How do you solve the following system using substitution?: #3x=2y-5, y-5x=-1#

1 Answer
Oct 21, 2015

Answer:

#{(x=1), (y = 4) :}#

Explanation:

Your system of equations looks like this

#{(3x = 2y - 5), (y - 5x = -1) :}#

The first thing to do here is rewrite the first equation to match the form of the second equation, i.e. get the #x# and #y# terms together on one side of the equation.

You can do that by adding #-2y# to both sides of the equation

#3x - 2y = color(red)(cancel(color(black)(2y))) - color(red)(cancel(color(black)(2y))) = -5#

#3x - 2y = -5#

The system of equations now looks like this

#{(3x - 2y = -5), (-5x + y = -1) :}#

To solve this by using the substitution method, pick one of the two variables, solve one of the two equations for that variable, and use the resulting expression to find the other variable.

Simply put, let's say that you pick #y# and solve the second equation for #y#

#-5x + y = -1#

#color(red)(cancel(color(black)(5x))) - color(red)(cancel(color(black)(5x))) + y = -1 + 5x#

#y = 5x - 1" " " " color(red)("*")#

Now use this value of #y# in the first equation to get the value of #x#

#3x - 2 * overbrace((5x-1))^(color(blue)(=y)) = -5#

#3x - 10x + 2 = -5#

#-7x = -7 implies x = ((-7))/((-7)) = color(green)(1)#

Now that you know what the value of #x# is, use equation #color(red)("*")# to find the value of #y#

#y = 5 * (1) - 1 = color(green)(4)#

The solution set for this system of equations is

#{(x=1), (y = 4) :}#