Question

Simplify the algebraic fraction :$\frac{\mathrm{x}(\mathrm{x}-1)}{4}-\frac{2(\mathrm{x}+1)}{3}$

Answer 1

To simplify: $\frac{\mathrm{x}(\mathrm{x}-1)}{4}-\frac{2(\mathrm{x}+1)}{3}$

Convert the given algebraic fractions into like fractions by obtaining L.C.M of their denominators, we get

L.C.M of $4$and $3=12$

$\frac{\mathrm{x}(\mathrm{x}-1)}{4}=\frac{\mathrm{x}(\mathrm{x}-1)\times 3}{4\times 3}=\frac{3\mathrm{x}(\mathrm{x}-1)}{12}=\frac{3{\mathrm{x}}^2-3\mathrm{x}}{12}$

$\frac{2(\mathrm{x}+1)}{3}=\frac{2(\mathrm{x}+1)\times 4}{3\times 4}=\frac{8(\mathrm{x}+1)}{12}=\frac{8\mathrm{x}+8}{12}$

$\frac{\mathrm{x}(\mathrm{x}-1)}{4}-\frac{2(\mathrm{x}+1)}{3}$ $=\frac{3{\mathrm{x}}^2-3\mathrm{x}}{12}-\frac{8\mathrm{x}+8}{12}$

$=\frac{3{\mathrm{x}}^2-3\mathrm{x}-8\mathrm{x}-8}{12}$

$=\frac{3{\mathrm{x}}^2-11\mathrm{x}-8}{12}$

Thus, $\frac{\mathrm{x}(\mathrm{x}-1)}{4}-\frac{2(\mathrm{x}+1)}{3}$$=\frac{3{\mathrm{x}}^2-11\mathrm{x}-8}{12}$