How do you solve the following linear system $- 4 x + 9 y = 5, x - 4 y = - 3$ $-4x +9y = 5 , x - 4y = -3$ ?

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Answer 1 · 2016-06-15T11:50:38

$x = 1; y = 1$ $x=1; y=1$

First, let's obtain x in the second equation:
$x = 4 y - 3$ $x=4y-3$

then let's substitute its value in the first one:
$- 4 (4 y - 3) + 9 y = 5$ $-4(4y-3)+9y=5$

then, by simplifying:
$- 16 y + 12 + 9 y = 5$ $-16y+12+9y=5$
$- 16 y + 9 y = - 12 + 5$ $-16y+9y=-12+5$
$- 7 y = - 7$ $-7y=-7$
$y = 1$ $y=1$

and now let's substitute the value of y in
$x = 4 y - 3$ $x=4y-3$

so that
$x = 4 \cdot 1 - 3$ $x=4*1-3$
$x = 1$ $x=1$

How do you solve the following linear system -4x +9y = 5 , x - 4y = -3 −4x+9y=5,x−4y=−3-4x +9y = 5 , x - 4y = -3 ?