## Non-linear equations for IITJEE Mathematics training

(Non-linear) equations involving three or more variables can only be solved in special cases. We shall here consider some of the most useful methods of solution.

Example 1.

Solve the following system of equations:

$x+y+z=13$ $\ldots \ldots \ldots$ Equation I

$x^{2}+y^{2}+z^{2}=65$ $\ldots \ldots \ldots$ Equation II

$xy=10$ $\ldots \ldots\ldots$ Equation III

Solution I:

From II and III, we get $(x+y)^{2} + z^{2}=85$

Plug in $u=x+y$; then this equation becomes $u^{2}+z^{2}=85$.

Also, from I, we get $u+z=13$

hence, we obtain $u=7 \hspace{0.1in}or \hspace{0.1in}u=6$, $z=6 \hspace{0.1in} or \hspace{0.1in}7$.Thus,we have

$x+y=7$, $xy=10$; and $x+y=6$, $xy=10$. Hence, the solutions are

$x=5, \hspace{0.1in} or \hspace{0.1in} 2$; $y=2, \hspace{0.1in} or \hspace{0.1in} 5$; $z=6$;

or,

$x=3 \pm \sqrt{-1}$; $y=3 \mp \sqrt{-1}$; $z=7$.

Example 2:

Solve the following system of equations:

$(x+y)(x+z)=30$ Equation I

$(y+z)(y+x)=15$ Equation II

$(z+x)(z+y)=18$ Equation III

Solution: 2:

Write u, v, w for $y+z$, $z+x$, $x+y$ respectively; thus

$vw=30$, $uw=15$, $vu=18$…..call this equations A

Multiplying these equations together, we have

$u^{2}v^{2}w^{2}=30 \times 15 \times 18 = 8100$. Hence, $uvw=\pm 90$.

Combining this result with each of the equations in A, we have

$u=3, v=6. w=5$; or, $u=-3, v=-6, w=-5$; therefore, we get two sets of linear equations, as follows:

$y+z=3$

$z+x=6$

$x+y=5$

for which the solution set is: $x=4, y=1, z=2$; and the other set is:

$y+z=-3$

$z+x=-6$

$x+y=-5$

for which the solution set if $x=-4, y=-1, z=-2$.

Example 3:

Solve the following system of equations:

$y^{2}+yz+z^{2}=49$…call this equation (1)

$z^{2}+xz+x^{2}=19$…call this equation (2)

$x^{2}+xy+y^{2}=39$…call this equation (3)

Subtracting (2) from (1), we get the following:

$y^{2}-x^{2}+z(y-x)=30$, that is, $(y-x)(x+y+z)=30$…call this equation (4).

Similarly, from (1) and (3), we get the following:

$(z-x)(x+y+z)=10$…call this equation (5).

Hence, from equations (4) and (5), by division, we get the following:

$\frac{y-x}{z-x}=3$, hence, $y=3x-2z$

Substituting in Equation (3), we obtain $x^{2}-3xz+3z^{2}=13$.

From (2), we get the following: $x^{2}+xz+z^{2}=19$

Solving these homogeneous equations (hint: put $z=mx$, where m is parameter to be found), we get the following:

$x=\pm 2$, $z=\pm 3$; and therefore, $y=\pm 5$

or, $x=\pm \frac{11}{\sqrt{7}}$; $z=\pm \frac{1}{\sqrt{7}}$; and therefore, $y=\mp \frac{19}{\sqrt{7}}$.

Example 4:

Solve the following system of equations:

$x^{2}-yz=a^{2}$

$y^{2}-zx=b^{2}$

$z^{2}-xy=c^{2}$

Solution 4:

Multiply the equations by y, z, and x respectively, and add; then,

$c^{2}x+a^{2}y+b^{2}z=0$…call this equation I

Multiply the equations by z, x and y respectively; and add; then

$b^{2}x+c^{2}y+a^{2}z=0$…call this equation II

From (I) and (II), by cross multiplication,

$\frac{x}{a^{4}-b^{2}c^{2}}=\frac{y}{b^{4}-c^{2}a^{2}}=\frac{x}{c^{4}-a^{2}b^{2}}=k$, suppose.

Substitute in any one of the given equations; then,

$k^{2}(a^{6}+b^{6}+c^{6}-3a^{2}b^{2}c^{2})=1$; hence, we get the following solution:

$\frac{x}{a^{4}-b^{2}c^{2}}=\frac{y}{b^{4}-c^{2}a^{2}}=\frac{z}{c^{4}-a^{2}b^{2}}=\pm \frac{1}{\sqrt{a^{6}+b^{6}+c^{6}-3a^{2}b^{2}c^{2}}}$.

More stuff in the pipeline,

Nalin Pithwa

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