How do you solve the following linear system: #3x + 5y = -1, 2x − 5y = 16#?

3 Answers
Apr 6, 2018

Answer:

#x = 3, y = -2#

Explanation:

Starting with the first equation, we need to make either x or y the subject (I chose x)

#3x + 5y = -1 #

#x = (-5y-1)/3#

#2x - 5y = 16#

#2*(-5y-1)/3 - 5y = 16#

#(-10y-2)/3 - 5y = 16#

#-10y-2 - 15y = 48#

#-25y = 50#

#y = -2#

Now we simply substitute #(y = -2)# into the second equation...

#2x-5y = 16#

#2x-5(-2)=16#

#2x + 10 = 16#

#2x = 6#

#x = 3#

Finally, we should also verify both answers by substituting both of them into the other equation again.

#3x + 5y = -1#

#3(3) + 5(-2) = -1

#9 - 10 = -1#

#-1=-1#, so therefore our answers are correct!

Apr 6, 2018

Answer:

See a solution process below:

Explanation:

Step 1) First, solve each equation for #5y#:

  • Equation 1:

#3x + 5y = -1#

#3x - color(red)(3x) + 5y = -1 - color(red)(3x)#

#0 + 5y = -1 - 3x#

#5y = -1 - 3x#

  • Equation 2:

#2x - 5y = 16#

#2x - 5y + color(red)(5y) - color(blue)(16) = 16 - color(blue)(16) + color(red)(5y)#

#2x - 0 - color(blue)(16) = 0 + 5y#

#2x - 16 = 5y#

#5y = 2x - 16#

Step 2) Because both equations are equal on the left side we can now equate the right sides of the equation and solve for #x#:

#-1 - 3x = 2x - 16#

#-1 + color(blue)(16) - 3x + color(red)(3x) = 2x + color(red)(3x) - 16 + color(blue)(16)#

#15 - 0 = (2 + color(red)(3))x - 0#

#15 = 5x#

#15/color(red)(5) = (5x)/color(red)(5)#

#3 = (color(red)(cancel(color(black)(5)))x)/cancel(color(red)(5))#

#3 = x#

#x = 3#

Step 3) Substitute #3# for #x# in either of the solved equations in Step 1 and solve for #y#:

#5y = 2x - 16# becomes:

#5y = (2 xx 3) - 16#

#5y = 6 - 16#

#5y = -10#

#(5y)/color(red)(5) = -10/color(red)(5)#

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

#y = -2#

The Solution Is:

#x = 3# and #y = -2#

Or

#(3, -2)#

Apr 6, 2018

Answer:

#x = 3 and y =-2#

Explanation:

Notice that the terms in #y# in the two equations are are additive inverses. The sum of additive inverses is #0#
Add the equations together to eliminate the #y# terms.

#color(white)(xxxxxxxx)3xcolor(blue)(+5y) = -1" " A#
#color(white)(xxxxxxxx)2xcolor(blue)(-5y) = 16" "B#

#A+B:color(white)(xxx)5xcolor(blue)(+0y) = 15#
#color(white)(xxxxxxxxx)xcolor(white)(xxx) = 3#

Substitute in A:
#color(white)(xxxxxxxx)3(3)+5y = -1" " A#
#color(white)(xxxxxxxxxx)9+5y = -1#
#color(white)(xxxxxxxxxxxxx)5y = -10#
#color(white)(xxxxxxxxxxxxxx)y = -2#

Check in B:
#color(white)(xxxxx)2(3)-5(-2) = 16" "B#
#color(white)(xxxxxxxxxx)6+10 = 16#
#color(white)(xxxxxxxxxxxxx)16 = 16#

The values are correct.