How do you solve $[(2x), (2x+3y)]=[(y), (12)]$ ?

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Answer 1 · 2017-08-05T01:36:06

$x=3/2 " and " y=3$

Here we have the system

$[(2x),(2x+3y)]=[(y),(12)] ->_(R2-R1)$

We can subtract the first equation from the second equation to cancel out the $2x$ factor

$[(0),(0+3y)]=[(0),(12-y)]$

Which then leaves us with the equation

$3y=12-y$

$<=>$ Add $y$ to both sides

$4y=12$

$<=>$ Divide both sides by $4$

$y=3$

Then, since we know that $y=2x$ from the first equation, we can plug in

$2x=3$

$<=>$ Divide both sides by $2$

$x=3/2$

We can check out answer by plugging in $(3/2,3)$ into the second equation

$2(3/2)+3(3)=12$

$<=>$

$3+9=12$

$<=>$

$12=12$

How do you solve [(2x), (2x+3y)]=[(y), (12)][(2x), (2x+3y)]=[(y), (12)]?