## System of Equations — IITJEE Foundation Maths

Some problems are presented below as warm-up exercises:

Question 1:

Solve : $x+y = 15$ and $xy=36$

Question 2:

Solve: $x-y=12$ and $xy=85$

Question 3:

Solve: $x^{2}+y^{2}=74$ and $xy=35$

Question 4:

Solve : $x^{2}+y^{2}=185$ and $x+y=17$

Some other standard set of systems of equations and techniques for solving them are presented below:

Any pair of equations of the form

$x^{2} \pm pxy + y^{2}=a^{2}$….Equation I

$x \pm y = b$…Equation II

where p is any numerical quantity, can be reduced to one of the cases in the warm-up exercises; for, by squaring II and combining with I, an equation to find xy is obtained, the solution can then be completed by the aid of equation II.

Example 1:

Solve $x^{3}-y^{3}=999$…Equation 1

and $x-y=3$…Equation 2.

By division, $x^{2}+xy+y^{2}=333$…Equation 3.

From Equation 2, $x^{2} - 2xy + y^{2} = 9$;

by subtraction, $3xy=324$ and $xy=108$…Equation 4.

From Equation (2) and (4), $x=12, y=9$ or $x=9, y = -12$

Example 2:

Solve: $x^{4}+x^{2}y^{2}+y^{4}=2613$…Equation I

and $x^{2}+xy+y^{2}=67$…Equation 2

Dividing (1) by (2), $x^{2}-xy+y^{2}=39$…Equation 3.

From Equation (2) and (3), by addition, $x^{2}+y^{2}=33$; by subtraction $xy=14$, hence, we get the following: $x = \pm 7, x = \pm 2$, or $y = \pm 2, y = \pm 7$

Example 3:

Solve: $\frac{1}{x} - \frac{1}{y} = \frac{1}{3}$…Equation (1)

and $\frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{5}{9}$…Equation (2)

From (1), by squaring, $\frac{1}{x^{2}} - \frac{2}{xy} + \frac{1}{y^{2}} = \frac{1}{9}$,

and by subtraction, $\frac{2}{xy} = \frac{4}{9}$

adding to (2), $\frac{1}{x^{2}} + \frac{2}{xy} + \frac{1}{y^{2}} = 1$

and hence, $\frac{1}{x} + \frac{1}{y} = \pm 1$.

Combining with (1), $\frac{1}{x} = \frac{2}{3}$, or $-\frac{1}{3}$; $\frac{1}{y} = - \frac{1}{3}$, or $-\frac{2}{3}$; so, the solutions are: $x = \frac{3}{2}, -3$ and $y =3, -\frac{3}{2}$.

The following method of solution may always be used when the equations are of the same degree and homogeneous.

Example:

Solve $x^{2}+ xy + 2y^{2} = 74$ …Equation 1

and $2x^{2} + 2xy + y^{2} = 73$…Equation 2.

Put $y = mx$, and substitute in both equations. Thus,

$x^{2}(1+m+2m^{2}) =74$…Equation 3.

$x^{2}(2 + 2m + m^{2}) =73$…Equation 4.

By division, $\frac{1+m+2m^{2}}{2+2m+m^{2}} = \frac{74}{73}$

$73 + 73m + 146m^{2} = 148 + 148m + 74m^{2}$

Hence, $72m^{2} -75m - 75=0$; or $24m^{2} - 25m - 25 = 0$; hence, $(8m + 5)(3m - 5) = 0$; so $m = -\frac{5}{8}$ or $\frac{5}{3}$.

Case (i):

Take $m= -\frac{5}{8}$ and substitute in either (3) or (4).

From (3): $x^{2}(1- \frac{5}{8} + \frac{50}{64}) = 74$

$x^{2} = 64$, so $x = \pm 8$; so $y = mx = - \frac{5}{8}x = \mp 5$.

Case (ii):

Take $m = \frac{5}{3}$, then from (3) $x^{2}(1 + \frac{5}{3} + \frac{50}{9}) = 74$, so we get $x^{2} = 9$, so $x = \pm 3$; and hence, $y = mx = \frac{5}{3}x = \pm 5$.

When one of the equations is of first degree and the other of a higher degree, we may from the simple equation, find the value of one of the unknowns in terms of the other, and substitute in the second equation.

Example:

Solve $3x - 4y = 5$…Equation (1)

and $3x^{2} - xy - 3y^{2} = 21$…Equation (2)

From (1), we have $x = \frac{5+4y}{3}$

and substituting in (2), $3\frac{(5+4y)^{2}}{9} - \frac{y(5+4y)}{3} - 3y^{2}= 21$

$75 + 120y +48y^{2} - 15y - 12y^{2} - 27y^{2} = 189$

$9y^{2} + 105y -114 = 0$

$3y^{2} + 35y - 38 = 0$

$(y-1)(3y+31) = 0$

Hence, $y = 1, -\frac{38}{3}$, and $x = 3, -\frac{137}{9}$.

The examples we have presented above will be sufficient as a general explanation of the methods to be employed; but, in some cases, special artifices are necessary.

Example:

Solve: $x^{2} + 4xy + 3x = 40 - 6y -4y^{2}$…Equation (1)

and $2xy - x^{2} = 3$…Equation (2)

From (1) and (2), we have $x^{2} + 4xy + 4y^{2} + 3x + 6y = 40$;

that is, $(x+2y)^{2} + 3(x+2y) - 40 = 0$;

or, $((x+2y)+8)((x+2y)-5) = 0$, hence, $x+2y = -8, 5$

Case:

Combining $x+2y =5$ with (2), we obtain $2x^{2} - 5x + 3=0$, so $x = 1, \frac{3}{2}$; and by substituting in $x + 2y = 5$, we get $y = 2, -\frac{7}{4}$.

Case:

Combining  $x+2y = -8$ with (2), we obtain $2x^{2} + 8x + 3 = 0$; hence, $x = \frac{-4 \pm \sqrt{10}}{2}$, and $y = \frac{-12 \mp \sqrt{10}}{4}$

Example:

Solve: $x^{2}y^{2} - 6x = 34 - 3y$…Equation (1)

and $3xy + y = 2((9+x)$…Equation (2)

From (1), we have $x^{2}y^{2} -6x + 3y = 34$;

From (2), we have $9xy - 6x + 3y = 54$;

By subtraction, $x^{2}y^{2} - 9xy + 20 = 0$; that is, $(xy - 5)(xy -4) = 0$; so $xy = 5, 4$.

Case:

Substituting $xy = 5$ in (2), gives $y -2x = 3$. From these equations, we obtain $x =1$ or $-\frac{5}{2}$; $y = 5$, or $-2$.

Case:

Substituting $xy =4$ in (2), gives $y - 2x = 6$. From these equations, we obtain $x = \frac{-3 \pm \sqrt{17}}{2}$, and $y = 3 \pm \sqrt{17}$.

Hope you enjoyed it,

Nalin Pithwa.