How do you solve $x + 4 y = - 1, - 3 x - 14 = y$ $x+4y=-1, -3x-14=y$ by substitution?

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How do you solve $x + 4 y = - 1, - 3 x - 14 = y$ $x+4y=-1, -3x-14=y$ by substitution?

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Answer 1 · 2016-12-07T02:15:46

$x = - 5$ $x=-5$ and $y = 1$ $y=1$

Explanation:

$x + 4 y = - 1$ $x+4y=-1$
$- 3 x - 14 = y$ $-3x-14=y$

From the second equation, we know the value of $y$ $y$ as $(- 3 x - 14)$ $color(red)((-3x-14))$ . So in the first equation, substitute $y$ $y$ with $(- 3 x - 14)$ $color(red)((-3x-14))$ .

$x + 4 y = - 1$ $x+4y=-1$

$x + 4 (- 3 x - 14) = - 1$ $x+ 4color(red)((-3x-14))=-1$

Open the brackets and simplify. The product of positive and negative is a negative.

$x - 12 x - 56 = - 1$ $x-12x-56=-1$

$- 11 x - 56 = - 1$ $-11x-56=-1$

Add $56$ $56$ to both sides.

$- 11 x = 55$ $-11x=55$

Divide both sides by $11$ $11$ .

$- x = 5$ $-x=5$ or $x = - 5$ $x=-5$

In the first equation, substitute $x$ $x$ with $- 5$ $color(blue)(-5)$ .

$x + 4 y = - 1$ $x+4y=-1$

$- 5 + 4 y = - 1$ $color(blue)(-5)+4y=-1$

Add $5$ $5$ to both sides.

$4 y = 4$ $4y=4$

Divide both sides by $4$ $4$ .

$y = 1$ $y=1$

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How do you solve $x + 4 y = - 1, - 3 x - 14 = y$ $x+4y=-1, -3x-14=y$ by substitution?

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Explanation:

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How do you solve x+4y=-1, -3x-14=yx+4y=−1,−3x−14=yx+4y=-1, -3x-14=y by substitution?

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How do you solve $x + 4 y = - 1, - 3 x - 14 = y$ $x+4y=-1, -3x-14=y$ by substitution?