$7 x-23$

$\=a_1\left(x-4\right)\+a_2\left(x-3\right)$

$\=\left(a_1\+a_2\right) x\+\left(-4 a_1-3 a_2\right)$

$a_1\+a_2$

$\=7$

$-4 a_1-3 a_2$

$\=-23$

Solution by Context Panel

$\frac{7x-23}{x^2-7 x\+12}$$\stackrel{}{ \to }$$\frac{2}{x-3}\+\frac{5}{x-4}$

Make $\frac{7 x-23}{\left(x-3\right) \left(x-4\right)}\=\frac{a_1}{x-3}\+\frac{a_2}{x-4}$ an identity

$\frac{7 x-23}{\left(x-3\right) \left(x-4\right)}\=\frac{a_1}{x-3}\+\frac{a_2}{x-4}$$\stackrel{\text{solve (identity)}}{\to }$$\left\{a_1\=2\,a_2\=5\right\}$

Solution by Task Template

Tools≻Tasks≻Browse: Algebra≻Partial Fractions≻Stepwise

Table 6.4.2(a)   Task template for an interactive stepwise partial-fraction decomposition

Interactive solution from first principles

$\frac{7 x-23}{\left(x-3\right) \left(x-4\right)}$$\stackrel{\text{assign to a name}}{\to }$$f$

$\frac{a_1}{x-3}\+\frac{a_2}{x-4}$$\stackrel{\text{assign to a name}}{\to }$$g$

$f-g$

$\frac{7x-23}{\left(x-3\right)\left(x-4\right)}-\frac{a_1}{x-3}-\frac{a_2}{x-4}$

$\stackrel{\text{simplify}}{\=}$

$\frac{\left(-a_1-a_2+7\right)x+4a_1+3a_2-23}{\left(x-3\right)\left(x-4\right)}$

$\stackrel{\text{numerator}}{\to }$

$-a_{1}x-a_{2}x+7x+4a_1+3a_2-23$

$\stackrel{\text{collect w.r.t. x}}{\=}$

$\left(-a_1-a_2+7\right)x+4a_1+3a_2-23$

$\stackrel{\text{coefficients in x}}{\to }$

$4a_1+3a_2-23,-a_1-a_2+7$

$\stackrel{\text{solve}}{\to }$

$\left\{a_1=2\,a_2=5\right\}$

$\stackrel{\text{substitute into}}{\to }$

$\frac{2}{x-3}+\frac{5}{x-4}$

Table 6.4.2(b)   Interactive solution from first principles

$f≔\frac{7x-23}{x^2-7 x\+12}\:$

$\mathrm{convert}\left(f\,\mathrm{parfrac}\,x\right)$

$\frac{2}{x-3}\+\frac{5}{x-4}$

Table 6.4.2(c)   Use of the convert command to obtain a partial-fraction decomposition

Coded solution from first principles

$q_1≔f\=g$

$\frac{7x-23}{\left(x-3\right)\left(x-4\right)}\=\frac{a_1}{x-3}\+\frac{a_2}{x-4}$

$q_2≔\mathrm{simplify}\left(q_1\cdot \mathrm{denom}\left(f\right)\right)$

$7x-23\=x{a}_1\+x{a}_2-4a_1-3a_2$

$q_3≔\mathrm{map}\left(\mathrm{coeff}\,q_2\,x\,1\right)$

$7\=a_1\+a_2$

$q_4≔\mathrm{map}\left(\mathrm{coeff}\,q_2\,x\,0\right)$

$-23\=-4a_1-3a_2$

$q_5≔\mathrm{solve}\left(\left\{q_3\,q_4\right\}\right)$

$\left\{a_1\=2\,a_2\=5\right\}$

$\mathrm{eval}\left(q_1\,q_5\right)$

$\frac{7x-23}{\left(x-3\right)\left(x-4\right)}\=\frac{2}{x-3}\+\frac{5}{x-4}$

Table 6.4.2(d)   Coded solution from first principles

$\mathrm{solve}\left(\mathrm{identity}\left(q_1\,x\right)\,\left\{a_1\,a_2\right\}\right)$

$\left\{a_1\=2\,a_2\=5\right\}$