ON THE SOLUTION OF SIMPLE EQUATIONS, CONTAINING TWO OR MORE UNKNOWN QUANTITIES. 145. A single equation, containing two unknown quantities, admits of an infinite number of solutions; for if we assign any arbitrary value to one of the unknown quantities, the equation will determine the corresponding value of the other unknown quantity. Thus, in the equation y=x+10, each value which we may assign to x will, when augmented by 10, furnish a corresponding value of y. Thus, if x=2, y=12; if x=3, y=13, and so on. An equation of this nature is called an indeterminate equation, and since the value of y depends upon that of x, y is said to be a function of x. In general, every quantity, whose value depends upon one or more quantities, is said to be a FUNCTION of these quantities. Thus, in the equation y=ax+b, we say that y is a function of x, and that y is expressed in terms of x, and the known quantities a, b. If, however, we have two equations between two unknown quantities, and if these equations hold good together, then it will be seen presently that we can combine them in such a manner as to obtain determinate values for each of the unknown quantities; that is to say, each of the unknown quantities will have but a single value, which will satisfy the equations. The equations in this case are called determinate. In general, in order that questions may admit of determinate solutions, we must have as many separate equations as there are unknown quantities; a group of equations of this nature is called a system of simultaneous equations. If the number of equations exceed the number of unknown quantities, unless the equations in excess conform to the values of the unknown quantities determined by the others, the equations are said to be incompatible. Thus, if we have x+y=10 and x-y=6, the only values of x and y which will satisfy both these equations are 8 for x, and 2 for y. Now, if we were to add another equation to these, it must conform to these values, and could not be written in any form at pleasure. Thus, we might for a third equation say xy=16; but we could not write ry=100, for this third equation would be incompatible with the other two.* *Equations may be incompatible when the number does not exceed the number of unknowns, as the following problem will show: A sportsman was asked how many birds he had taken. He replied, if 5 be added to the third of those I took last year, it will make the half of the number taken this year. But if from three times this last half 5 be taken, you will have precisely the number taken last. year. How many did he take in each year? Let x= the number this year, and y= the number last year. an absurd equality, whence we conclude that there exist no values of x and y which satisfy the two equations. This is because the conditions of the problem are inconsistent with each other. When, however, the two equations are derived from the same problem, and its conditions are not contradictory, values for r and y will always be found to satisfy them. 146. In order to solve a system of two simple equations containing two u known quantities, we must endeavor to deduce from them a single equation containing only one unknown quantity; we must, therefore, make one of the unknown quantities disappear, or, as it is termed, we must eliminate it. The equation thus obtained, containing one unknown quantity only, will give the value of the unknown quantity which it involves, and, substituting the value of this unknown quantity in either of the equations containing the two unknown quantities, we shall arrive at the value of the other unknown quantity. The process which most naturally suggests itself for the elimination of one of the unknown quantities, is to derive from one of the two equations an expression for that unknown quantity in terms of the other unknown quantity, and then substitute this expression in the other equation. We shall see that the elimination may be effected by different methods, which are more or less simple according to the nature of the question proposed. EXAMPLE I. Let it be proposed to solve the system of equations 147. FIRST METHOD. From equation (1) we find the value of y in terms of x, which gives y=x+6; substituting the expression r+6 for y in equation (2), it becomes x+6+x=12, from which we find the determinate value x=3; since we have already seen that y=x+6, we find also the determinate value y=3+6 or 9. Thus it appears, that although each of the above equations, considered separately, admits of an infinite number of solutions, yet the system of equations admits only one common solution, x=3, y=9. 148. SECOND METHOD.-Derive from each equation an expression for y in terms of x, we shall then have y= x+6 These two values of y must be equal to one another, and, by comparing them, we shall obtain an equation involving only one unknown quantity, viz., x+6=12-x. Whence x=3. Substituting the value of x in the expression y=x+6, we find y=9. The substitution of 3, the value of x, in the second expression, y=12—x, leads necessarily to the same value of y; thus, 12-3=9, for we derived the value of x from the equation r+6=12—x. 149. THIRD METHOD.