Please i really need this !!! how do you solve the system of equations x-2y=18 , 3x-2y=-10?

Basically, you want to rearrange one equation to give you x= or y=, then you substitute one of those into the OTHER equation.

This will make more sense when I do it.

I will rearrange the 3x-2y to give me y=
#3x-2y=-10#
#-2y=-10-3x#
#y=5+3/2x#

Now you substitute this 'y' into the other equation so that

#x-2(5+3/2x)=18#

Expand and simplify

#x-10-3x=18#
#-2x-10=18#
#-2x=28#
#x=-14#

Use this x value and sub. it into an equation to solve for y

#3(-14)-2y=-10#
#-42-2y=-10#
#-2y=-32#
#y=-16#

the two simultaneous equations are to be inspected for a minute

and see whether one of the unknown can be removed by taking

difference or addition of the two equations.

see the coefficients of x and y in the two equations.

we find that y has identical coefficient of -2 in the equation

so if we take a difference of the two on the left as well as right

side one will be left with only y in the resulting difference

suppose i subtract the first from the 2nd equation .
then we get

#3x -2y - (x-2y) = -10 - (18) #

2x = -28

x= -14

If x is known using any one equation to get y

#-14 - 2y = 18 , -2y = 18 + 14 # = 32

then y = -16.