Method 4

Solution of Simultaneous Equations

This method is based on the Sutra "Patavartya Yogayet" which means ' transpose and apply ' .

Explaining the method by an esample .

Example 1 : - Find the values of x and y .
2x + 3y = 9
3x + y = 4

Answer : -

Step 1 : -

First finding the value of x .
Cross multipling the cofficients of y and constants and subtracting rightward .

2.x + 3 y 9

3.x + 1 y 4

( 3 * 4 ) - ( 9 * 1 ) (Subtract from left to right )

= 3 which is numerator for x .

Step 2 : -

Now cross multipling the cofficients of x and y and subtracting leftward .

2.x + 3.y 9

3.x + 1.y 4

( 3 * 3 ) - ( 2 * 1 ) (Subtract from right to left )
= 9 - 2
= 7 which is the denominator .
x = 3 / 7

Step 3 : -

Finding the value of y .
Now cross multipling the cofficients of x and constants and subtracting leftward .

3.y + 2.x 9

1.y + 3.x 4

( 9 * 3 ) - ( 2 * 4 )
(Subtract from right to left Starting from constants)
= 27 - 8
= 19 which is numerator .

Denominator is same as step 2 .

So y = -19 / 7

Hence
x = 3 / 7
y = -19 / 7

Algebraic Proof : -

Let there be two equations
a.x + b.y = m ...........( i )
c.x + d.y = n ...........( ii )
Multiply ( i ) by d and ( ii ) by b .

a.d.x + b.d.y = md
c.b.x + d.b.y = nb (Subtracting)
(a.d - c.b).x = m.d - n.b

Hence x = (md - nb)
(a.d - b.c)
x = ( b.n - m.d)
( b.c - a.d)

which is same as step ( i ).

Now multipling ( i ) by c and ( ii ) by a And subtracting ( ii ) from ( i )

We get
y = ( m.c - n.a )
( b.c - a.d )
which is same as step 2 .

Exercises : -

Find the value of x and y in the following equations : -
1.) 2x + y = 5
3x – 4y = 2
2.) 3x – 4y = 7
5x + y = 4
3.) 4x + 3y = 8
6x - y = 1
4.) x + 3y = 7
2x + 5y = 11
5.) 3x + y = 5
4x – y = 9