How do you solve the following system?: $x - 3y = 0 , 3x + y = 7$

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Answer 1 · 2018-05-03T21:32:57

$x=21/10$ and $y=7/10$

$x-3y=0$
$3x+y=7$

Let's solve for $x$ in the first equation so we can use substitution

$x=3y$

Now plug that into the second equation

$3(3y)+y=7$

$9y+y=7$

$10y=7$

$y=7/10$

Plug that back into the first equation:

$x=3(7/10)$

$x=21/10$

$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$

To check our work, let's substitute $21/10$ and $7/10$ for $x$ and $y$ and solve. If we get the same answer as the equation, we are correct

$3x+y=7$

$3(21/10)+(7/10)$ should equal $7$

$63/10+7/10$

$70/10$

$7$ equals $7$

We were right!

How do you solve the following system?: x - 3y = 0 , 3x + y = 7 x - 3y = 0 , 3x + y = 7