QQuestion
Using properties of proportion, find x:
\frac{6\mathrm{x}^2+3\mathrm{x}-5}{3\mathrm{x}-5}=\frac{9\mathrm{x}^2+2\mathrm{x}+5}{2\mathrm{x}+5}, x ≠ 0.
✓Answer
Given:
\frac{6\mathrm{x}^2+3\mathrm{x}-5}{3\mathrm{x}-5}=\frac{9\mathrm{x}^2+2\mathrm{x}+5}{2\mathrm{x}+5}Applying componendo and dividendo:
\frac{(6\mathrm{x}^2+3\mathrm{x}-5)+(3\mathrm{x}-5)}{(6\mathrm{x}^2+3\mathrm{x}-5)-(3\mathrm{x}-5)}=\frac{(9\mathrm{x}^2+2\mathrm{x}+5)+(2\mathrm{x}+5)}{(9\mathrm{x}^2+2\mathrm{x}+5)-(2\mathrm{x}+5)} \frac{6\mathrm{x}^2+6\mathrm{x}-10}{6\mathrm{x}^2}=\frac{9\mathrm{x}^2+4\mathrm{x}+10}{9\mathrm{x}^2}Since x ≠ 0:
\frac{6\mathrm{x}^2+6\mathrm{x}-10}{6}=\frac{9\mathrm{x}^2+4\mathrm{x}+10}{9}⇒ 9(6x² + 6x − 10) = 6(9x² + 4x + 10)
⇒ 54x² + 54x − 90 = 54x² + 24x + 60
⇒ 30x = 150
⇒ x = 5
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