solution:     Let the two supplementary angles be x°and y°.

x° + y°  =  180°

x°  =  y° + 18°

substituting the value of x:

y° + 18° + y°  =  180°       ⇒           2 y° + 18°  =  180°

y°  =  81°                            ⇒           x°  =  180°  –  81°  =  99°

So the two angles are 99°and 81°.

( iii )The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.

Solution :    Let the cost of  1  bat  be  Rs .x  and  l  ball  be  Rs. y.

7 x + 6 y  =  3800

3 x + 5 y  =  1750

By solving we get      x  =  Rs.500 ,     y  =  Rs.50.

( iv ).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 km. How much does a person have to pay for travelling a distance of 25 km.

Solution:   Let fixed charge be Rs x and charge per km be Rs y.
⇒   x + 10y  =  105              ……….               ( i )
⇒   x + 15 y  =  155             ………..              ( ii ).
Now From equation  (ii)
⇒   15 y  =  155  −  x
⇒      y  =   ( 155  −  x )/ 15             ………    ( iii )
Substituting y from equation (iii) in equation (i)
⇒   x + 10 ( 155  −  x )​/ 15  =  105
⇒  15 x + 1550  −  10 x  =  1575              ⇒  5 x  =  1575  −  1550
⇒   5 x  =  25                     ⇒  x  =  5
Substituting  x in equation (ii)
⇒  5 + 15 y  =  155
⇒  15 y  =  155  −  5
⇒  15 y  =  150
⇒  y  =  10
Hence    x  =  5    and  y  =  10

For  25 km  a person has to pay   =   5 + 10  ×  25  =   Rs 255.

( v ). A fraction becomes 9/11  , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes  5/6.  Find the fraction.

Solution:  Let numerator be x and denominator be y.
If 2 is added to both numerator and denominator, the fraction becomes 9 / 11.

x + 2 / y + 2   =   9 / 11
11( x + 2 )   =   9( y + 2 )          ⇒         11 x + 22   =   9 y + 18

9 y  –  11 x   =   22  –  18          ⇒         9 y   –   11 x   =   4
9 y   =   4 + 11 x                            ⇒         y   =   ( 4 + 11 x ) / 9    …………….  (i)

If 3 is added to both numerator and denominator, the fraction becomes 5 / 6.
x + 3 / y + 3   =   5 / 6
6 ( x + 3 )   =   5( y + 3 )        ⇒          6 x + 18   =   5 y + 15
5 y   =   6 x + 18  –  15            ⇒           5 y   =   6 x + 3

y   =   ( 6 x + 3 ) / 5   …………….  (ii)

From (i) and (ii)  we get

( 4 + 11 x ) / 9   =   ( 6 x + 3 ) / 5
5( 4 + 11 x )   =   9( 6 x + 3 )        ⇒           20 + 55 x   =   54 x  +  27
55 x  –  54 x   =   27  –  20           ⇒           x   =   7
y   =   ( 4 + 11( 7 ) ) / 9   =   ( 4 + 77 ) / 9   =   81 / 9   =   9
∴    x   =   7   and   y   =   9
The fraction   =   7 / 9

( vi )Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages.

Solution:  Let ‘x’ be the present age of Jacob and ‘y’  be the age of his son

As per question

x + 5  =  3( y + 5 )           …..   ( i )

x  −  5  =  7( y  −  5 )        …… ( ii )

Adding ( i ) and ( ii )

2 x  =  3 y + 15 + 7 y  −  35          ⇒            2 x  =  10 y  −  20

x  =  5 y  −  10         …………   ( iii )

Putting ( iii )   in   ( i )

5 y  −  5  =  3 y + 15      ⇒   2 y  =  20    ⇒      y  =  10

Putting in ( iii ) we get

x  =  5  ×  10  −  10

x  =  40

∴   present age of Jacob is 40 years and his son is 10 years.

1.Solve the following pair of linear equations by the substitution method.

