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 · 2016-03-27T10:43:19

$x = 2$ $x=2$ and $y = - 1$ $y=-1$

The two linear equations are

$x + y = 1$ $x+y=1$ ..............(A)

$2 x + 3 y = 1$ $2x+3y=1$ ..............(B)

Multiplying (A) by $2$ $2$ and subtracting it from (B), we get

$(2 x + 3 y) - (2 x + 2 y) = 1 - 2$ $(2x+3y)-(2x+2y)=1-2$ or

$2 x + 3 y - 2 x - 2 y = - 1$ $2x+3y-2x-2y=-1$ or

$y = - 1$ $y=-1$

Now putting this in (A), we get

$x + (- 1) = 1$ $x+(-1)=1$ or $x = 1 + 1 = 2$ $x=1+1=2$

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