Question

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

Answer 1

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 { 3y + 3 = 6x} = 6y + 6 = 12x`

Multiply the second equation by -3

`-3{ 2y -4x = 6} = -6y + 12x = -18`

Now add the two equations

`6y + 6 = 12 x`
`+-6y + 12x = -18`
`= 0y +6 + 12 x = 12x - 18`

Now solve for x

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

`6 + 12 -12x = 12x -12x - 18`

`6 = -18` which is not true.

There are no real solutions to this set of equations.

Answer 2

see a solution step process below;

Explanation:

We have this two system of equations;

`3y + 3 = 6x - - - eqn1`

`2y - 4x = 6 - - - eqn2`

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

Using Substitution Method..

From `eqn1`

`3y + 3 = 6x`

Making `y` to subject formula;

`3y + 3 - 3 = 6x - 3 ->"adding -3 to both sides"`

`3y = 6x - 3`

`y = (6x - 3)/3 ->"dividing both sides by 3"`

`y = (6x)/3 -3/3`

`y = 2x - 1 - - - eqn3`

Now substitute `eqn3` into `eqn2`

`2y - 4x = 6`

`2(2x - 1) - 4x = 6`

`4x - 2 - 4x = 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`