( a ) x  +  y   =   15   :   x  – y   =  8

( b ) 6 x  –  3 y   =   9  :    9 x  –  6 y   =   18

1. Solve  3 x + 4 y   =   14 and 2 x  –  4 y   = –  24   and hence find the value of ‘m’ for which y   =   m x  +  3.

1. Form the pair of linear equations for the following problems and find their solution by substitution method.

( a )The difference between two numbers is 36 and one number is three times the other. Find them.

( b )The larger of two supplementary angles exceeds the smaller by 22 degrees. Find them.

( c ) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 15 km, the charge paid is Rs 150 and for a journey of 18 km, the charge paid is Rs 175. What are the fixed charges and the charge per km. How much does a person have to pay for travelling a distance of 35 km.

Elimination method of finding solution of a pair of Linear Equations:

Consider   x  +  2 y   =   8   and   2 x  –  3 y   =   2

Step 1: Make the coefficients of any variable the same by multiplying the equations with constants.               Multiplying the first equation by 2, we get,

2 x  +  4 y   =   16

Step 2: Add or subtract the equations to eliminate one variable, giving a single variable equation.
Subtract second equation from the previous equation
2 x  +  4 y   =   16
2 x   –  3 y   =    2
(–)      ( +)         (–0
———————
0( x )  +  7 y   =  14
Step 3: Solve for one variable and substitute this in any equation to get the other variable.

y   =   2,

x   =   8  –  2 y             ⇒ x   =   8  –  4                  ⇒ x   =   4

( 4, 2 ) is the solution.

EXERCISE  3.4

1.Solve the following pair of linear equations by the elimination method and the substitution method :

x +  y   =   5  and    2 x  –  3 y   =   4

Solution:           x  +  y   =  5 and 2 x  – 3 y   =   4
By elimination method:
x  +  y   =  5                  ……. ( i )
2 x  – 3 y   =   4           …….. ( ii )
Multiplying equation  ( i )  by  ( ii ),  we get
2 x  +  2 y   =   10        ……….. ( iii )
2 x  – 3 y   =   4 … ( ii )
Subtracting equation ( ii ) from equation ( iii ), we get
5 y   =   6        ⇒            y   =   6 / 5
Putting the value in equation ( i ), we get
x   =   5   –   ( 6 / 5 )   =   19 / 5

Hence, x   =   19 / 5   and   y   =   6 / 5

By substitution method:     x  +  y   =   5         … ( i )
Subtracting y from  both sides, we get
x   =   5   –   y                        … ( iv )
Putting the value of x in equation ( ii ) we get
2( 5  –  y )  –  3 y   =   4      ⇒            –  5 y   =     –   6
y   =     –  6 /   –  5   =   6 / 5
Putting the value of y in equation ( iv ) we get
x   =   5  –  6 / 5        ⇒      x   =   19 / 5
Hence, x   =   19 / 5  and  y   =   6 / 5 .

(ii).     3 x +  4 y   =   10   and   2 x  –  2 y   =   2

Solution :       3 x  +  4 y   =   10 and 2 x  –  2 y   =   2
By elimination method:
3x  +  4 y   =   10                …. ( i )
2 x  –  2 y   =   2                  … ( ii )
Multiplying  equation   ( ii )   by   2,   we   get
4 x  –  4 y   =   4                … ( iii )
3 x  +  4 y   =   10               … ( i )
Adding equation   ( i )   and   ( iii ), we get
7 x  +  0   =   14
Dividing both side by 7,   we get
x   =   14 / 7   =   2
Putting in equation ( i ), we get
3 x  +  4 y   =   10      ⇒       3( 2 )  +  4 y   =   10
6  +  4 y   =   10         ⇒      4 y   =   10  –  6
4 y   =    4                    ⇒         y   =   4 / 4   =   1

