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.

 

 

 

 

 

 

 

 

 

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