`star` Cross - Multiplication Method

● We introduce one more algebraic method to solve a pair of linear equations which for many reasons is a very useful method of solving these equations. Before we proceed further, let us consider the following situation.

● The cost of 5 oranges and 3 apples is Rs. 35 and the cost of 2 oranges and 4 apples is Rs. 28. Let us find the cost of an orange and an apple.

● Let us denote the cost of an orange by Rs. x and the cost of an apple by Rs. y. Then, the equations formed are :

`5x + 3y = 35,` i.e.,` 5x + 3y – 35 = 0`............ (1)

`2x + 4y = 28,` i.e., `2x + 4y – 28 = 0`.......... (2)

● Let us use the elimination method to solve these equations.

Multiply Equation (1) by 4 and Equation (2) by 3. We get

`(4)(5)x + (4)(3)y + (4)(–35) = 0`............ (3)

`(3)(2)x + (3)(4)y + (3)(–28) = 0`............ (4)

Subtracting Equation (4) from Equation (3), we get

`[(5)(4) – (3)(2)]x + [(4)(3) – (3)(4)]y + [4(–35) – (3)(–28)] = 0`

Therefore, `x = ( - [ (4) (-35) - (3) (-28 ) ] )/( (5) (4) - (3) (2) )`

i.e., `x = ( (3) (-28 ) - (4) ( -35) )/( (5) (4) - (2) ( 3) )` ................ (5)

● If Equations (1) and (2) are written as `a_1 x + b_1 y + c_1 = 0` and `a_2 x + b_2 y + c_2 = 0`, then we have

`a_1 = 5, b_1 = 3, c_1 = –35, a_2 = 2, b_2 = 4, c_2 = –28`.

● Then Equation (5) can be written as `color{orange}{x = (b_1 c_2 - b_2 c_1 )/( a_1 b_2 - a_2 b_1 )}`

● Similarly, you can get `color{orange}{y = (c_1 a_2 - c_2 a_1 )/( a_1 b_2 - a_2 b_1)}`

By simplyfing Equation (5), we get

`x = (-84 +140 )/(20-6)=4`

Similarly, ` y = ( (-35) (2) - (5) ( -28 ) )/(20- 6 ) = (-70+140 )/14 =5`

● Therefore, `x = 4, y = 5` is the solution of the given pair of equations.

Then, the cost of an orange is Rs. 4 and that of an apple is Rs, 5.

`"Verification :"` Cost of 5 oranges + Cost of 3 apples = ` 20 + 15 = 35.` Cost of 2 oranges + Cost of 4 apples `= 8 + 20 = 28.`

● The cost of 5 oranges and 3 apples is Rs. 35 and the cost of 2 oranges and 4 apples is Rs. 28. Let us find the cost of an orange and an apple.

● Let us denote the cost of an orange by Rs. x and the cost of an apple by Rs. y. Then, the equations formed are :

`5x + 3y = 35,` i.e.,` 5x + 3y – 35 = 0`............ (1)

`2x + 4y = 28,` i.e., `2x + 4y – 28 = 0`.......... (2)

● Let us use the elimination method to solve these equations.

Multiply Equation (1) by 4 and Equation (2) by 3. We get

`(4)(5)x + (4)(3)y + (4)(–35) = 0`............ (3)

`(3)(2)x + (3)(4)y + (3)(–28) = 0`............ (4)

Subtracting Equation (4) from Equation (3), we get

`[(5)(4) – (3)(2)]x + [(4)(3) – (3)(4)]y + [4(–35) – (3)(–28)] = 0`

Therefore, `x = ( - [ (4) (-35) - (3) (-28 ) ] )/( (5) (4) - (3) (2) )`

i.e., `x = ( (3) (-28 ) - (4) ( -35) )/( (5) (4) - (2) ( 3) )` ................ (5)

● If Equations (1) and (2) are written as `a_1 x + b_1 y + c_1 = 0` and `a_2 x + b_2 y + c_2 = 0`, then we have

`a_1 = 5, b_1 = 3, c_1 = –35, a_2 = 2, b_2 = 4, c_2 = –28`.

