How do you solve the system of equations #x-6y=-3# and #2x+3y=9#?

1 Answer

Answer:

Add or subtract to find one variable then substitute to find the other.

Explanation:

Both equations are in the same form so you can use addition or subtraction to remove one of the variables.

In this case, the easiest thing would be to multiply the first equation by #2# (each term must be multiplied) then subtract the second equation. We do this to remove the #x# variable so we can solve for #y#.

Step 1 - multiply the first equation by #2#:

#(x - 6y = -3) xx 2 => 2x - 12y = -6#

Step 2 - subtract the second equation from the first

#" " 2x - 12y = -6#
#- (2x + 3y = 9)" "# (remember to subtract each term from the like term)

# 0x - 15y = -15#

Step 3 - solve for #y#

#-15y = -15 => y = 1#

Step 4 - substitute #y = 1# into one of the equations to solve for #x#

#x - 6(1) = -3#

#x - 6 = -3#

#x = -3 + 6#

#x = 3#

Step 4 - check your work by substituting values for #x# and #y# into the other equation

#2(3) +3(1) = 9#

#6 + 3 = 9 " "#(true)

The solution to the two equations is #x = 3, y = 1#.