How do you solve the following linear system: $3 x + 5 y = - 1, 3 x + y = - 2$ $3x + 5y = -1, 3x+y=-2$ ?

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How do you solve the following linear system: $3 x + 5 y = - 1, 3 x + y = - 2$ $3x + 5y = -1, 3x+y=-2$ ?

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Answer 1 · 2015-11-14T12:17:33

$(x, y) = (- \frac{3}{4}, \frac{1}{4})$ $(x,y)=(-3/4,1/4)$

Explanation:

Given:
[1] $XXX 3 x + 5 y = - 1$ $color(white)("XXX")3x+5y=-1$
[2] $XXX 3 x + y = - 2$ $color(white)("XXX")3x+y=-2$

Subtract [2] from [1]
[3] $XXX 4 y = 1$ $color(white)("XXX")4y = 1$
Divide [3] by 4
[4] $XXX y = \frac{1}{4}$ $color(white)("XXX")y = 1/4$

Substitute $\frac{1}{4}$ $1/4$ for $y$ $y$ in [2]
[5] $XXX 3 x + \frac{1}{4} = - 2 S u b t r a c t$ $color(white)("XXX")3x+1/4 =-2 Subtract$ 1/4 $from both sides of [5] [6]$ color(white)("XXX")3x = -9/4 $Divide [6] by$ 3 $[7]$ color(white)("XXX")x = -3/4#

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How do you solve the following linear system: $3 x + 5 y = - 1, 3 x + y = - 2$ $3x + 5y = -1, 3x+y=-2$ ?

1 Answer

Explanation:

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