1/2(x+1)+1/3(x+4)=1/6(x-1)

1/2(x+1)+1/3(x+4)=1/6(x-1)

 

This is the question you have asked

 

\(\frac{(x+1)}{2}+\frac{(x+4)}{3}=\frac{(x-1)}{6}\)

 

 

The easiest way to solve it is to multiply both sides by the smallest common denominator, which is 6.

That will get rid of all the fractions for you.

Give it a go and if you want more help post what you have done.

\(\frac{x+1}{2}+\frac{x+4}{3}=\frac{x-1}{6}\\ \text{That's what Melody have answered.}\\ \frac{1}{2(x+1)}+\frac{1}{3(x+4)}=\frac{1}{6(x-1)}\\ \text{And I am going to answer this.}\)

\(\frac{1}{2(x+1)}+\frac{1}{3(x+4)}=\frac{1}{6(x-1)}\\ \frac{3x+12+2x+2}{2(3)(x+1)(x+4)}=\frac{1}{6}\times \frac{1}{x-1}\\ \frac{5x+14}{(x+1)(x+4)}=\frac{1}{x-1}\\ (5x+14)(x-1)=(x+1)(x+4) \\5x^2+9x-14=x^2+5x+4\\ 4x^2+4x-16=0\\ x^2+x-4=0\\ x=\frac{-1\pm\sqrt{1^2-4(1)(-4)}}{2(1)}\\ \;\;=\frac{-1\pm\sqrt{17}}{2}\)

\(\text{The answer is}\mathbf{\frac{-1+\sqrt{17}}{2}}\text{and}\mathbf{\frac{-1-\sqrt{17}}{2}}\)

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