How do you solve the system $4x + y = 6$ and $6x + 3y = 6$ ?

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Answer 1 · 2015-05-15T22:44:32

Try to eliminate $y$ by expressing it in terms of $x$ in two ways from the two original equations.

With the first equation, subtract $4x$ from both sides to get:

$y = 6 - 4x$

With the second equation, first divide through by 3 to get

$2x+y = 2$

Then subtract $2x$ from both sides to get

$y = 2 - 2x$

Now we have two expressions both equal to $y$ , so we can put them together:

$6 - 4x = y = 2 - 2x$

So $6 - 4x = 2 - 2x$

Add $4x$ to both sides to get

$6 = 2 + 2x$

Subtract 2 from both sides to get

$4 = 2x$

Then divide both sides by 2 to get

$2 = x$ , that is $x = 2$ .

Then looking back at one of our previous equations:

$y = 2 - 2x = 2 - 2*2 = 2-4 = -2$

Answer 2 · 2015-05-15T22:44:48

You first express $y$ as function of $x$ , then solve for $x$

$4x+y=6->y=6-4x$

Substitute this in the other equation:
$6x+3y=6->6x+3*(6-4x)=6->$
$6x+18-12x=6->-6x=-12->x=2$

Now put this $x$ into the first equation:
$y=6-4x=6-4*2=-2$

Answer:
$x=2$
$y=-2$

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