How do you solve the following system: $- 6 x + y = - 8, 6 y + 4 x = 12$ $-6x + y = -8, 6y + 4x = 12$ ?

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How do you solve the following system: $- 6 x + y = - 8, 6 y + 4 x = 12$ $-6x + y = -8, 6y + 4x = 12$ ?

1 Answer

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Answer 1 · 2018-05-05T07:53:39

$y = 1$ $y=1$
$x = \frac{3}{2}$ $x=3/2$

Explanation:

$- 6 x + y = - 8$ $-6x+y=-8$ --- (1)
$6 y + 4 x = 12$ $6y+4x=12$ --- (2)

From (1),
$- 6 x + y = - 8$ $-6x+y=-8$
$y = 6 x - 8$ $y=6x-8$ --- (3)

Sub (3) into (2)
$6 (6 x - 8) + 4 x = 12$ $6(6x-8)+4x=12$
$36 x - 48 + 4 x = 12$ $36x-48+4x=12$
$40 x = 60$ $40x=60$
$x = \frac{3}{2}$ $x=3/2$ --- (4)

Sub (4) into (3)
$y = 6 \cdot \frac{3}{2} - 8$ $y=6*3/2-8$
$y = 1$ $y=1$

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How do you solve the following system: $- 6 x + y = - 8, 6 y + 4 x = 12$ $-6x + y = -8, 6y + 4x = 12$ ?

1 Answer

Explanation:

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Impact of this question

Science

Math

Humanities

... and beyond

How do you solve the following system: −6x+y=−8,6y+4x=12-6x + y = -8, 6y + 4x = 12 ?

1 Answer

Explanation:

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Impact of this question

How do you solve the following system: $- 6 x + y = - 8, 6 y + 4 x = 12$ $-6x + y = -8, 6y + 4x = 12$ ?