Question

Solve the following pair of linear equation by using cross multiplication method:-
x+y=14
x-y=4
Guys... Please solve this!!! Please help...!!

Answer 1

Given : x + y = 14    

Here a1 = 1, b1 = 1, c1 = -14 ------ (1)

Given x - y = 4    

Here a2 = 1, b2 = -1, c2 = -4------ (2)

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Cross multiplication method:

$=> \frac{x}{b1c2 - b2c1} = \frac{y}{c1a2 - c2a1} = \frac{1}{a1b2 - a2b1}$

$=> \frac{x}{(1)(-4) - (-1)(-14)} = \frac{y}{(-14)(1) - (-4)(1)} = \frac{1}{(1)(-1) - (1)(1)}$

$=> \frac{x}{-4 - 14} = \frac{y}{-14 + 4} = \frac{1}{-1 - 1}$

Now,

(1)

$=> \frac{x}{-18} = \frac{1}{-2}$

$=> -2x = -18$

$=> x = 9$

(ii)

$=> \frac{y}{-10} = \frac{1}{-2}$

$=> -2y = -10$

$=> y = 5$

Therefore, the values are :

$=> \boxed{x = 9, y = 5}}$

Hope this helps!

Answer 2

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