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

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

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Answer 1 · 2018-03-19T13:12:38

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

$2 {3 y + 3 = 6 x} = 6 y + 6 = 12 x$ $2 { 3y + 3 = 6x} = 6y + 6 = 12x$

Multiply the second equation by -3

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

Now add the two equations

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

Now solve for x

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

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

$6 = - 18$ $6 = -18$ which is not true.

There are no real solutions to this set of equations.

Answer 2 · 2018-03-19T13:39:48

see a solution step process below;

Explanation:

We have this two system of equations;

$3 y + 3 = 6 x - - - e q n 1$ $3y + 3 = 6x - - - eqn1$

$2 y - 4 x = 6 - - - e q n 2$ $2y - 4x = 6 - - - eqn2$

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

Using Substitution Method..

From $e q n 1$ $eqn1$

$3 y + 3 = 6 x$ $3y + 3 = 6x$

Making $y$ $y$ to subject formula;

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

$3 y = 6 x - 3$ $3y = 6x - 3$

$y = \frac{6 x - 3}{3} \to dividing both sides by 3$ $y = (6x - 3)/3 ->"dividing both sides by 3"$

$y = \frac{6 x}{3} - \frac{3}{3}$ $y = (6x)/3 -3/3$

$y = 2 x - 1 - - - e q n 3$ $y = 2x - 1 - - - eqn3$

Now substitute $e q n 3$ $eqn3$ into $e q n 2$ $eqn2$

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

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

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

$- 2 = 6$ $-2 = 6$

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

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

2 Answers

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