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

2 Answers
Nov 22, 2015

Answer:

#x=-1/2#, #y=3 #

Explanation:

Substitute in the second equation the expression for #y# given by the first, i.e. #y=-2x+2#, obtaining

#6x+2(-2x+2)=3#

Expanding, you have

#6x-4x+4=3 \iff 2x=-1 \iff x=-1/2#.

Once #x# is known, you can obtain #y# from the foretold relation

#y=-2x+2 \iff y=-2(-1/2)+2 = 1+2=3#

Nov 22, 2015

Answer:

#color(blue)("Solution: "y=1 ,color(white)(xxx) x=-1/2)#

Explanation:

Given: #y=-2x+2 .............................(1)#
#color(white)(xxxx) 6x+2y=3..................................(2)#

I am going change equation (2) into the format of #y=#something

Then I can equate both equations to each other through #y#
removing that variable. We will then only have #x#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)("Consider equation (2) "color(green)("Method shown in detail")#

Subtract #color(blue)(6x)# from both sides isolating the y-term

#color(brown)((6x+2y)color(blue)(-6x) =(3)color(blue)(-6x))#

#color(green)("The brackets are there only to show what is being altered or")#
#color(green)("grouped to make things easier to see what is happening")#

#(6x-6x)+2y = 3 -6x#

But #6x-6x =0# giving

#0+2y=3-6x#

#2y=3-6x#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Divide both sides by 2 so that we have y on its own")#
Not that #divide 2 -> times 1/2#

#color(brown)((2y) color(blue)( times 1/2) = (3-6x)color(blue)(times 1/2)#

#2/2 times y = 3/2 -6/2x#

#y=-3x+3/2.........................................(2_a)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Combining equations the remove the variable "y)#

#Equation (1) = y = Equation(2_a)#

#-2x+2 = y= -3x+3/2#

#-2x+2 = -3x+3/2#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Collecting like terms")color(green)(" Less detail now")#

#3x-2x=3/2-2#

#x=-1/2.........................................(3)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Substitute (3) into (1) to find y")#

#y=-2x+2 -> y=(-2)(1/2)+2#

#y=1#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Solution: "y=1 , x=-1/2)#