How do you solve the following system: #-5x + 3y= 6, 4x + 2y = 10 #?

Arrange the equations so that the factors align, with all of the unknowns on one side:
#−5x + 3y = 6#
#4x + 2y = 10#
Then algebraically combine them. You can use multiplicative factors to an entire equation if the coefficients are not already equal. Then we can simply subtract one equation from the other to get a single equation in only the 'x' variable.
#(−5x + 3y = 6)*2#
#( 4x + 2y = 10)*3#

(1) #−10x + 6y = 12#
(2) # 12x + 6y = 30# Subtract (1) from (2):
# 22x = 18 #
#x = (9/11)#

Substitute this value back into one equation to solve for 'y', then use the other equation to check the final values for correctness.
#−5(9/11) + 3y = 6# ; #-(45/11) + 3y = 6# ; #3y = (111/11)# ; # y = (37/11)#
CHECK:
#4x + 2y = 10# ; #4*(9/11) + 2*(37/11) = 10# #(36/11) + (74/11) = 10# ; #(110/11) = 10# ; #10 = 10# CORRECT!

Let's use elimination:

#color(white)(-) -5x+3y=6#
#color(white)(-)#
#color(white)(-)color(white)(-)4x+2y=10#

Our first step is to change one of our varables to match another. I'm going to multiply #4x+2y=10# by #1.5# to change it to #6x+3y=15#

#color(white)(-) -5x+3y=6#
#color(black)(-)#
#color(white)(-)color(white)(-)6x+3y=15#
.........................................................
#color(white)(-)-11x=-9#

#-11x=-9#

divide by #-11#

#x=9/11#

Now we can solve for #y#:
#4(9/11)+2y=10#

#36/11+2y=10#

subtract #36/11# on both sides

#2y=74/11#

divide by #2# on both sides

#y=37/11#

Just to double check our work, let's solve the first equation, replacing #x# and #y# with #9/11# and #37/11#:

#-5x+3y=6#

#-(9/11)+3(37/11)=6#

#-45/11+111/11#

#6=6#

We were right! #x=9/11#, #y=37/11#