How do you solve the system #x= 3y-1# and #x+2y=9#?

2 Answers
Apr 16, 2018

Answer:

Arrange the equations. You get #x=5# and #y=2#

Explanation:

#x-3y = -1#
#-x-2y = -9#

(after multiplying the 2nd equation by -1#

Now add these:
#-5y=-10#

#y=2#

Put this value in the 1st original equation:
#x-(3times2) = -1#
#x-6 = -1#
#x=-1+6#
#x=5#

Apr 16, 2018

Answer:

#(x,y)to(5,2)#

Explanation:

#x=3y-1to(1)#

#x+2y=9to(2)#

#"rearrange equation "(2)" to give x in terms of y"#

#rArrx=9-2yto(3)#

#"since "(1)" and "(3)" both give x in terms of y we"#
#"can equate the right sides"#

#rArr3y-1=9-2y#

#"add 2y to both sides"#

#3y+2y-1=9cancel(-2y)cancel(+2y)#

#rArr5y-1=9#

#"add 1 to both sides"#

#rArr5y=10rArry=2#

#"substitute "y=2" in either "(1)" or "(3)#

#(1)tox=6-1=5#

#"solution is "(x,y)to(5,2)#