How do you solve the following system of equations?: #x+y= -2 , x-2y=13#?

2 Answers
Mar 23, 2018

Answer:

#x = 3#
#y = -5#

Explanation:

since we have two unknowns we need minimum of two equations to solve for them , and we have that too

lets see the first equation
#x+y =-2#
the first objective is to write either x in terms of y or y in terms of x
so i add -y to both sides , and we get
#x = -2-y#
take this value of x and plug it in equation 2
so
this #x-2y=13#
becomes
#(-2-y)-2y =13#
solve for y to get a value
#-2 -3y = 13 #
#-3y = 15#
#-y = 5#
or
# y = -5#
since we got y value , we can plug it back in the first equation to get x value
#x + (-5) = -2# or #x = 5-2 = 3#
So , #x=3 , y=-5#

Mar 23, 2018

Answer:

#x = 3#
#y = -5#

Explanation:

You add/subtract the two equations with each other, using the #=# sign as a point of reference:

#x+y = -2#

would be the first equation. We can label it as equation #(1)#

#x-2y = 13#

would be the second equation. We can label it as equation #(2)#

Now, we can subtract equation #(2)# from equation #(1)# [the idea here is to leave us with only one variable, that way we can solve the equation as usual]

#(1) - (2)# would be

#x-x + y-(-2y) = -2-13#

#0 + 3y = -15#

so

#y = -5#

Now we can take #y# and use it in equation #(1)#;

#x+(-5) = -2#

#x = 3#

So

#x = 3 and y = -5#