● Then Equation (5) can be written as `color{orange}{x = (b_1 c_2 - b_2 c_1 )/( a_1 b_2 - a_2 b_1 )}`

● Similarly, you can get `color{orange}{y = (c_1 a_2 - c_2 a_1 )/( a_1 b_2 - a_2 b_1)}`

By simplyfing Equation (5), we get

`x = (-84 +140 )/(20-6)=4`

Similarly, ` y = ( (-35) (2) - (5) ( -28 ) )/(20- 6 ) = (-70+140 )/14 =5`

● Therefore, `x = 4, y = 5` is the solution of the given pair of equations.

Then, the cost of an orange is Rs. 4 and that of an apple is Rs, 5.

`"Verification :"` Cost of 5 oranges + Cost of 3 apples = ` 20 + 15 = 35.` Cost of 2 oranges + Cost of 4 apples `= 8 + 20 = 28.`

● Let us now see how this method works for any pair of linear equations in two variables of the form

`a_1 x + b_1 y + c_1 = 0` .........(1)

and `a_2 x + b_2 y + c_2 = 0`.......... (2)

● To obtain the values of x and y as shown above, we follow the following steps:

Step 1 : Multiply Equation (1) by `b_2` and Equation (2) by `b_1` , to get

`b_2 a_1 x + b_2 b_1 y + b_2 c_ 1 = 0`............ (3)

`b_1 a_2 x + b_1 b_2 y + b_1 c_2 = 0` ...........(4)

Step 2 : Subtracting Equation (4) from (3), we get:

`(b_2 a_1 – b_1 a_2) x + ( b_2 b_1 – b_1 b_2) y + (b_2 c_1– b_1 c_2) = 0`

i.e., `(b_2 a_1 – b_1 a_2) x = b_1 c_2 – b_2 c_1`

So, `x= (b_1 c_2 - b_2 c_1 )/( a_1 b_2 - a_2 b_1)` , provided `a_1 b_2– a_2 b_1≠ 0`........ (5)

Step 3 : Substituting this value of x in (1) or (2), we get

` y = (c_1 a_2 - c_2 a_1)/( a_1 b_2 - a_2 b_1 )` ........(6)

● Now, two cases arise :

Case 1 : `a_1 b_2 – a_2 b_1 ≠ 0`. In this case `a_1/a_2 ≠ b_1/b_2` . Then the pair of linear equations has a unique solution.

Case 2 :` a_1 b_2 – a_2 b_1 = 0`. If we write `a_1/a_2 = b_1/b_2 = k` , then `a_1 = k a_2 , b_1 = k b_2` .

Substituting the values of a1 and b1 in the Equation (1), we get

`k (a_2 x + b_2 y) + c_1 = 0`........... (7)

It can be observed that the Equations (7) and (2) can both be satisfied only if

`c_1 = k c_2` , i.e., `c_1/c_2 = k` .

● If `c_1 = k c_2`, any solution of Equation (2) will satisfy the Equation (1), and vice versa. So, if `a_1/a_2 = b_1/b_2 = c_1/c_2 = k` , then there are infinitely many solutions to the pair of linear equations given by (1) and (2).

● If `c_1 ≠ k c_2` , then any solution of Equation (1) will not satisfy Equation (2) and vice versa. Therefore the pair has no solution.

We can summarise the discussion above for the pair of linear equations given by (1) and (2) as follows:

(i) When `color{orange}{a_1/a_2 ≠ b_1/b_2}` , we get a `"unique"` solution.

(ii) When ` color{orange}{a_1/a_2 = b_1/b_2 = c_1/c_2}` , there are `"infinitely"` many solutions.

