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

2 Answers
Mar 19, 2018

Answer:

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.

Mar 19, 2018

Answer:

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#