How do you solve the system of equations #x+ y = 3# and #x = 3y - 5#?

3 Answers
Nov 15, 2016

Answer:

#x=1#
#y=2#

Explanation:

You must make an equation that has only one variable in it so you can solve for that variable. By finding the variable you can use it to find the other one.

Let's solve for #y#

#x+y=3->y=3-x#

substitute #(3-x)# instead of #y# in #x=3y-5" "# we get:

#x=3(3-x)-5#

#x=9-3x-5#

#4x=4#

"#x=1#"

Now we need to find #y#, we know that #x+y=3#, and we know that

#x=1#, so we substitute #1# instead of #x#

#x+y=3->1+y=3#

"#y=2#"

Nov 15, 2016

Answer:

#x=1# and #y=2#

Explanation:

  1. #x+y=3#
  2. #x=3y-5#

From the first equation, we can determine a value for #x#.

#color(red)(x+y=3)#

#color(red)(x=3-y)#

Substituting #x# with #color(red)((3-y))# in the second equation, we get:

#x=3y-5#

#3-y=3y-5#

Add #5# to both sides.

#8-y=3y#

Add #y# to each side.

#8=4y#

Divide both sides by #4#.

#2=y# or #y=2#

Substituting #y# with #2# in the first equation, we get:

#x+y=3#

#x+2=3#

Subtract #2# from each side.

#x=1#

Nov 15, 2016

Answer:

#x =1 and y = 2#

Explanation:

Note the each equation has a single #x# term.

Therefore we can have:

#color(blue)(x = 3y-5)" and "color(red)(x = 3-y)#

However, the #x# is the same in each equation: #x = x#

#color(white)(xxxxxxxxxxxxxxxx)color(blue)(x)=color(red)(x)#

#color(white)(xxxxxxxxx):. color(blue)(3y-5)" = "color(red)(3-y)#

#color(white)(xxxxxxxxxxxxx)3y+y= 3+5#

#color(white)(xxxxxxxxxxxxxxxx)4y= 8#

#color(white)(xxxxxxxxxxxxxxxxx)y= 2#

There are now 2 ways to find a value for #x#, given that #y=2#

#color(blue)(x = 3(2)-5)" and "color(red)(x = 3-(2))#

#x = 6-5 = 1" and "x = 3-2=1#

#x=1#