Question

Solve each of the following systems of equations by the method of cross-multiplication:
(a+2b)x+(2a-b)y = 2
(a – 2b)x + (2a + b)y = 3

Answer 1

The value of x and y for system of equation is $\dfrac{1}{2a}$ - $\dfrac{1}{5b}$  and  $\dfrac{1}{10b}$ + $\dfrac{1}{a}$  

Step-by-step explanation:

Given as :

The system of equation are

( a + 2 b ) x + ( 2 a - b ) y = 2                       ...........1

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

Adding the equation 1 and equation 2

x [ ( a + 2 b ) + ( a - 2 b ) ] + y [ ( 2 a - b ) + ( 2 a + b ) ] = 2 + 3

Or, x ( 2 a ) + y ( 4 a ) = 5

i.e   2 a x + 4 a y = 5                               ..............3

Subtracting the equation 1 from equation 2

x [ ( a - 2 b ) - ( a + 2 b ) ] + y [ ( 2 a + b ) - ( 2 a - b ) ] = 3 - 2

Or, x ( - 4 b ) + y ( 2 b ) = 1

i.e - 4 b x + 2 b y = 1                              .................4

Solving eq 3 and 4

a ( - 4 b x + 2 b y ) + 2 b ( 2 a x + 4 a y )  = a × 1 + 2 b × 5

Or, x ( - 4 b a + 4 b a ) + y ( 2 a b + 8 a b ) = a + 10 b

Or,  0 + y ( 10 a b ) = a + 10 b

∴    y = $\dfrac{a+10b}{10ab}$

i.e  y = $\dfrac{1}{10b}$ + $\dfrac{1}{a}$

Again

Put the value of y in eq 4

 - 4 b x + 2 b y = 1  

i.e   4 b x =  2 b y - 1  

Or, 4 b x =  2 b ( $\dfrac{1}{10b}$ + $\dfrac{1}{a}$ ) - 1

Or, 4 b x = $\dfrac{1}{5}$ + $\dfrac{2b}{a}$ - 1

or, 4 b x = $\dfrac{2b}{a}$ - $\dfrac{4}{5}$

∴          x = $\dfrac{1}{2a}$ - $\dfrac{1}{5b}$

Hence, The value of x and y for system of equation is $\dfrac{1}{2a}$ - $\dfrac{1}{5b}$ and $\dfrac{1}{10b}$ + $\dfrac{1}{a}$  Answer