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 $y = 2, -\frac{7}{4}$ .