How Cheenta works to ensure student success?
Explore the Back-Story

Test of Mathematics Solution Subjective 71 - Real solutions

This is a Test of Mathematics Solution Subjective 71 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

Consider the following simultaneous equations in x and y :
${\displaystyle{4x + y + axy = a}}$
${\displaystyle{x - 2y -xy^2 = 0}}$
where ${\displaystyle{a}}$ is a real constant. Show that these equations admit real solutions in x and y.

Solution

${\displaystyle{4x + y + axy = a}}$          ...  (i)
${\displaystyle{x - 2y -xy^2 = 0}}$           ... (ii)
${\displaystyle{x - 2y -xy^2 = 0}}$
${\Rightarrow}$ ${\displaystyle{x = {\frac{2y}{1-y^2}}}}$
${\displaystyle{x + y + axy = a}}$
${\Rightarrow}$ ${\displaystyle{\frac{2y}{1-y^2} + 4 + a{\frac{2y^2}{1-y^2}}}}$ ${\displaystyle{= a}}$ [ replacing x in terms of y. ]
${\Rightarrow}$ ${\displaystyle{3y - y^3 + 3ay^2 -a = 0}}$
This is a 3rd degree polynomial over y.
Now as we know any odd degree polynomial has at least one real root.That is why y has also at least one real root.Now for that particular root we get ${\displaystyle{x = {\frac{2y}{1-y^2}}}}$ also real.
Conclusion: ${\displaystyle{x + y + axy = a}}$ & ${\displaystyle{x - 2y -xy^2 = 0}}$ admits real solutions in x and y.

This is a Test of Mathematics Solution Subjective 71 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

Consider the following simultaneous equations in x and y :
${\displaystyle{4x + y + axy = a}}$
${\displaystyle{x - 2y -xy^2 = 0}}$
where ${\displaystyle{a}}$ is a real constant. Show that these equations admit real solutions in x and y.

Solution

${\displaystyle{4x + y + axy = a}}$          ...  (i)
${\displaystyle{x - 2y -xy^2 = 0}}$           ... (ii)
${\displaystyle{x - 2y -xy^2 = 0}}$
${\Rightarrow}$ ${\displaystyle{x = {\frac{2y}{1-y^2}}}}$
${\displaystyle{x + y + axy = a}}$
${\Rightarrow}$ ${\displaystyle{\frac{2y}{1-y^2} + 4 + a{\frac{2y^2}{1-y^2}}}}$ ${\displaystyle{= a}}$ [ replacing x in terms of y. ]
${\Rightarrow}$ ${\displaystyle{3y - y^3 + 3ay^2 -a = 0}}$
This is a 3rd degree polynomial over y.
Now as we know any odd degree polynomial has at least one real root.That is why y has also at least one real root.Now for that particular root we get ${\displaystyle{x = {\frac{2y}{1-y^2}}}}$ also real.
Conclusion: ${\displaystyle{x + y + axy = a}}$ & ${\displaystyle{x - 2y -xy^2 = 0}}$ admits real solutions in x and y.