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Question Number 104696 by nimnim last updated on 23/Jul/20

Answered by Rasheed.Sindhi last updated on 23/Jul/20

$$A^{\prime }sageatpresentxyears$$$$B^{\prime }sageatpresentyyears$$$$x-7=5{(y-7)}^2.......\left(i\right)$$$$y+3=\frac{2}{5}(x+3).........\left(ii\right)$$$$\left(ii\right)\Rightarrow x=\frac{5}{2}(y+3)-3$$$$=\frac{5y+9}{2}$$$$\left(i\right)\Rightarrow \frac{5y+9}{2}-7=5(y^2-14y+49)$$$$5y+9-14=10y^2-140y+490$$$$10y^2-145y+495=0$$$$2y^2-29y+99=0$$$$y=\frac{29\pm \sqrt{841-792}}{4}$$$$y=\frac{29\pm 7}{4}=9,\stackrel{\times }{5\frac{1}{2}}$$$$5\frac{1}{2}isfalse,because7yearsago$$$$theagebenegative.Negative$$$$ageisnotacceptable.$$$$y=9:x=\frac{5\left(9\right)+9}{2}=27$$$$A^{\prime }sage=27years$$$$B^{\prime }sage=9years$$$$$$

Commented by 1549442205PVT last updated on 23/Jul/20

$$\mathrm{Solve}\mathrm{quadratic}\mathrm{equation}\mathrm{false}$$$$\mathrm{applying}\mathrm{root}\mathrm{formular}\mathrm{y}=\frac{\mathrm{b}\pm \sqrt{\mathrm{\Delta }}}{2\mathrm{a}}$$$$\mathbf{a}=2\Rightarrow 2\mathbf{a}=4!$$

Commented by Rasheed.Sindhi last updated on 23/Jul/20

$$Thankyousir.I^{\prime }mgoingto$$$$correctit.$$

Commented by nimnim last updated on 23/Jul/20

$$ThankyouSir.$$

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