Question

Solve the following equation.
${x^2} + y\left( {x + 1} \right) = 17,{y^2} + x\left( {y + 1} \right) = 13$

Answer 1

Hint – In this particular type of question use the concept that from any one of the given equations calculate the value of x in terms of y or y in terms of x and substitute the calculated value in another given equation and solve, so use these concepts to reach the solution of the question.

Complete step by step solution:
Given equation:
${x^2} + y\left( {x + 1} \right) = 17$..................... (1)
${y^2} + x\left( {y + 1} \right) = 13$...................... (2)
Now from equation (1) calculate the value of y we have,
${x^2} + y\left( {x + 1} \right) = 17$
$ \Rightarrow y\left( {x + 1} \right) = 17 - {x^2}$
$ \Rightarrow y = \dfrac{{17 - {x^2}}}{{x + 1}}$...................... (3)
Now substitute this value in equation (2) we have,
$ \Rightarrow {y^2} + x\left( {y + 1} \right) = 13$
$ \Rightarrow {\left( {\dfrac{{17 - {x^2}}}{{x + 1}}} \right)^2} + x\left( {\left( {\dfrac{{17 - {x^2}}}{{x + 1}}} \right) + 1} \right) = 13$
Now simplify the above equation using ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ so we have,
$ \Rightarrow \dfrac{{289 + {x^4} - 34{x^2}}}{{{{\left( {x + 1} \right)}^2}}} + x\left( {\dfrac{{17 - {x^2} + x + 1}}{{x + 1}}} \right) = 13$
$ \Rightarrow 289 + {x^4} - 34{x^2} + \left( {x + 1} \right)x\left( {17 - {x^2} + x + 1} \right) = 13{\left( {x + 1} \right)^2}$
Now simplify the above equation using ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ so we have,
$ \Rightarrow 289 + {x^4} - 34{x^2} + \left( {{x^2} + x} \right)\left( {18 - {x^2} + x} \right) = 13\left( {{x^2} + 1 + 2x} \right)$
$ \Rightarrow 289 + {x^4} - 34{x^2} + 18{x^2} - {x^4} + {x^3} + 18x - {x^3} + {x^2} = 13{x^2} + 13 + 26x$
$ \Rightarrow 276 - 28{x^2} - 8x = 0$
$ \Rightarrow 28{x^2} + 8x - 276 = 0$
Now divide by 4 throughout we have,
$ \Rightarrow 7{x^2} + 2x - 69 = 0$
Now this is a quadratic equation so apply quadratic formula, the solution in x is given as
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, where, a = 7, b = 2, c = -69
Now substitute the values we have,
$ \Rightarrow x = \dfrac{{ - 2 \pm \sqrt {{2^2} - 4\left( 7 \right)\left( { - 69} \right)} }}{{2\left( 7 \right)}} = \dfrac{{ - 2 \pm \sqrt {1936} }}{{14}} = \dfrac{{ - 2 \pm 44}}{{14}}$
$ \Rightarrow x = \dfrac{{ - 2 + 44}}{{14}},\dfrac{{ - 2 - 44}}{{14}}$
$ \Rightarrow x = \dfrac{{42}}{{14}} = 3,\dfrac{{ - 46}}{{14}} = - 3.2857$
Now from equation (3) when, x = 3
$ \Rightarrow y = \dfrac{{17 - {3^2}}}{{3 + 1}} = \dfrac{{17 - 9}}{4} = \dfrac{8}{4} = 2$
Now from equation (3) when, x = -3.2857
$ \Rightarrow y = \dfrac{{17 - {{\left( { - 3.2857} \right)}^2}}}{{ - 3.2857 + 1}} = \dfrac{{17 - 10.796}}{{ - 2.2857}} = \dfrac{{6.204}}{{ - 2.2857}} = - 2.714$
So the solution of the equation is,
$ \Rightarrow $ (x, y) = (3, 2) and (x, y) = (-3.2857, -2.714)
So this is the required answer.

Note – Whenever we face such types of question there are lots of method to solve these given equation such as substitution method, elimination method, cross multiplication method, graphical method so we can solve these given equation by any one of these method, here we use substitution method to solve these equations as above and simplify, we will get the required solution for x and y.