What is 1+(0.732(z+0.1))/((z-1)(z-0.9512)) ?

The solution to 1+(0.732(z+0.1))/((z-1)(z-0.9512)) is (z^2-1.2192z+1.0244)/((z-1)(z-0.9512))

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simplify 1+(0.732(z+0.1))/((z-1)(z-0.9512))

simplify

1 + \frac{0.732 ( z + 0.1 )}{( z - 1 ) ( z - 0.9512 )}

\frac{z ^{2} - 1.2192 z + 1.0244}{( z - 1 ) ( z - 0.9512 )}

Solution steps

1 + \frac{0.732 ( z + 0.1 )}{( z - 1 ) ( z - 0.9512 )}

Apply the fraction rule:

1 + \frac{b}{c} = \frac{c + b}{c}

= \frac{( z - 1 ) ( z - 0.9512 ) + 0.732 ( z + 0.1 )}{( z - 1 ) ( z - 0.9512 )}

Simplify

(z - 1) (z - 0.9512) + 0.732 (z + 0.1) : z^{2} - 1.2192 z + 1.0244

= \frac{z ^{2} - 1.2192 z + 1.0244}{( z - 1 ) ( z - 0.9512 )}

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What is 1+(0.732(z+0.1))/((z-1)(z-0.9512)) ?
The solution to 1+(0.732(z+0.1))/((z-1)(z-0.9512)) is (z^2-1.2192z+1.0244)/((z-1)(z-0.9512))