How do you solve #1/3(9x-2) = 1/2(8x-6)#?

Question

1928 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-10T07:22:37

#x=7/3#

#color(blue)(1/3(9x-2)=1/2(8x-6)#

Use distributive property

#color(brown)(a(b+c)=ab+ac#

#rarr(1/3 *9x)-(1/3 *2)=(1/2*8x)-(1/2*6)#

Remove the brackets and solve

#rarr1/cancel3^1 *cancel9^3x-1/3 *2=1/cancel2^1*cancel8^4x-1/cancel2^1*cancel6^3#

#rarr3x-2/3=4x-3#

Subtract #3x# both sides

#rarrcancel(3x)-2/3-cancel(3x)=cancel(4x)-3-cancel(3x)#

#rarr-2/3=x-3#

Add #3# both sides

#rarr-2/3+3=xcancel(-3+3#

#color(green)(rArrx=7/3#

Answer 2 · 2016-04-10T07:31:47

#x=7/3#

#1#. Start by factoring out #2# from #(8x-6)#.

#1/3(9x-2)=1/2(8x-6)#

#1/3(9x-2)=1/2*2(4x-3)#

#2#. Simplify the right side of the equation.

#1/3(9x-2)=(4x-3)#

#3#. Multiply both sides of the equation by #3# to get rid of the denominator.

#color(red)3[1/3(9x-2)]=color(red)3(4x-3)#

#4#. Simplify the left side of the equation.

#9x-2=3(4x-3)#

#5#. Expand the bracket.

#9x-2=12x-9#

#6#. Solve for #x#.

#3x=7#

#color(green)(|bar(ul(color(white)(a/a)x=7/3color(white)(a/a)|)))#