How do you solve #2/3x+1=5/6x+3/4#?

2 Answers
May 16, 2017

#x=3/2#

Explanation:

When you have an equation with fractions, you can rid of the denominators straight away and then carry on to solve the equation.

Multiply each term by the LCM of the denominators so that you can cancel them. In this case it is #12#

#2/3x+1 = 5/6x+3/4" "xx LCM# ( in this case #LCM = color(blue)(12)#)

#(color(blue)(12xx)2x)/3+color(blue)(12xx)1 = (color(blue)(12xx)5x)/6+(color(blue)(12xx)3)/4#

Cancel the denominators, then simplify each term.

#(color(blue)(cancel12^4xx)2x)/cancel3+color(blue)(12xx)1 = (color(blue)(cancel12^2xx)5x)/cancel6+(color(blue)(cancel12^3xx)3)/cancel4#

#8x +12 = 10x +9" "# There are now no fractions!

#12-9 = 10x -8x" "larr# re-arrange the terms

#color(white)(........)3 = 2x#

#color(white)(........)x = 3/2 #

May 16, 2017

#"answer : "x=1.5#

Explanation:

#2/3x+1=5/6x+3/4#

#"First,let us rearrange the equation given."#

#1-3/4=5/6x-2/3x#

#"let us equalize the denominator of terms in the left side of equation."#

#color(red)(4/4)*1-3/4=5/6 x-2/3x#

#4/4-3/4=5/6x-2/3 x#

#(4-3)/4=5/6 x-2/3 x#

#1/4=5/6x-2/3 x#

#"let us equalize the denominator of terms in the right side of equation."#

#1/4=5/6 x-color(red)(2/2)*2/3x#

#1/4=5/6x-4/6x#

#1/4 =(5x-4x)/6#

#1/4=x/6#

#a/b=c/d" "rArr" "a*d=b*c#

#1*6=4*x#

#6=4x#

#"let us divide both sides by 4."#

#6/4=(cancel(4)x)/cancel(4)#

#x=6/4#

#x=3/2#

#x=1.5#