Hence,   x   =   2,     y   =   1

(iii).   3 x  – 5 y  –  4   =   0 and   9 x   =   2 y  +  7

Solution :      By elimination method:
3 x  –  5 y  –  4   =   0
3 x  –  5 y   =   4                 ………  ( i )
9 x   =   2 y  +  7
9 x  –  2 y   =   7                ……. … ( ii )
Multiplying equation  ( i )  by  3,  we get
9 x  –  15 y   =   11             …… ( iii )
9 x  –  2 y   =   7                 ….. ( ii )
Subtracting equation ( ii ) from equation ( iii ), we get
–  13y   =   5
y   =     –  5 / 13
Putting value in equation ( i ), we get
3 x  –  5 y   =   4 … ( i )        ⇒         3 x   –   5 (   –  5 / 13 )   =   4
Multiplying by 13 we get
39 x  +  25   =   52        ⇒      39 x   =   27
x   =  27 / 39   =   9 / 13
Hence    x   =   9 / 13    and   y   = –  5 / 13

By substitution method:
3 x  –  5 y   =   4 … ( i )
Adding 5 y both side we get
3 x   =   4  +  5 y
Dividing by 3 we get
x   =   ( 4  +  5 y  ) / 3 … ( iv )
Putting this value in equation ( ii ) we get
9 x  –  2 y   =   7 … ( ii )
9 ( ( 4  +  5 y  ) / 3 )  –  2 y   =   7
Solve it we get
3( 4  +  5 y  )  –  2 y   =   7
12  +  15 y  –  2 y   =   7
13y   =     –   5
y   =     –  5 / 13
x   =   4  +  5 (    –  5 / 13 ) /  3
=   4   –   25 / 13  /  3
=   4   ×   13   –   25 / 13  /  3
=   27 / 13  ×  3
=   27 / 39
=   9 / 13
Hence we get x   =   9 / 13 and y   =     –   5 / 13 again.

(iv).    x / 2 +  2 y / 3   =     –   1 and x  –  y / 3   =   3

Solution:By elimination method
x  –  y / 3   =   3 … ( ii )
Multiplying equation ( i ) by 2, we get
x  +  4 y / 3   =     –   2 … ( iii )

x  –  y / 3   =   3 … ( ii )
Subtracting equation ( ii ) from equation ( iii ), we get
5 y / 3   =     –  5
Dividing by 5 and multiplying by 3, we get
y   =     –  15 / 5
y   =     –   3
Putting this value in equation ( ii ), we get
x  –  y / 3   =   3 … ( ii )
x  –  (   –  3 ) / 3   =   3
x  +  1   =   3
x   =   2
Hence our Answers is x   =   2 and y   =     −  3.
By substitution method
x  –  y / 3   =   3 … ( ii )
Add y / 3 both side, we get
x   =   3  +  y / 3 … ( iv )
Putting this value in equation ( i ) we get
x / 2  +  2 y / 3   =     –   1 … ( i )
( 3 +  y / 3 ) / 2  +  2 y / 3   =     –  1
3 / 2  +  y / 6  +  2 y / 3   =     –   1
Multiplying by 6, we get
9  +  y  +  4 y   =     –   6
5 y   =     –  15
y   =     –   3

Hence our Answers is x   =   2 and y   =     −  3.

1. Form the pair of linear equations in the following problems, and find their solutions ( if they exist ) by the elimination method :
( i ) If we add 1 tothe numerator and subtract 1 fromthe denominator, a fraction reduces to 1. It becomes  1 / 2 if we only add 1 to the denominator. What is the fraction.

Solution:   Let numerator   =   x    and denominator   =   y
∴  The fraction    =   x / y
If we add 1  to the numerator and subtract 1   from the denominator,the  fraction reduces to 1

Cross multiplying we get
x  +  1    =   y  –  1
x  –   y   =      –  2         ……..    ( i )

It becomes   1 / 2 if we only add 1 to the denominator.
Cross multiplying   we get
2 x     =   y  +  1
2 x  –   y   =    1        ……..   ( ii )
x  –   y   =      –  2         ……..( i )
Subtracting equation ( i ) from equation ( ii ) we get

x   =   3

Substituting this value in equation ( i ) we get
3 –  y   =     –          ⇒        –   y   =     –   5
∴    y   =   5
Hence our fraction is  3/5

( ii ) Five  years ago, Nuri was thrice as old as  Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu.

