How do you solve the following system: $4 x - 5 y = - 1, 4 x + 3 y = - 1$ $4x-5y= -1, 4x+3y= -1$ ?

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How do you solve the following system: $4 x - 5 y = - 1, 4 x + 3 y = - 1$ $4x-5y= -1, 4x+3y= -1$ ?

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Answer 1 · 2016-01-14T08:40:16

I found:
$x = - \frac{1}{4}$ $x=-1/4$
$y = 0$ $y=0$

Explanation:

I would try multiplying the first equation by $- 1$ $-1$ and then add (in columns) to the second to get:
${(color(red)(-4x-5y=1)),(4x+3x=-1):}$ add them
$0-2y=0$
So:
$y=0$
Substitute this back into the first equation:
$4x-0=-1$
$x=-1/4$

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How do you solve the following system: $4 x - 5 y = - 1, 4 x + 3 y = - 1$ $4x-5y= -1, 4x+3y= -1$ ?

1 Answer

Explanation:

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How do you solve the following system: $4 x - 5 y = - 1, 4 x + 3 y = - 1$ $4x-5y= -1, 4x+3y= -1$ ?