Question

Question 3 Solve the following pair of linear equations by the substitution and cross-multiplication methods:
8x+5y=9
3x+2y=4

Class 10 - Math - Pair of Linear Equations in Two Variables Page 62

Answer 1

Cross multiplication method :

8x + 5y = 9
3x + 2y = 4

= 8x + 5y - 9 = 0
3x + 2y - 4 = 0

x/ (- 20 + 18 )
y/ ( -32 + 27)
1/ ( 16 - 15 )

x(-2) y( -5 ) 1(1)

x = - 2

or

y = 5

Answer 2

Substitution method:

The general form of a pair of linear equations
 
a1x + b1y + c1 = 0 , &        a2x + b2y + c2 = 0

1. The first step to solve a pair of linear equations by the substitution method is to solve one equation for either of the variables.
2. The choice of equation or variable in a given pair does not affect the solution for the pair of equations.
3. In the next step, we’ll substitute the resultant value of one variable obtained in the other equation and solve for the other variable.
4. In the last step, we can substitute the value obtained of the variable in any one equation to find the value of the second variable.

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Cross - multiplication method:

 The general form of a pair of linear equations
 
a1x + b1y + c1 = 0 , &        a2x + b2y + c2 = 0.

When a1 / a2 ≠ b1 / b2, the pair of linear equations will have a unique solution.


To solve this pair of equations for x and y using cross-multiplication,
we’ll arrange the variables x and y and their coefficients a1, a2, b1 and b2, and the constants c1 and c2  as shown below



⇒ x =  b 1 c 2 - b 2 c 1  / a 1 b 2 - a 2 b 1  

⇒ y =  c 1 a 2 - c 2 a 1  / a 1 b 2 - a 2 b 1

The above equation is generally written as :

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
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Solution:

By substitution method

8x +5y = 9 … (i)
3x +2y = 4 … (ii)

From equation (ii), we get
x = 4-2y/3 … (iii)

Putting this value in equation (i), we get
8(4-2y/3) + 5y = 9

(32 – 16y +15y) / 3= 9 

(32 – y)= 9 ×3

-y= 27-32

–y = -5

y = 5 …....................... (iv)

Putting this value in equation (ii), we get
3x + 10 = 4
x = -2

Hence, x = -2, y = 5

By cross multiplication  we get

8x + 5y -9 = 03x + 2y – 4 = 0

a1=8, b1= 5, c1=-9
a2=3, b2=2, c2=-4
 
Formula for cross multiplication 
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

x/-20-(-18) / = y/-27-(-32) = 1/16-15
x/-2 = y/5 = 1/1
x/-2 = 1             &      y/5 = 1

x = -2 and y = 5

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hope this will help you.......