How do you solve #3a+b=4# and #a-2b=6# using substitution?

1 Answer
Mar 3, 2018

Answer:

#(a,b)to(2,-2)#

Explanation:

#3a+b=4to(1)#

#a-2b=6to(2)#

#"from equation "(2)" we obtain "a=6+2bto(3)#

#color(blue)"substitute "a=6+2b" into equation "(1)#

#rArr3(6+2b)+b=4#

#rArr18+6b+b=4#

#rArr18+7b=4#

#"subtract 18 from both sides"#

#cancel(18)cancel(-18)+7b=4-18#

#rArr7b=-14#

#"divide both sides by 7"#

#(cancel(7) b)/cancel(7)=(-14)/7#

#rArrb=-2#

#color(blue)"substitute "b=-2" in equation "(1)#

#3a-2=4#

#rArr3a=4+2=6#

#rArra=6/3=2#

#"the solution is "a=2,b=-2#