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Solve the following pairs of equations.

$2 x + 3 y = 2 4 x − 9 y = − 1$

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The solution of the given system of equation is

x = 4

and

y = 9

The solution of the given system of equation is

x = − 4

and

y = 9

The solution of the given system of equation is

x = 4

and

y = − 9

The solution of the given system of equation is

x = − 4

and

y = − 9

answer is A.

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Detailed Solution

Given equations are,

2 x + 3 y = 2............. (1) 4 x − 9 y = − 1............. (2)

We know that if the two linear equations are

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

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

then the 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

.
Let us assume

1 x = m, 1 y = n

So, equation (1) and (2) becomes

2 m + 3 n = 2

......(3)

4 m − 9 n = − 1

......(4)
Convert equation (3) and (4) in linear standard form.

a m + b n + c = 0

2 m + 3 n = 2

2 m + 3 n − 2 = 0

.......(5)

4 m − 9 n = − 1

4 m − 9 n + 1 = 0

......(6)
Compare equation (5)

2 m + 3 n − 2 = 0

with standard linear equation. We get,

a 1 = 2, b 1 = 3 and c 1 = − 2.

Now, compare equation (6),

4 m − 9 n + 1 = 0

with standard linear equation. We get,

a 2 = 4, b 2 = − 9 and c 2 = 1.

Using the formula of cross multiplication.

m b 1 c 2 − b 2 c 1 = n c 1 a 2 − c 2 a 1 = 1 b 2 a 1 − b 1 a 2

Put the value of

a 1 = 2, b 1 = 3, c 1 = − 2, a 2 = 4, b 2 = − 9

and

c 2 = 1

in above formula.

m 3 (1) − (− 9) (− 2) = n − 2 (4) − 1 (2) = 1 (− 9) (2) − 3 (4)

m 3 − 18 = n − 8 − 2 = 1 − 18 − 12

m − 15 = n − 10 = 1 − 30

Taking first and third term

m − 15 = 1 − 30

m = − 15 − 30

m = 12

Now, taking second and third term.

n − 10 = 1 − 30

n = 13

since,

1 x = m

1 x = 12

x = 2

x = 22

x = 4
And

1 y = n

1 y = 13

y = 3

y = 32

y = 9
Therefore, the solution of the given equations is

x = 4

and

y = 9

.
The correct option is (1).

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Solve the following pairs of equations. 2 x + 3 y =2 4 x −9 y =−1

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Detailed Solution

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Solve the following pairs of equations.

$2 x + 3 y = 2 4 x − 9 y = − 1$