Solution:    Let the present age of Nuri    =   x year
And present age of Sonu   =   y year
Five years ago
Age of Nuri   =   x  –  5 years
Age of Sonu   =   y  –  5 years
Nuri was thrice as old as Sonu .So  the equation will be
x  –   5    =    3 ( y  –  5 )
x  –  5    =    35  –  15
x  –  3 y    =      –  15   +  5
x  –  3 y    =     –   10                        ………..( i )
Ten years later,
Age of Nuri    =   x  +  10
Age of Sonu   =   y  +  10
Nuri will be twice as old as Sonu.
x +  10   =   2( y + 10 )
x  +  10    =   2 y  +  20
x  –  2 y    =   10                           ………..( ii )
x  –  3 y    =  –   10                        ………..(i)
Subtracting equation ( i ) from equation ( ii )  we get
y    =   20
Plug this value in equation first we get
x    –   3(20)   =     –  10          ⇒          x   =   60  –  10
x    =   50

Hence age  of  Nuri   =   50 years and age of Sonu    =   20 years

( iii ) The sum of the digits of a two   digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Solution:   Let unit digit    =   x
Tens digit    =   y
Number will 10 times the tens digit  +  unit times the unit digit
Hence number will   10 y  +  x

Sum of digits are 9
So that x  +  y    =   9        ………….( i )
nine times this number is twice the number obtained by reversing the order of the digits

9 ( 10 y  +  x  )    =    2 ( 10 x  +  y  )
90 y  +  9 x    =   20 x  +  2 y
88 y  –  11 x    =   0
Divide by 11 we get
8 y    –   x    =   0     …………..( ii )
x  +  y    =   9        ………….( i )

9 y   =   9
y   =   9 / 9   =   1
Put  this value in equation (i) we get
x +  y   =   9        ⇒          x  +  1   =   9
x   =   8
So our original number is       10 y   +  x      =   10 (1)  +  8    =   18

( iv ) Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.

Solution:    The number of Rs 100 notes    =  y
According to the given information
Meena went to a bank to withdraw Rs 2000
So cost of 50 rupee notes    =   50 x
And cost of 100 rupee notes    =   100 y
Total cost is Rs 2000
∴    50 x  +  100 y    =   2000
Divide by 50 we get
x  +  2 y   =   40               ………( i )

Meena got 25 notes in all so that
x  +   y    =   25                ………( ii )
x  +  2 y   =   40               ………( i )

Subtracting equation ( i ) from ( ii) we get
–  y   =     –   15
y   =   15
Put this value in equation first we get
x  +  2 (15)   =    40      ⇒        x  +   30    =   40
x   =   10
Hence 50 rupee notes are 10 and 100 Rupee notes are 15

( v )  A  lending library has a fixed  charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Solution:  Let fixed change   =  Rs x
And charge for extra day    =   Rs  y
Saritha paid Rs 27 for a book kept for seven days
So first three days she paid  Rs x   and for remaining 4 days she will pay 4 y
So total payment will be
x  +  4 y   =   27               …….. ( i )
Similarly
For while Susy paid Rs 21 for the book she kept for five days
x  +  2 y    =   21               ……..( ii )
x  +  4 y   =   27               …….. ( i )
Subtracting equation ( i ) from equation ( ii ), we get
–  2 y    =     –   6
y   =   3
Plug this value in equation ( 1 ) we get
x  +  4 y   =   27               …….. ( 1 )
x  +  4 ( 3 )    =   27        ⇒            x   =   27  –  12
x   =   15
Hence fixed cost is Rs 15 and cost for extra each day is Rs 3.

1.Solve the following pair of linear equations by the elimination method and the substitution method :

( a ) 2 x  +  3 y   =   15    and  x  –  3y   =   14

( b )4 x  +  5 y   =   20   and  x  –  y   =   12

2.Form the pair of linear equations in the following problems, and find their solutions ( if they exist ) by the elimination method :

( a )The sum of the digits of a two    digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

( b )Arish went to a bank to withdraw Rs 5000. She asked the cashier to give her Rs 100 and  Rs 200 notes only. Meena got 35 notes in all. Find how many notes of Rs 100 and Rs  100 she received

Cross  –  multiplication Method of finding solution of a pair of Linear Equations

