How do you solve the system of equations $x + 3 y = - 7$ $x+3y=-7$ and $- x - 3 y = 1$ $-x-3y=1$ ?

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How do you solve the system of equations $x + 3 y = - 7$ $x+3y=-7$ and $- x - 3 y = 1$ $-x-3y=1$ ?

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Answer 1 · 2017-04-25T00:46:02

$N o$ $No$ $s o l u t i o n$ $solution$

Explanation:

$1 . x + 3 y = - 7$ $1. x + 3y = -7$
$2 . - x - 3 y = 1$ $2.-x-3y = 1$

Consider $1 .$ $1.$
$x = - 7 - 3 y$ $x = -7 - 3y$

Substitute the value of x from $1 .$ $1.$ to $2 .$ $2.$ :
$- (- 7 - 3 y) - 3 y = 1$ $-(-7-3y) -3y = 1$
$7 + 3 y - 3 y = 1$ $7 + 3y - 3y = 1$
$7 = 1$ $7 = 1$

This means that there is no solution to the system, thus these lines must be parallel. If you convert both $1 .$ $1.$ and $2 .$ $2.$ to the form $y = m x + b$ $y = mx + b$ , you will see the $m$ $m$ values, or the slope of both lines is the same, thus they will never intersect, hence no solution.

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How do you solve the system of equations $x + 3 y = - 7$ $x+3y=-7$ and $- x - 3 y = 1$ $-x-3y=1$ ?

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Explanation:

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How do you solve the system of equations x+3y=−7x+3y=-7 and −x−3y=1-x-3y=1?

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How do you solve the system of equations $x + 3 y = - 7$ $x+3y=-7$ and $- x - 3 y = 1$ $-x-3y=1$ ?