-Since the coefficients of y are equal in the two equations, it is manifest that we may eliminate y by subtracting the two equctions from each other, which gives (y+x)-(y-x)=12-6. Whence 2x=6 Having thus obtained the value of x, we may deduce that of y by making r=3 in either of the proposed equations; we can, however, determine the value of y directly, by observing that, since the coefficients of x in the proposed equations are equal, and have opposite signs, we may eliminate x by adding the two equations together, which gives (y-x)+(y+x)=12+6. Whence 2y=18 If we examine the three above methods, we shall perceive that they consist in expressing that the unknown quantities have the same values in both equations. These methods have derived their names from the processes employed to effect the elimination of the unknown quantities. The first is called the method of elimination by substitution. The third is called the method of elimination by addition and subtraction. The rule for the first is to find the value of one of the unknown quantities in one of the equations, and substitute it in the other equation. For the second, is to find the value of the same unknown quantity in each of the two given equations, and set these values equal. And for the third, is to make the coefficient of the unknown quantity to be eliminated the same in the two equations, and add or subtract as the case may require. Add, if the signs of the equal terms are different, and if they are alike, subtract. By either of these rules a single equation, containing but one unknown quantity, is obtained. Substituting the value of y in terms of x in equation (2), it becomes an equation containing r alone, which, when solved, gives x=2. This value of x, substituted in either of the equations (1) or (2), gives Whence x=2. Substituting this value for r in either of the preceding expressions for y, we find 3o. Eliminating by subtraction. y=3. In order to eliminate y, we perceive that if we could deduce from the proposed equations two other equations in x and y, in which the coefficients of y should be equal, the elimination of y would be effected by subtracting one of these new equations from the other. It is easily seen that we shall obtain two equations of the form required if we multiply all the terms of each equation by the coefficient of y in the other. Multiplying, therefore, all the terms of equation (1) by 4, and all the terms of equation (2) by 3, they become 8x+12y=52 15x+12y=66. Subtracting the former of these equations from the latter, we find Whence 7x=14. x=2. In like manner, in order to eliminate x, multiply the first of the proposed equations by 5, and the second by 2, they will then become 10x+15y=65 Subtracting the latter of these two equations from the former, 7y=21. Whence y=3. In order to solve a system of three simple equations between three unknown quantities, we must first eliminate one of the unknown quantities by one of the methods explained above; this will lead to a system of two equations, containing only two unknown quantities; the value of these two unknown quantities may be found by any of the methods described in the last article, and substituting the value of these two unknown quantities in any one of the original equations, we shall arrive at an equation which will determine the value of the third unknown quantity. Substituting this value of z in equations (2) and (3), they become 2x+2y+2(16—3x-2y)=18 . . . (5) ? ... 2x+2y+ (16-3x-2y)=14... (6) S these last two equations contain x and y only, and, if treated according to any of the above methods, will give us x=2, y=3. Substituting these values of x and y in any one of the equations (1), (2), (3), (4), we find 20. Eliminating by comparison. z=4. In order to eliminate z, derive from each of the three proposed equations a value of z in terms of x and y; we then have z=16-3x-2y x-y equating the first of these values of z with the second and with the third in succession, we arrive at a system of two equations : 16-3x-2y=14—2x-2y containing x and y only; these equations give x=2, y=3;. these values of x and y, when substituted in any of the three expressions for z, give 3°. Eliminating by subtraction. z=4. In order to eliminate z between equations (1) and (2), 3x+2y+x=16 2x+2y+2x=18; we perceive that, in order to reduce these equations to two others in which the coefficients of z shall be the same, it will be sufficient to divide the two members of the second equation by 2, for we thus have x+y+z=9. Subtracting this from the first equation, 3x+2y+z=16, we find an equation between two unknown quantities, 2x+y=7. In order to eliminate z between equations (1) and (3), 3x+2y+2=16 the substitution of this value of x in equation (a) gives y=3, (a). and the substitution of these values of x and y in any of the proposed equations gives z=4. The particular form of the proposed equations enables us to simplify the above calculation; for if we subtract equation (3) from equations (1) and (2) in succession, we have (3x+2y+ z)−(2x+2y+z)=16-14, whence x=2 and substituting these values of r and z in any of the proposed equations, we find |