How do you solve the following system: $-x+6y=12, 3x+4y=-10$ ?

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How do you solve the following system: $-x+6y=12, 3x+4y=-10$ ?

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Answer 1 · 2016-09-08T13:04:04

$x=-54/11$

$y=13/11$

Explanation:

$-x+6y=12$ and $3x+4y=-10$

or

$x=6y-12$

or

$x=6(y-2)$

Multiplying $-x+6y=12$ by $3$

We get

$-3x+18y=36$

By adding $-3x+18y=36$ with $3x+4y=-10$

We get

$-3x+18y+3x+4y=36-10$

or

$22y=26$

or

$y=26/22$

or

$y=13/11$ ----------------------Ans $1$

By plugging the value $y=13/11$ in the equation $x=6(y-2)$

We get

$x=6(13/11-2)$

or

$x=6 times (13-22)/11$

or

$x=6times (-9/11)$

or

$x=-54/11$ ---------------------Ans $2$

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How do you solve the following system: $-x+6y=12, 3x+4y=-10$ ?

1 Answer

Explanation:

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How do you solve the following system: -x+6y=12, 3x+4y=-10 -x+6y=12, 3x+4y=-10 ?

1 Answer

Explanation:

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How do you solve the following system: $-x+6y=12, 3x+4y=-10$ ?