How do you solve the simultaneous equations #3x-4y=11# and #5x+6y=12#?

1 Answer
Aug 1, 2015

Answer:

#x=3#; #y=-1/2#

Explanation:

You can solve this system of equations by using the multiplication method.

The first thing that you need to do is pick a variable to eliminate first, then figure out the least common multiple (LCM) of its coefficients.

Let's say that you want to eliminate #x# and solve for #y# first. The two coefficients of #x# are #3# and #5#, which means that they're LCM will be equal to #15#.

So, multiply the first equation by #5# and the second equation by #-3# to get

#5 * (3x - 4y) = 5 * 11#

#15x - 20y = 55#

and

#(-3) * (5x + 6y)= -3 * 12#

#-15x - 18y = -36#

Add the left side and the right side of these two equations separately to get

#color(red)(cancel(color(black)(15x))) - 20y - color(red)(cancel(color(black)(15x))) - 18y = 55 - 36#

#-38y = 19 => y = 19/(-38) = color(green)(-1/2)#

Now use this value of #y# in one of the two equations to determine the value of #x#.

#3x - 4(-1/2) = 11#

#3x + 2 = 11 => x = (11-2)/3 = color(green)(3)#

The solutions to this system of equations are

#{(x=3), (y=-1/2) :}#