How do you solve the system #y= 1/2x + 1# and #y= -2x + 6#?

2 Answers
Mar 3, 2017

Answer:

#x=3#,#y=0#
#Sol^n= (3,0)#

Explanation:

#y= 1/2x+1#

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

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

Substitute eq ii in eq i

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

#-4x+12=-2x+6#

#2x=6#

#ul(x=3)#

Put #x=3# in eq ii

#y=-6+6#

#ul(y=0)#

#Sol^n= (3,0)#

Mar 3, 2017

Answer:

#2,2)#

Explanation:

#color(red)(y)=1/2x+1to(1)#

#color(red)(y)=-2x+6to(2)#

Since both equations are expressed with #color(red)(y)# as the subject, we can equate the right sides.

#rArr1/2x+1=-2x+6#

To eliminate the fraction, multiply ALL terms on both sides by 2, the denominator of the fraction.

#(cancel(2)^1xx x/cancel(2)^1)+(2xx1)=(2xx-2x)+(2xx6)#

#rArrx+2=-4x+12#

add 4x to both sides.

#x+4x+2=cancel(-4x)cancel(+4x)+12#

#rArr5x+2=12#

subtract 2 from both sides.

#5xcancel(+2)cancel(-2)=12-2#

#rArr5x=10#

divide both sides by 5

#(cancel(5) x)/cancel(5)=10/5#

#rArrx=2#

To find the corresponding value of y, substitute x = 2 into either (1) or (2). I've chosen equation (2)

#x=2toy=(-2xx2)+6=-4+6=2#

#"solution is "(2,2)#