How do you solve the following linear system: $y = - 3 x + 8, 2 x + y = 15$ $y = - 3x + 8, 2x + y = 15$ ?

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Answer 1 · 2015-11-14T09:38:31

$x = - 7$ $x=-7$
$y = 29$ $y=29$

$y = - 3 x + 8$ $y=-3x+8$ --------------------(1)
$2 x + y = 15$ $2x+y=15$ ---------------------(2)

Substitute > $y = - 3 x + 8$ $y=-3x+8$ in equation (2)

$2 x + (- 3 x + 8) = 15$ $2x+(-3x+8)=15$
$2 x - 3 x + 8 = 15$ $2x-3x+8=15$
$- x = 15 - 8 = 7$ $-x=15-8=7$
$x = - 7$ $x=-7$

Put $x = - 7$ $x=-7$ in equation (1)

$y = - 3 (- 7) + 8$ $y = -3(-7)+8$
$y = 21 + 8 = 29$ $y =21+8=29$
$y = 29$ $y=29$

How do you solve the following linear system: y = - 3x + 8, 2x + y = 15 y=−3x+8,2x+y=15 y = - 3x + 8, 2x + y = 15 ?