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Solve the following systems of equations by the method of cross-multiplication: $2 (a x − b y) + a + 4 b = 0 2 (b x + a y) + b − 4 a = 0$

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the solution of the given system of equation is

x = − 12

and

y = 2

the solution of the given system of equation is

x = 12

and

y = 2

the solution of the given system of equation is

x = − 12

and

y = 4

the solution of the given system of equation is

x = − 32

and

y = 2

answer is A.

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Detailed Solution

Given linear equations are,

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

We know that if the two linear equations are of the form,

a 1 x + b 1 y + c 1 = 0 a 2 x + b 2 y + c 2 = 0

Then Cross multiplication formula is,

x b 1 c 2 − b 2 c 1 = y c 1 a 2 − c 2 a 1 = 1 b 2 a 1 − b 1 a 2

Convert given equation in linear standard form,

a 1 x + b 1 y + c 1 = 0

2 a x − 2 b y + a + 4 b = 0

...............(1)

2 b x + 2 a y + b − 4 a = 0

… … . (2)

Compare equation (1),

2 a x − 2 b y + a + 4 b = 0

with standard linear equation

a 1 x + b 1 y + c 1 = 0

We get,

a 1 = 2 a b 1 = − 2 b c 1 = a + 4 b

Now, compare equation (2),

2 b x + 2 a y + b − 4 a = 0

with standard linear equation

a 2 x + b 2 y + c 2 = 0

We get,

a 2 = 2 b b 2 = 2 a c 2 = b − 4 a

Applying formula of cross multiplication

x b 1 c 2 − b 2 c 1 = y c 1 a 2 − c 2 a 1 = 1 b 2 a 1 − b 1 a 2

.
Substitute the values of

a 1 = 2 a, b 1 = − 2 b, c 1 = a + 4 b, a 2 = 2 b, b 2 = 2 a

and

c 2 = b − 4 a

in above formula,

x − 2 b (b − 4 a) − (2 a) (a + 4 b) = y (a + 4 b) 2 b − (b − 4 a) 2 a = 1 2 a (2 a) − (− 2 b) (2 b)

x − 2 b 2 + 8 a b − 2 a 2 − 8 a b = y 2 a b + 8 b 2 − 2 a b + 8 a 2 = 1 4 a 2 + 4 b 2

x − 2 b 2 − 2 a 2 = y 8 a 2 + 8 b 2 = 1 4 a 2 + 4 b 2

Considering first and third term,

x − 2 b 2 + 2 a 2 = 1 4 a 2 + 4 b 2

x = − 2 b 2 + a 2 4 a 2 + b 2

x = − 12

Now, taking second and third term,

y 8 a 2 + 8 b 2 = 1 4 a 2 + 4 b 2

y = 8 a 2 + b 2 4 a 2 + b 2

y = 2

Hence, the solution of the given system of equation is

x = − 12

and

y = 2.

The correct option is (1).

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Solve the following systems of equations by the method of cross-multiplication: 2(ax−by)+a+4b=0 2(bx+ay)+b−4a=0

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Solve the following systems of equations by the method of cross-multiplication: $2 (a x − b y) + a + 4 b = 0 2 (b x + a y) + b − 4 a = 0$