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

May 15, 2015

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

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

$y = 6 - 4 x$

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

$2 x + y = 2$

Then subtract $2 x$ from both sides to get

$y = 2 - 2 x$

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

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

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

Add $4 x$ to both sides to get

$6 = 2 + 2 x$

Subtract 2 from both sides to get

$4 = 2 x$

Then divide both sides by 2 to get

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

Then looking back at one of our previous equations:

$y = 2 - 2 x = 2 - 2 \cdot 2 = 2 - 4 = - 2$

May 15, 2015

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

$4 x + y = 6 \to y = 6 - 4 x$

Substitute this in the other equation:
$6 x + 3 y = 6 \to 6 x + 3 \cdot \left(6 - 4 x\right) = 6 \to$
$6 x + 18 - 12 x = 6 \to - 6 x = - 12 \to x = 2$

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

$x = 2$
$y = - 2$