Question

Solve this eq by substitution method

Answer 1

Answer:

x = - 1 / 2  and y = 2

Step-by-step explanation:

Given :

2 ( a x - b y ) + a + 4 b = 0

Rewrite as

2 a x - 2 b y + a + 4 b = 0

Now taking a and 2 b common we get

a ( 2 x + 1 ) - 2 b ( y - 2 ) = 0

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

Another equation given :

2 ( b x + a y ) + b - 4 a = 0

Rewrite as

2 b x + 2 a y  + b - 4 a = 0

Again taking a and 2 b common we get

b ( 2 x + 1 ) + 2 a ( y - 2 ) = 0

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

Now we can see in both ( i )  and  ( ii ) equation there is  2 x + 1 and  y - 2

Let ,  2 x + 1  = α  and  y - 2 =  β

Now we have :

a α = 2 b β  ... ( iii )   and  b α  = - 2 a β ( iv)

Now multiply by a in ( iii )  and b  in ( iv ) we get

a α = 2 a b β

b α  = - 2 a β

Adding both we get

a α + b α = 0

α ( a + b ) = 0

a + b = 0

a = - b

Putting a = - b in ( iii )  and ( iv )

a α = 2 b β   and  b α  = - 2 a β

- b α = 2 b β ⇒  α = - 2 β ( v )

b α  = - 2 ( - b )  β ⇒  α = 2 β ( vi )

From ( v )  and  ( vi )

- 2 β = -2 β

β = 0 and  α = 0

Since α and β are zero .

Putting value which we have let :

2 x + 1  = α  and   y - 2 =  β

2 x + 1 = 0

x = - 1 / 2

y - 2 =  0

y = 2

Thus we get answer x = - 1 / 2 and y = 2 .

Answer 2

$\underline{\large{\sf Answer:}}$

here we have given Two equations,

2(ax - by) + a + 4b = 0 -------(i)

2(bx + ay)+( b - 4a) = 0 --------(ii)

from equation (1) we get,

$\sf 2ax-2by+a+4b = 0$---(iii)

$\implies \sf 2ax+a -2by+4b = 0$

$\implies \sf a(2x+1)-2b(y-2) = 0$

$\implies \sf a(2x+1)=2b(y-2)$

$\implies \sf (2x+1)=(\frac{2b(y-2)}{a})$

----------(iv)

Now, take equation (ii)

$\sf 2(bx + ay)+ b - 4a = 0$

$\sf 2bx + 2ay + b - 4a =0$

$\sf 2bx + b + 2ay - 4a =0$

$\implies \sf b(2x + 1)+2ay-4a=0$

put the value of (2x+1) from equation (iv)

$\implies \sf b(\frac{2b(y-2)}{a})+2ay - 4a = 0$

$\implies \sf 2b^2(y-2)+2a^2y-4a^2=0\times a$

$\implies \sf 2b^2y-4b^2+2a^2y-4a^2=0$

$\implies \sf 2b^2y+2a^2y-4b^2-4a^2=0$

$\implies \sf y(2b^2+2a^2)-2(2b^2+2a^2)=0$

$\implies \sf y(2b^2+2a^2)=2(2b^2+2a^2)$

$\implies \boxed{\sf y = 2}$

Now, put the value of y in equation (iii)

$\sf 2ax-2by+a+4b=0$

$\implies \sf 2ax -2b(2)+a+4b=0$

$\implies \sf 2ax -4b + a + 4b =0$

$\implies \sf 2ax +a=0$

$\implies \sf 2ax = -a$

$\implies \sf 2x =-1$

$\implies \boxed{\sf x=(\frac{-1}{2})}$

Hence the values of x and y are,

$\boxed{\sf x =-(\frac{1}{2})}$and$\boxed{\sf y=2}$