How do you solve #-\frac { 2} { 3} q + 2q = \frac { 6} { 5} q + \frac { 6} { 5}#?

3 Answers
Oct 4, 2017

#q=9#

Explanation:

#-2/3xxq+2q=6/5xxq+6/5#

#4/3xxq=6/5xxq+6/5#

#4/3xxq-6/5xxq=6/5#

#2/15xxq=6/5#

#q=6/5xx15/2#

#q=9#

Oct 4, 2017

#q=9#

Explanation:

If you make all the denominators the same you can then just look at the numerators to solve this.

#color(white)("ddd")-(2q)/3color(white)("dd")+color(white)("ddd")2qcolor(white)("dddd")=color(white)("dd")(6q)/5color(white)("ddd")color(white)("dd")+6/5#

#-[(2qxx5)/(3xx5)]+[(2qxx15)/(1xx15)]=[(6qxx3)/(5xx3)]+[(6xx3)/(5xx3)]#

#color(white)("dddd")-(10q)/15color(white)("dd")+color(white)("ddd")(30q)/15color(white)("dd")=color(white)("ddd")(18q)/15color(white)("d")+color(white)("d")18/15#

The purist would now say: "Multiply both sides by 15"

#-10q+30q=18q+18#

subtract #18q# from both sides

#-10q+30q-18a=18#

#2q=18#

Divide both sides by 2

#q=18/2#

#q=(18-:2)/(2-:2) = 9/1=9#

Oct 4, 2017

#color(magenta)(q=9#

Explanation:

#-2/3q+2q=6/5q+6/5#

multiply both sides by#3#

#:.-2q+6q=18/5q+18/5#

multiply both sides by#5#

#:.-10q+30q=18q+18#

#:.20q=18q+18#

#:.20q-18q=18#

#:.2q=18#

#:.color(magenta)(q=9#

~~~~~~~~~~~~~~~~~~

check:-

substitute #color(magenta)(q=9#

#-2/3(color(magenta)9)+2(color(magenta)9)=6/5(color(magenta)9)+6/5#

#:.-18/3+18=54/5+6/5#

multiply both sides by#3#

#:.-18+54=162/5+18/5#

multiply bothd sides by#5#

#:.-90+270=162+18#

#:.color(magenta)(180=180#