How do you solve #3x-y=1# and #x+y=3#?

2 Answers
Aug 3, 2015

Answer:

I found:
#x=1#
#y=2#

Explanation:

I would isolate #x# from the second equation and substitute it into the first:
#x=3-y#
into the first:
#3(color(red)(3-y))-y=1#
#9-3y-y=1#
#-4y=-8#
#y=2#
substitute this back into the second equation:
#x=3-2=1#

Aug 3, 2015

Answer:

#color(red)( x=1,y=2)#

Explanation:

Gió's method uses the method of substitution.

Here's how to do it by the method of elimination.

Step 1. Enter the equations.

[1] #3x-y=1#
[2] #x+y=3#

Step 2. Add Equations 1 and 2.**

#4x=4#

[3] #x=1#

Step 3. Substitute Equation 3 in Equation 2.

#x+y=3#
#1+y=3#

#y=2#

Solution: #x=1,y=2#

Check: Substitute the values of #x# and #y# in Equation 1.

#3x-y=1#
#3(1)-2=1#
#3-2=1#
#1=1#

It checks!

The solution is correct.