(iii) When `color{orange}{a_1/a_2 = b_1/b_2 ≠ c_1/c_2}` , there is `"no solution."`

● Note that you can write the solution given by Equations (5) and (6) in the following form :

`x/(b_1 c_2 - b_2 c_1) = y/(c_1 a_2 - c_2 a_1) = 1/(a_1 b_2 - a_2 b_1)`.........(8)

● In remembering the above result, the following diagram may be helpful to you :

● The arrows between the two numbers indicate that they are to be multiplied and the second product is to be subtracted from the first. For solving a pair of linear equations by this method, we will follow the following steps :

Step 1 : Write the given equations in the form (1) and (2).

Step 2 : Taking the help of the diagram above, write Equations as given in (8).

Step 3 : Find x and y, provided `a_1 b_2 – a_2 b_1 ≠ 0`

Step 2 above gives you an indication of why this method is called the

cross-multiplication method.

`a_1 x + b_1 y + c_1 = 0` .........(1)

and `a_2 x + b_2 y + c_2 = 0`.......... (2)

● To obtain the values of x and y as shown above, we follow the following steps:

Step 1 : Multiply Equation (1) by `b_2` and Equation (2) by `b_1` , to get

`b_2 a_1 x + b_2 b_1 y + b_2 c_ 1 = 0`............ (3)

`b_1 a_2 x + b_1 b_2 y + b_1 c_2 = 0` ...........(4)

Step 2 : Subtracting Equation (4) from (3), we get:

`(b_2 a_1 – b_1 a_2) x + ( b_2 b_1 – b_1 b_2) y + (b_2 c_1– b_1 c_2) = 0`

i.e., `(b_2 a_1 – b_1 a_2) x = b_1 c_2 – b_2 c_1`

So, `x= (b_1 c_2 - b_2 c_1 )/( a_1 b_2 - a_2 b_1)` , provided `a_1 b_2– a_2 b_1≠ 0`........ (5)

Step 3 : Substituting this value of x in (1) or (2), we get

` y = (c_1 a_2 - c_2 a_1)/( a_1 b_2 - a_2 b_1 )` ........(6)

● Now, two cases arise :

Case 1 : `a_1 b_2 – a_2 b_1 ≠ 0`. In this case `a_1/a_2 ≠ b_1/b_2` . Then the pair of linear equations has a unique solution.

Case 2 :` a_1 b_2 – a_2 b_1 = 0`. If we write `a_1/a_2 = b_1/b_2 = k` , then `a_1 = k a_2 , b_1 = k b_2` .

Substituting the values of a1 and b1 in the Equation (1), we get

`k (a_2 x + b_2 y) + c_1 = 0`........... (7)

It can be observed that the Equations (7) and (2) can both be satisfied only if

`c_1 = k c_2` , i.e., `c_1/c_2 = k` .

● If `c_1 = k c_2`, any solution of Equation (2) will satisfy the Equation (1), and vice versa. So, if `a_1/a_2 = b_1/b_2 = c_1/c_2 = k` , then there are infinitely many solutions to the pair of linear equations given by (1) and (2).

● If `c_1 ≠ k c_2` , then any solution of Equation (1) will not satisfy Equation (2) and vice versa. Therefore the pair has no solution.

We can summarise the discussion above for the pair of linear equations given by (1) and (2) as follows:

(i) When `color{orange}{a_1/a_2 ≠ b_1/b_2}` , we get a `"unique"` solution.

(ii) When ` color{orange}{a_1/a_2 = b_1/b_2 = c_1/c_2}` , there are `"infinitely"` many solutions.

(iii) When `color{orange}{a_1/a_2 = b_1/b_2 ≠ c_1/c_2}` , there is `"no solution."`

● Note that you can write the solution given by Equations (5) and (6) in the following form :

`x/(b_1 c_2 - b_2 c_1) = y/(c_1 a_2 - c_2 a_1) = 1/(a_1 b_2 - a_2 b_1)`.........(8)

● In remembering the above result, the following diagram may be helpful to you :

● The arrows between the two numbers indicate that they are to be multiplied and the second product is to be subtracted from the first. For solving a pair of linear equations by this method, we will follow the following steps :

Step 1 : Write the given equations in the form (1) and (2).

