How do solve the following linear system?: $- 4 x + 4 = - y, 3 x + 6 y + 13 = 0$ $-4x + 4 = -y, 3x + 6y + 13 = 0$ ?

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How do solve the following linear system?: $- 4 x + 4 = - y, 3 x + 6 y + 13 = 0$ $-4x + 4 = -y, 3x + 6y + 13 = 0$ ?

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Answer 1 · 2017-05-31T15:47:28

$x = \frac{11}{27} and y = - 2 \frac{17}{27}$ $x = 11/27 and y = -2 17/27$

Explanation:

It might be better to work on the first equation a little and multiply by $- 1$ $-1$ so the $y$ $y$ term is positive. Then we have an equation which is suitable for the substitution method:

$- 4 x + 4 = - y \to (\times - 1) \to 4 x - 4 = y$ $-4x +4 = -y" "rarr (xx-1)" "rarrcolor(blue)(4x-4 = y)$

Substitute $(4 x - 4)$ $color(blue)((4x-4))$ for $y$ $y$ in the other equation:

$3 x + 6 y + 13 = 0$ $3x+6color(blue)( y) +13=0$

$3 x + 6 (4 x - 4) + 13 = 0 \leftarrow$ $3x+6color(blue)((4x-4)) +13=0" "larr$ only $x$ $x$ terms

$3 x + 24 x = - 13 + 24$ $3x+24x = -13+24$

$27 x = 11$ $27x = 11$

$x = \frac{11}{27}$ $x = 11/27$

Now use $x = \frac{11}{27}$ $x=11/27$ to find a value for $y$ $y$

$y = 4 x - 4 = 4 (\frac{11}{27}) - 4$ $y =4x-4 = 4(11/27)-4$

$y = - 2 \frac{10}{27}$ $y= -2 10/27$

Checking into the second equation confirms these results:
$0 = 0$ $0=0$

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How do solve the following linear system?: $- 4 x + 4 = - y, 3 x + 6 y + 13 = 0$ $-4x + 4 = -y, 3x + 6y + 13 = 0$ ?

1 Answer

Explanation:

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Impact of this question

Science

Math

Humanities

... and beyond

How do solve the following linear system?: −4x+4=−y,3x+6y+13=0-4x + 4 = -y, 3x + 6y + 13 = 0 ?

1 Answer

Explanation:

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Impact of this question

How do solve the following linear system?: $- 4 x + 4 = - y, 3 x + 6 y + 13 = 0$ $-4x + 4 = -y, 3x + 6y + 13 = 0$ ?