For the pair of linear equations

a 1 +  b 1 +  c 1  =  0

a 2x   +  b 2 y  +  c 2  =  0,
x and y can be calculated as

EXERCISE 3.5

1.Which of the following pairs of linear equations has unique solution, no solution or infinitely many solutions. In case there is unique solution, find it by using cross  –  multiplication method.
( i ) x  –  3 y  –  3   =   0,         3 x  –  9 y  –  2   =   0
( ii )
2 x  +  y   =   5,                  3 x  +  2 y   =   8
( iii ) 3 x  –  5 y   =   20,         6 x  –  10 y   =   40
( iv ) x  –  3 y  –  7   = 0,        3 x  –  3 y  –  15   =   0

Solution:( i ) Given, x  –  3 y  –  3   =  0   and  3 x  –  9 y   –  2   =  0

a 1 / a 2  =  1 / 3 ,         b 1 / b 2  = –  3 / –  9  =  1 / 3,         c 1 / c 2  =  –  3 / –  2   =  3 / 2

( a 1 / a 2 )   =   ( b 1 / b 2 )   ≠   ( c 1 / c 2 )

Since, the given set of lines are parallel to each other they will not intersect each other and therefore there is no solution for these equations.

( ii ) Given, 2 x  +  y   =   5 and 3 x  + 2 y   =   8

a 1 / a 2   =   2 / 3 ,          b 1 / b 2   =   1 / 2 ,         c 1 / c 2   =  –  5 / –  8

( a 1 / a 2 )   ≠   ( b 1 / b 2 )

Since they intersect at a unique point these equations will have a unique solution by cross multiplication method:

x / ( b 1 c 2  –  c 1 b 2 )   =   y / ( c 1 a 2  –  c 2 a1 )   =   1 / ( a 1b 2  –  a 2b 1 )

x / (   –  8  –  (   –  10 ) )   =   y / ( 15 + 16 )   =   1 / ( 4  –  3 )

x / 2   =   y / 1   =   1

x   =   2 and y   =  1

( iii ) Given, 3 x  –  5 y   =   20 and 6 x  –  10 y   =   40

( a 1 / a 2 )   =   3 / 6   =   1 / 2

( b 1 / b 2 )   =     –  5 /   –  10   =   1 / 2

( c 1 / c 2 )   =   20 / 40   =   1 / 2

a 1 / a 2   =   b 1 / b 2   =   c 1 / c 2

Since the given sets of lines are overlapping each other there will be infinite number of solutions for this pair of equation.

( iv ) Given, x  –  3 y  –  7   =   0 and 3 x  –  3 y  –  15   =   0

( a 1 / a 2 )   =   1 / 3

( b 1 / b 2 )   =     –  3 /   –  3   =   1

( c 1 / c 2 )   =     –  7 /   –  15

a 1 / a 2   ≠   b 1 / b 2

Since this pair of lines are intersecting each other at a unique point, there will be a unique solution.

By cross multiplication,

x / ( 45  –  21 )   =   y / (   –  21 + 15 )   =   1 / (   –  3 + 9 )

x / 24   =   y /    –  6   =   1 / 6

x / 24   =   1 / 6 and y /   –  6   =   1 / 6

x   =   4 and y   =   1.

2.( i ) For which values of a and b does the following pair of linear equations have an infinite number of solutions.
2 x  +  3 y   =   7
( a  –  b ) x  +  ( a  +  b ) y   =   3 a  +  b− 2

( ii ) For which value of k will the following pair of linear equations have no Solution

3x  + y   =   1 ; ( 2k  – 1 ) x  +  ( k  – 1 ) y   =   2k

1. Solve the following pair of linear equations by the substitution and cross – multiplication methods:

8x  +  5 y   =   9

3x  +  2 y   =   4

Solution. Substitution Method:

8 x  +  5 y   =   9                   … ( i )

3 x  +  2 y   =   4                  … ( ii )

1. Form the pair of linear equations in the following problems and find their solutions ( if they exist ) by any algebraic method:

( i ) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of food per day.

