How do solve the following linear system?: $x + y = 12, 3 x + y = - 20$ $x + y= 12 , 3x + y = -20$ ?

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How do solve the following linear system?: $x + y = 12, 3 x + y = - 20$ $x + y= 12 , 3x + y = -20$ ?

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Answer 1 · 2016-03-23T18:47:14

line up the system of equations making sure that the variables are in the same relative positions.
$\frac{x + y = 12 . e q .1}{3 x + y = - 20 . e q .2}$ $(x+y=12. eq.1)/(3x+y=-20 . eq.2)$

subtract equation 2 from equation 1
this will eliminate the y variable since they both have the same coefficient and same sign.

$- 2 x = 32$ $-2x=32$

solve for x
x = -16

now substitute the value for x into either the first or second equation (doesn't matter). A good choice would be to pick the simplest equation of the two and this looks like equation 1.
By substituting x = -16 into eqn,1, you will now be able to solve for the y value.

Answer 2 · 2016-03-23T18:50:25

$x = - 16$ $x=-16$
$y = 28$ $y=28$

Explanation:

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$x = - 16$ $x=-16$
$y = 28$ $y=28$

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How do solve the following linear system?: $x + y = 12, 3 x + y = - 20$ $x + y= 12 , 3x + y = -20$ ?

2 Answers

Explanation:

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Science

Math

Humanities

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How do solve the following linear system?: x+y=12,3x+y=−20 x + y= 12 , 3x + y = -20 ?

2 Answers

Explanation:

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How do solve the following linear system?: $x + y = 12, 3 x + y = - 20$ $x + y= 12 , 3x + y = -20$ ?