How do you solve #x+2y=8# and #x-2y=4#?

1 Answer
Jul 9, 2015

#x=6# and #y=1#

Explanation:

Hi there,
The given system of an equation can be solved by using various method such as elimination, substitution, by graphing, trial and error methods.

Here we will use elimination method to solve the given system of an equation.

We have,
#x+2y = 8#................equation 1
#x-2y=4#..................equation 2

Note: here we have equal coefficient with same sign for x and opposite sign for y. Hence it can be solve either by addition or subtraction.

Adding equation 1 and 2, we get

#x+2y = 8#
#x-2y=4#
#----#
#2x = 12#

Now divide both side by 2, we get
or, #x = 12/2 = 6#

Plug the value of x= 6 in equation 1, we get
#6+2y=8#
Subtract 6 on both side, we get
or, #2y=8-6#
or. #2y=2#
Divide each side by 2, we get
or, #y = 2/2=1#
Therefore, the final answer is #x=6# and #y=1#

To check the answer, plug #x=6# and #y=1# in above equation and check weather it satisfies or not.

Thanks