# How do you solve 3y+3= 6x and 2y-4x= 6?

Mar 19, 2018

One method is to add the equations so that one variable is eliminated and then solve for the second variable.

#### Explanation:

3 and 2 have a common multiple of 6 .
Multiply the first equation by 2

$2 \left\{3 y + 3 = 6 x\right\} = 6 y + 6 = 12 x$

Multiply the second equation by -3

$- 3 \left\{2 y - 4 x = 6\right\} = - 6 y + 12 x = - 18$

$6 y + 6 = 12 x$
$\pm 6 y + 12 x = - 18$
$= 0 y + 6 + 12 x = 12 x - 18$

Now solve for x

$+ 6 + 12 x = 12 x - 18$ subtract 12x from both sides.

$6 + 12 - 12 x = 12 x - 12 x - 18$

$6 = - 18$ which is not true.

There are no real solutions to this set of equations.

Mar 19, 2018

see a solution step process below;

#### Explanation:

We have this two system of equations;

$3 y + 3 = 6 x - - - e q n 1$

$2 y - 4 x = 6 - - - e q n 2$

We can either use subsitution or elimination method to solve..

Using Substitution Method..

From $e q n 1$

$3 y + 3 = 6 x$

Making $y$ to subject formula;

$3 y + 3 - 3 = 6 x - 3 \to \text{adding -3 to both sides}$

$3 y = 6 x - 3$

$y = \frac{6 x - 3}{3} \to \text{dividing both sides by 3}$

$y = \frac{6 x}{3} - \frac{3}{3}$

$y = 2 x - 1 - - - e q n 3$

Now substitute $e q n 3$ into $e q n 2$

$2 y - 4 x = 6$

$2 \left(2 x - 1\right) - 4 x = 6$

$4 x - 2 - 4 x = 6$

$- 2 = 6$

This shows that the equation is not valid, hence it is unsolvable cause there would be no accurate value for $x$ and $y$