Solution:    Let fixed monthly charge   =   Rs x and let charge of food for one day   =   Rs y

According to given conditions,

x  + 20 y  =   1000                     … ( i ),

and x  +  26 y   =   1180            … ( ii )

Subtracting equation ( i ) from equation ( ii ), we get

6 y   =   180                  ⇒ y   =   30

Putting value of y in ( i ), we get

x  +  20 ( 30 )   =   1000

x   =   1000  –  600   =   400

Therefore, fixed monthly charges   =   Rs 400 and, charges of food for one day   =   Rs 30

( ii ) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction.

Solution:   Let numerator   =   x     and let denominator   =   y

According to given conditions

(x  − 1)/y  =  1/3    and      x / ( y + 8)  =  1/4

⇒  3 x  –  3   =   y                        … ( i )

⇒  4x   =   y  +  8                          … ( ii )

Subtracting equation ( i ) from ( ii ), we get

4 x  –  y   −   ( 3 x   −   y )   =   8  –  3

x   =   5

Putting value of x in ( i ), we get

3 ( 5 )  –  y   =   3              ⇒ 15  –  y   =   3

⇒ y   =   12

∴ numerator   =   5 and, denominator   =   12

The  fraction   =    5 /12 .

( iii ) Yash scored 40 marks in a test, getting 3 marks for each right answers and losing 1 mark for each wrong answers. Had 4 marks been awarded for each correct answers and 2 marks been deducted for each incorrect answers, then Yash would have scored 50 marks. How many questions were there in the test.

Solution:  Let number of correct answers   =   x            and let number of wrong answers   =   y

According to given conditions,

3  x  –  y   =   40                  … ( i )

And, 4 x   −   2 y   =   50     … ( ii )

From equation ( i ), y   =   3 x   −   40

Putting this in ( ii ), we get

4 x  –  2 ( 3 x   −   40 )   =   50         ⇒   4 x   −   6 x  +  80   =   50

⇒   −  2 x   =     −  30                         ⇒    x   =   15

Putting value of x in ( i ), we get

3 ( 15 )  –  y   =   40                  ⇒ 45  –  y   =   40                ⇒ y   =   45  –  40  ⇒   5

Therefore, number of correct answers    =  x  =  15    and number of wrong solution  =   y   =  5

Total questions   =   x  +  y   =   15  +  5   =   20

( iv ) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars.

Solution: Let speed of car which starts from part  A   =   x km / hr

Let speed of car which starts from part B   =   y km / hr

According to given conditions,

( v ) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Solution:  Let length of rectangle   =   x units   and let breadth of rectangle   =   y units

Area   =  x y square units. According to given conditions,

x y  –  9   =   ( x   −   5 ) ( y  +  3 )

⇒ x y  –  9   =   x y  +  3 x   −   5 y  –  15                  ⇒    3 x   −   5 y   =   6 … ( 1 )

And, x y  +  67   =   ( x  +  3 ) ( y  +  2 )                   ⇒    x y  +  67   =   x y  +  2 x  +  3 y  +  6

⇒ 2 x  +  3 y   =   61 … ( 2 )

From equation ( i ),    3 x   =   6  +  5 y

x   =  6 + 5 y /3

Putting this in ( ii ), we get

2   +  3 y   =   61                 ⇒    12  +  10 y  +  9 y   =   183

⇒ 19 y   =   171                   ⇒    y   =   9 units

Putting value of y in ( ii ), we get

2 x  +  3 ( 9 )   =   61                  ⇒  2 x   =   61  –  27   =   34

x   =   17 units

∴    length   =   17 units   and   breadth   =   9 units

1.Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.

( a ) 3 x  –  5 y   =  20 :   3 x  +  2 y   =   40

( b ) 3 x  –  7 y  –  21   =   0:6 x  –  9 y  –  18   =   0

1. For which values of a and b does the following pair of linear equations have an infinite number of solutions?

3 x  +  5 y   =   10

( a  –  b ) x  +  ( a  +  b ) y   =   3 a  +  b  –  2

1. Solve the following pair of linear equations by the substitution and cross – multiplication methods :

6 x  +  8 y   =   12

4 x  +  5 y   =   16.