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Two fractions are such that their product is 4 and sum is148/35. Find the two fractions.1). 6/15, 10/32). 6/5, 10/33). 7/2, 8/74). 10/7, 14/5 |
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Answer» LET the two fractions be ‘x’ and ‘y’. ⇒ xy = 4and x + y = 148/35 ⇒ x = 4/y ⇒ (4/y) + y = 148/35 $(\begin{array}{l} \Rightarrow {\rm{\;}}\frac{{4 + {y^2}}}{y}\; = \;\frac{{148}}{{35}}\\ \Rightarrow {\rm{\;}}140 + 35{y^2}\; = \;148y\\ \Rightarrow {\rm{\;}}35{y^2} - 148y + 140\; = \;0\\ \Rightarrow {\rm{\;}}y\; = \;\frac{{148\; \PM \SQRT {{{148}^2} - 4 \times 35 \times 140} }}{{70}}\; = \;\frac{{148\; \pm \sqrt {2304} }}{{70}}\; = \;\frac{{148 \pm 48}}{{70}}\\ \Rightarrow {\rm{\;}}y\; = \;\frac{{100}}{{70}}\;or\frac{{196}}{{70}}\; = \;\frac{{10}}{7}\;or\frac{{14}}{5}\\ \Rightarrow {\rm{\;}}x\; = \;\frac{4}{y}\; = \;\frac{4}{{\frac{{10}}{7}}}\; = \;\frac{{28}}{{10}}\; = \;\frac{{14}}{5}\;or\;x\; = \;\frac{4}{y}\; = \;\frac{4}{{\frac{{14}}{5}}}\; = \;\frac{{20}}{{14}}\; = \;\frac{{10}}{7} \end{array})$ ∴ The two fractions are 10/7 and 14/5. |
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