How do you solve the system $2 y + 6 = x$ $2y+6=x$ and $y - 3 x = - 13$ $y-3x=-13$ using substitution?

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How do you solve the system $2 y + 6 = x$ $2y+6=x$ and $y - 3 x = - 13$ $y-3x=-13$ using substitution?

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Answer 1 · 2015-05-27T16:27:35

$2 y + 6 = x$ $2y+6=x$ ...............equation $(1)$ $(1)$

$y - 3 x = - 13$ $y-3x=-13$
$y = - 13 + 3 x$ $y = -13 +3x$ ........equation $(2)$ $(2)$

substituting equation $(2)$ $(2)$ in equation $(1)$ $(1)$
$2 y + 6 = x$ $2y+6=x$
$2 (- 13 + 3 x) + 6 = x$ $2(-13 +3x) +6 = x$
$- 26 + 6 x + 6 = x$ $- 26 +6x +6 = x$
$6 x - x = 20$ $6x -x = 20$
$x = \frac{20}{5} = 4$ $x = 20/5 = 4$

substituting this value of $x$ $x$ in equation 2
$y = - 13 + 3 x$ $y = -13 +3x$
$y = - 13 + 3 . (4)$ $y = -13 +3.(4)$
$y = - 13 + 12 = - 1$ $y = -13 +12 = -1$

the solutions for the system of equations are
$x = 4 and y = - 1$ $color(purple)(x=4 and y =-1$

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How do you solve the system $2 y + 6 = x$ $2y+6=x$ and $y - 3 x = - 13$ $y-3x=-13$ using substitution?

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How do you solve the system 2y+6=x2y+6=x and y−3x=−13y-3x=-13 using substitution?

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How do you solve the system $2 y + 6 = x$ $2y+6=x$ and $y - 3 x = - 13$ $y-3x=-13$ using substitution?