How do you solve the following linear system: # -x+2y=-6 , 5x-2y=-35 #?

1 Answer
Apr 26, 2018

Answer:

Isolate #x# , replace #x# with the expression equal to #x# , isolate #y#, and then use the value of #y# to solve for #x#.

Explanation:

Isolate #x# in the simplest manner possible:

Equation 1: #-x + 2y =-6#
Equation 2: #5x - 2y =-35#

  1. #-x + x + 2y = - 6 + x#
  2. # 2y + 6 = -6 + x +6 #
  3. #2y + 6 = x#

Substitute the expression in step 3 in for #x# in equation 2

#2y+6=x->5[2y+6]-2y=-35#

Distribute #5# onto #2y+6# to get the equation,

#10y+30-2y=-35#

Which becomes...

#8y = -65#

Now isolate #y# ...

#(8y)/8 = -65/8#

#y = -65/8#

... and use the value of #y# to find #x# !

#-x - 2((-65)/8) = -6#

#-2*-65/8 =130/8=16.25#
#-x=-6+130/8-> -x=82/8#

There aren't negative signs on both sides of the equation, so the value of #x# is in turn, positive, and the right side of the equation, becomes negative:

#x= -82/8, y=-65/8#

Since these are linear systems, these are the only solutions to the system. You can always rewrite both equations in terms of #y# and graph them. The x and y coordinates where the two lines intersect one another #(-82/8,-65/8)#, are the solutions to the system.

Graph these:
#y=5/2x+35/2#
#y = 1/2x-3#