How do you solve the following linear system: $x + y = 1, 2 x - 3 y = 12$ $x+y=1, 2x-3y=12$ ?

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How do you solve the following linear system: $x + y = 1, 2 x - 3 y = 12$ $x+y=1, 2x-3y=12$ ?

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Answer 1 · 2018-03-20T03:12:04

$x = 3 and y = - 2$ $x= 3 and y = -2$

Explanation:

Let $x + y = 1$ $x+y=1$ ----------equation (1),

and $2 x - 3 y = 12$ $2x-3y=12$ -----------equation (2).

Multiply equation (1) by 2 and subtract the resultant equation from (2). This is eliminate one variable, i.e. $x$ $x$ and we will get value of $y$ $y$ . Then substitute the obtained value of $y$ $y$ in any one original given equation to get value of $x$ $x$ :

(1) $\times 2$ $times 2$ - $(2) \Rightarrow$ $(2) =>$

$\Rightarrow 2 x + 2 y - (2 x - 3 y) = 2 - 12$ $=> 2x +2y - (2x -3y) = 2- 12$

$\Rightarrow 2 x + 2 y - 2 x + 3 y = - 10$ $=> 2x +2y -2x +3y = -10$

$\Rightarrow 5 y = - 10$ $=> 5y = -10$

$\Rightarrow y = - 2$ $=> y = -2$

Substituting $y$ $y$ in (1)

(1) $\Rightarrow x + (- 2) = 1$ $=> x + (-2) = 1$

$\Rightarrow x = 1 + 2 = 3$ $=> x = 1+2 =3$

$therefore x= 3 and y = -2$

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How do you solve the following linear system: $x + y = 1, 2 x - 3 y = 12$ $x+y=1, 2x-3y=12$ ?

1 Answer

Explanation:

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How do you solve the following linear system: x+y=1, 2x-3y=12x+y=1,2x−3y=12x+y=1, 2x-3y=12?

1 Answer

Explanation:

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How do you solve the following linear system: $x + y = 1, 2 x - 3 y = 12$ $x+y=1, 2x-3y=12$ ?