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5 / x + 6 + 2 / x2 + 7x + 6 = 3 / x = 1
- Thread starter dkarolasz
- Start date
Re: equations???
Hello, dkarolasz!
Since this is an equation, the work is even easier . . .
Multiply through by the common denominator: \(\displaystyle (x\,+\,1)(x\,+\,6)\)
\(\displaystyle \L(x\,+\,1)(x\,+\,6)\cdot\frac{5}{x\,+\,6}\,+\,(x\,+\,1)(x\,+\,6)\cdot\frac{2}{(x\,+\,1)(x\,+\,6)} \;=\;(x\,+\,1)(x\,+\,6)\cdot\frac{3}{x\,+\,1}\)
Reduce: \(\displaystyle \L\:5(x\,+\,1)\,+\,2\;=\;3(x\,+\,6)\)
. . . . . . . . \(\displaystyle \L 5x\,+\,5\,+\,2 \;=\;3x\,+\,18\)
. . . . . . . . . . . . . . . .\(\displaystyle \L 2x \;=\;11\)
. . . . . . . . . . . . . . . . .\(\displaystyle \L x \;=\;\frac{11}{2}\)
Hello, dkarolasz!
Since this is an equation, the work is even easier . . .
\(\displaystyle \L\frac{5}{x\,+\,6} \,+\,\frac{2}{x^2\,+\,7x\,+\,6}\;=\;\frac{3}{x\,+\,1}\)
Multiply through by the common denominator: \(\displaystyle (x\,+\,1)(x\,+\,6)\)
\(\displaystyle \L(x\,+\,1)(x\,+\,6)\cdot\frac{5}{x\,+\,6}\,+\,(x\,+\,1)(x\,+\,6)\cdot\frac{2}{(x\,+\,1)(x\,+\,6)} \;=\;(x\,+\,1)(x\,+\,6)\cdot\frac{3}{x\,+\,1}\)
Reduce: \(\displaystyle \L\:5(x\,+\,1)\,+\,2\;=\;3(x\,+\,6)\)
. . . . . . . . \(\displaystyle \L 5x\,+\,5\,+\,2 \;=\;3x\,+\,18\)
. . . . . . . . . . . . . . . .\(\displaystyle \L 2x \;=\;11\)
. . . . . . . . . . . . . . . . .\(\displaystyle \L x \;=\;\frac{11}{2}\)