How do you solve the system of equations #2.5x+y=-2# and #3x+2y=0#?

2 Answers
Jul 5, 2018

Solution: #x= -2, y=3#

Explanation:

#2.5 x+y= -2 or 5 x + 2 y = -4 ;(1)#

# 3 x +2 y =0 ; (2)# Subtracting equation (2) from equation (1)

we get, #(5 x + 2 y) - (3 x +2 y) = -4-0# or

#5 x + cancel(2 y) - 3 x - cancel(2 y) = -4# or

#2 x= -4 or x= -2#. Putting #x=-2# in equation (2) we get,

# 3* (-2) +2 y =0 or 2 y=6 or y = 3 :. (x= -2, y=3)#

Solution: #x= -2, y=3# [Ans]

Jul 5, 2018

x = -2, y = 3

Explanation:

Rearrange one of the equations to give an expression for one of the variables in terms of the other i.e. #y=ax+c# or #x=by+d#
(#a, b, c, d# are constants)

It doesn't matter which eqn we start with, so rearranging eqn1

#2.5x+y=-2#

#y=(-2-2.5x#)

Put this value for #y# into the other eqn (in this case into eqn2)

#3x+2y=0#

#3x+2*(-2-2.5x)=0#

multiply out and simplify

#3x +(-4-5x)=0#

#3x-5x-4=0#

#-2x=4#

#x=-2#

Now put this value for #x# into eqn1 or eqn2

eqn1

#2.5*(-2)+y=-2#

#-5+y=-2#

#y=-2+5 =3#

or eqn2

#3*(-2) + 2y = 0#

#-6 + 2y = 0#

#2y = 6#

#y = 3# (note: same answer whichever we choose)

We can start with either eqn and find either #x# or #y# in terms of the other, as long as we then use the other eqn to substitute our expression into to solve for the variable.