Question

Solve by factorization a/(ax-1)+b/(bx-1)=a+b

Polynomial factorization
ANSWERED
asked 2021-02-09
Solve by factorization
\(\displaystyle\frac{{a}}{{{a}{x}-{1}}}+\frac{{b}}{{{b}{x}-{1}}}={a}+{b}\)

Answers (1)

2021-02-10
Step 1: Given,
The expression \(\displaystyle\frac{{a}}{{{a}{x}-{1}}}+\frac{{b}}{{{b}{x}-{1}}}={a}+{b}\). We have to solve this by factorization.
Step 2: Calculation
\(\displaystyle\frac{{a}}{{{a}{x}-{1}}}+\frac{{b}}{{{b}{x}-{1}}}={a}+{b}\)
\(\displaystyle\frac{{{a}{\left({b}{x}-{1}\right)}+{b}{\left({a}{x}-{1}\right)}}}{{{\left({a}{x}-{1}\right)}{\left({b}{x}-{1}\right)}}}={a}+{b}\)
\(\displaystyle\frac{{{a}{b}{x}-{a}+{a}{b}{x}-{b}}}{{{\left({a}{x}-{1}\right)}{\left({b}{x}-{1}\right)}}}={a}+{b}\)
\(\displaystyle\frac{{{2}{a}{b}{x}-{\left({a}+{b}\right)}}}{{{\left({a}{x}-{1}\right)}{\left({b}{x}-{1}\right)}}}={a}+{b}\)
\(\displaystyle{2}{a}{b}{x}-{\left({a}+{b}\right)}={\left({a}+{b}\right)}{\left({a}{x}-{1}\right)}{\left({b}{x}-{1}\right)}\)
\(\displaystyle{2}{a}{b}{x}-{\left({a}+{b}\right)}={\left({a}+{b}\right)}{\left({a}{b}{x}^{{2}}-{a}{x}-{b}{x}+{1}\right)}\)
\(\displaystyle{\left({a}+{b}\right)}{a}{b}{x}^{{2}}-{\left({a}+{b}\right)}^{{2}}{x}+{\left({a}+{b}\right)}-{2}{a}{b}{x}+{\left({a}+{b}\right)}={0}\)
\(\displaystyle{\left({a}+{b}\right)}{a}{b}{x}^{{2}}-{\left[{\left({a}+{b}\right)}^{{2}}-{2}{a}{b}\right]}{x}+{2}{\left({a}+{b}\right)}={0}\)
If we have a quadratic equation \(\displaystyle{a}{x}^{{2}}+{b}{x}+{c}={0}{t}{h}{e}{n}{x}=\frac{{-{b}\pm\sqrt{{{b}^{{2}}-{4}{a}{c}}}}}{{{2}{a}}}\)
From the equation \(\displaystyle{\left({a}+{b}\right)}{a}{b}{x}^{{2}}-{a}+{b}{2}-{2}{a}{b}{x}+{2}{a}+{b}={0}\) we have
\(\displaystyle{x}=\frac{{{\left[{\left({a}+{b}\right)}^{{2}}-{2}{a}{b}\right]}\pm\sqrt{{{\left[{\left({a}+{b}\right)}^{{2}}-{2}{a}{b}\right]}^{{2}}-{8}{a}{b}{\left({a}+{b}\right)}^{{2}}}}}}{{{2}{a}{b}{\left({a}+{b}\right)}}}\)
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