Step 2 : Taking the help of the diagram above, write Equations as given in (8).

Step 3 : Find x and y, provided `a_1 b_2 – a_2 b_1 ≠ 0`

Step 2 above gives you an indication of why this method is called the

cross-multiplication method.

Q 3149778613

From a bus stand in Bangalore , if we buy 2 tickets to Malleswaram and

3 tickets to Yeshwanthpur, the total cost is Rs. 46; but if we buy 3 tickets to Malleswaram

and 5 tickets to Yeshwanthpur the total cost is Rs. 74. Find the fares from the bus stand

to Malleswaram, and to Yeshwanthpur.

Class 10 Chapter 3 Example 14

3 tickets to Yeshwanthpur, the total cost is Rs. 46; but if we buy 3 tickets to Malleswaram

and 5 tickets to Yeshwanthpur the total cost is Rs. 74. Find the fares from the bus stand

to Malleswaram, and to Yeshwanthpur.

Class 10 Chapter 3 Example 14

Let Rs. x be the fare from the bus stand in Bangalore to Malleswaram, and

Rs. y to Yeshwanthpur. From the given information, we have

`2x + 3y = 46`, i.e., `2x + 3y – 46 = 0` ........(1)

`3x + 5y = 74`, i.e., `3x + 5y – 74 = 0` .........(2)

To solve the equations by the cross-multiplication method, we draw the diagram as

given below.

Then

`x/( (3) ( -74 ) - (5) ( -46 ) ) = y/( (-46) (3) - (-74 ) (2) ) = 1/( (2) (5) - (3) (3) )`

i.e., `x/(-222 +230) = y/(-138 +148) = 1/(10-9)`

i.e, `x/8 = y/10 = 1/1`

i.e., `x/8 = 1/1` and `y/10 =1/1`

i.e., `x= 8 ` and ` y =10`

Hence, the fare from the bus stand in Bangalore to Malleswaram is ` 8 and the fare to

Yeshwanthpur is Rs. 10.

Verification : You can check from the problem that the solution we have got is correct.

Q 3159778614

For which values of p does the pair of equations given below has unique

solution?

4x + py + 8 = 0

2x + 2y + 2 = 0

Class 10 Chapter 3 Example 15

solution?

4x + py + 8 = 0

2x + 2y + 2 = 0

Class 10 Chapter 3 Example 15

Here `a_1 = 4, a_2 = 2, b_1 = p, b_2 = 2`.

Now for the given pair to have a unique solution : `a_1/a_2 ≠ b_1/b_2`

i.e., `4/2 ≠ p/2`

i.e., `p ≠ 4`

Therefore, for all values of p, except 4, the given pair of equations will have a unique

solution.

Q 3179778616

For what values of k will the following pair of linear equations have

infinitely many solutions?

`kx + 3y – (k – 3) = 0`

`12x + ky – k = 0`

Class 10 Chapter 3 Example 16

infinitely many solutions?

`kx + 3y – (k – 3) = 0`

`12x + ky – k = 0`

Class 10 Chapter 3 Example 16

Here, `a_1/a_2 =k/12 , b_1/b_2 = 3/k , c_1/c_2 = (k-3)/k`

For a pair of linear equations to have infinitely many solutions : `a_1/a_2 = b_1/b_2= c_1/c_2`

So, we need `k/12 = 3/k = (k-3)/k`

or, `k/12 = 3/k`

which gives `k^2 = 36`, i.e.,` k = ± 6`.

Also , `3/k = (k-3)/k`

gives `3k = k^2 – 3k`, i.e., `6k = k^2`

, which means k = 0 or k = 6.

Therefore, the value of k, that satisfies both the conditions, is k = 6. For this value, the

pair of linear equations has infinitely many solutions.