How do you solve #(x-3) /4 + x/2 =3#?

2 Answers
Jun 3, 2016

Answer:

#x=5#

Explanation:

#(x-3)/4+x/2=3#

First, we remove the fractions by multiplying the equation by the LCM of #4# and #2#, our two denominators:

LCM of #4# and #2# = #4#

Multiply the equation by #4#.

This means, the expression on the left side of the = sign will be multiplied by #4# and the expression on the right side of the = sign will be multiplied by #4#:

#[(x-3)/4+x/2]color(red)(xx4)=3color(red)(xx4)#

#[(x-3)/4color(red)(xx4)]+[x/2color(red)(xx4)]=12#

#[(x-3)/cancel4xxcancel 4]+[x/cancel2xxcancel4^2]=12#

#x-3+(x xx2)=12#

#x-3+2x=12#

#3x-3=12#

Next, add #3# to both sides of the equation:

#3x-3color(red)(+3)=12color(red)(+3)#

#3x=15#

Finally, divide both sides by #3#:

#(3x)/color(red)(3)=15/color(red)(3)#

#x=5#

You can check your answer by putting back the value #x=5# in the question:

#(x-3)/4+x/2=3#

Solving the left side:

#= (5-3)/4+5/2#

#= 2/4+5/2#

#=1/2+5/2#

#=(1+5)/2#

#=6/2=color(red)(3)#

Jun 3, 2016

Answer:

x = 5

Explanation:

To eliminate the fractions in this equation #color(blue)"multiply all terms on both sides"# by the L.C.M. (lowest common multiple) of 2 and 4 which is 4.

#rArr[cancel(4)^1 xx(x-3)/cancel(4)^1]+[cancel(4)^2xxx/cancel(2)^1]=4xx3#

The equation now simplifies to

x - 3 + 2x = 12

hence : 3x - 3 =12

and 3x = 15

divide both sides by 3

#(cancel(3)^1 x)/cancel(3)^1 =15/3#

#rArrx=5" is the solution"#