Question

Who can solve these equations?

Answer 1

let's hmmm hmmm multiply both sides by the LCD of all denominators, that way we do away with the denominators and see what "x" is

`\cfrac{x+1}{2}+\cfrac{x+2}{3}=3-\cfrac{x+3}{4}\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{12}}{12\left( \cfrac{x+1}{2}+\cfrac{x+2}{3} \right)=12\left( 3-\cfrac{x+3}{4} \right)} \\\\\\ (6x+6)+(4x+8)=36-(3x+9)\implies 10x+14=36-3x-9 \\\\\\ 10x+14=27-3x\implies 10x=13-3x\implies 13x=13 \\\\\\ x=\cfrac{13}{13}\implies x=1 \\\\[-0.35em] ~\dotfill`

`\cfrac{1-4x}{10}-\cfrac{2x+1}{2}=\cfrac{5x+1}{5}\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{10}}{10\left( \cfrac{1-4x}{10}-\cfrac{2x+1}{2}\right)=10\left( \cfrac{5x+1}{5} \right)} \\\\\\ (1-4x)-(10x+5)=10x+2\implies 1-4x-10x-5=10x+2 \\\\\\ -14x-4=10x+2\implies -4=24x+2\implies -6=24x \\\\\\ \cfrac{-6}{24}=x\implies -\cfrac{1}{4}=x`