Question

# (4/(k-1))-(1/(k+4))=0

Equations and inequalities
$$\displaystyle{\left(\frac{{4}}{{{k}-{1}}}\right)}-{\left(\frac{{1}}{{{k}+{4}}}\right)}={0}$$

2021-06-08
We are given: $$\displaystyle{\left(\frac{{4}}{{{k}-{1}}}\right)}-{\left(\frac{{1}}{{{k}+{4}}}\right)}={0}$$
Add $$\displaystyle\frac{{1}}{{{k}+{4}}}$$ to both sides: $$\displaystyle{\left(\frac{{4}}{{{k}-{1}}}\right)}={\left(\frac{{1}}{{{k}+{4}}}\right)}$$
Cross multiply: 4(k+4)=1(k-1)
4k+16=k-1
Subtract 16 from both sides: 4k=k-17
Subtract k from both sides: 3k=-17
Divide both sides by 3: $$\displaystyle{k}=-{\left(\frac{{17}}{{3}}\right)}$$