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

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Answer 1 · 2017-02-26T11:03:50

#x=-12#

#color(green)(x/4color(red)(xx1)" "=" "[2color(red)(xx1)]+[(x-3)/3color(red)(xx1)])#

I chose the common denominator (bottom number) to be 12

#color(green)(x/4color(red)(xx3/3)" "=" "[2color(red)(xx12/12)]+[(x-3)/3color(red)(xx4/4)])#

#color(green)(" "(3x)/12" "color(white)(.)=" "[24/12]color(white)(.)+" "[(4x-12)/12])#

There are now two ways to think about this.

I love the one that goes: now the denominators are all the same you can forget about them.

The other one is if you are a stickler for mathematical correctness:
Multiply both sides by 12

#color(green)(3x=24+4x-12" "->" "3x=4x+12)#

Subtract #3x# from both sides

#color(green)(0=x+12#

Subtract 12 from both sides

#color(green)(x=-12)#

Answer 2 · 2017-02-26T11:13:35

#"the answer is x=-12"#

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

#"let's make equal the denominators at the right side of equation."#

#x/4=color(red)(3/3)*2+(x-3)/3#

#x/4=6/3+(x-3)/3#

#x/4=(6+x-3)/3#

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

#"if "a/b=c/d" then " a*d=b*c#

#3*x=4*(3+x)#

#3x=4*3+4*x#

#3x=12+4x#

#4x-3x=-12#

#x=-12#