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

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Replaced a $+$ $+$ with $=$ $=$ for the first equation; hopefully, this is what was intended.

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Answer 1 · 2015-11-13T13:19:09

$x = - 2$ $x=-2$ , $y = - 1$ $y=-1$ .

From the second equation, we know that $y = 3 + x$ $y=3+x$ .

So, the first equation becomes

$4 x + 5 y = - 3 \to 4 x + 5 (3 + x) = - 3$ $4x+5color(green)(y)=-3 -> 4x+5color(green)((3+x))=-3$

Expand the left memeber: $4 x + 15 + 5 x = 9 x + 15$ $4x+15+5x = 9x+15$

So, $9 x + 15 = - 3 \Leftrightarrow 9 x = - 18 \Leftrightarrow x = - 2$ $9x+15=-3 \iff 9x = -18 iff x=-2$

Knowing this, we obtain $y$ $y$ , since $y = 3 + x = 3 - 2 = 1$ $y=3+x=3-2=1$

Answer 2 · 2015-11-13T13:20:23

$(x, y) = (- 2, 1)$ $(x,y)=(-2,1)$

Given
[1] $XXX 4 x + 5 y = - 3$ $color(white)("XXX")4x+5y=-3$
[2] $XXX - x + y = 3$ $color(white)("XXX")-x+y=3$

Multiply [2] by $4$ $4$
[3] $XXX - 4 x + 4 y = 12$ $color(white)("XXX")-4x+4y=12$

Add [1] and [3]
[4] $XXX 9 y = 9$ $color(white)("XXX")9y=9$

Divide [4] by $9$ $9$
[5] $XXX y = 1$ $color(white)("XXX")y=1$

Substitute $1$ $1$ for $y$ $y$ in [2]
[6] $XXX - x + (1) = 3$ $color(white)("XXX")-x+(1)=3$

Subtract $1$ $1$ from both sides of [6]
[7] $XXX - x = 2$ $color(white)("XXX")-x=2$

Multiply both sides of [7] by $(- 1)$ $(-1)$
[8] $XXX x = - 2$ $color(white)("XXX")x=-2$