How do solve the following linear system?: $4 x = - 11 + y, - x + 3 y = - 5$ $4x=-11+y , -x+3y=-5$ ?

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Answer 1 · 2017-09-07T17:37:39

From $- x + 3 y = - 5$ $-x+3y=-5$ , $x = 3 y + 5$ $x=3y+5$

Hence,

$4 \cdot (3 y + 5) = y - 11$ $4*(3y+5)=y-11$

$12 y + 20 = y - 11$ $12y+20=y-11$

$11 y = - 31$ $11y=-31$ or $y = - \frac{31}{11}$ $y=-31/11$

Thus, $x = 3 \cdot (- \frac{31}{11}) + 5 = - \frac{38}{11}$ $x=3*(-31/11)+5=-38/11$

1) I found $x$ $x$ in terms of $y$ $y$ from second equation.

2) I plugged $x$ $x$ into first one for findind $y$ $y$ .

3) I found $x$ $x$ .

How do solve the following linear system?: 4x=-11+y , -x+3y=-5 4x=−11+y,−x+3y=−5 4x=-11+y , -x+3y=-5 ?