CHAPTER IX. EQUATIONS OF THE FIRST DEGREE CONTAINING MORE THAN ONE UNKNOWN QUANTITY. 145. If we have a single equation containing two unknown quantities, then for every value which we please to ascribe to one of the unknown quantities, we can determine the corresponding value of the other, and thus find as many pairs of values as we please which will satisfy the equation. Thus, let 2x+4y=16. (1.) If y=1, we find x=6; if y=2, we find x=4, and so on; and each of these pairs of values, 1 and 6, 2 and 4, etc., substituted in equation (1), will satisfy it. Suppose that we have another equation of the same kind, as, for example, 5x+3y=19. (2.) We can also find as many pairs of values as we please which will satisfy this equation. But suppose we are required to satisfy both equations with the same set of values for x and y; we shall find that there is only one value of x and one value of y. For, multiply equation (1) by 3, and equation (2) by 4, Axiom 3, and we have 6x+12y=48, (3.) 20x+12y=76. (4.) Subtracting equation (3) from equation (4), Axiom 2, we have 14.x=28; (5.) whence (6.) Substituting this value of x in equation (1), we have 4+4y=16; (7.) whence (8.) Thus we see that if both equations are to be satisfied, x must equal 2, and y must equal 3. Equations thus related are called simultaneous equations. y=3. X=2. 146. Simultaneous equations are those which must be satisfied by the same values of the unknown quantities. When two or more simultaneous equations are given for solution, 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. 147. Elimination is the operation of combining two or more equations in such a manner as to cause one of the unknown quantities contained in them to disappear. There are three principal methods of elimination : 1st, by addition or subtraction; 2d, by substitution; 3d, by comparison. 148. Elimination by Addition or Subtraction.—Let it be proposed to solve the system of equations 5x+4y=35, (1.) 7x-3y=6. (2.) Multiplying equation (1) by 3, and equation (2) by 4, we have 15x+12y=105, (3.) 28x-12y=24. (4.) Adding (3) and (4), member to member (Axiom 1), we have 43x=129; (5.) whence (6.) We may now deduce the value of y by substituting the value of x in one of the original equations. Taking the first for ex. ample, we have 15+4y=35; whence 4y=20, and y=5. x=3. 149. In the same way, an unknown quantity may be eliminated from any two simultaneous equations. This method is expressed in the following RULE. Multiply or divide the equations, if necessary, in such a manner that one of the unknown quantities shall have the same coefficient in both. Then subtract one equation from the other if the signs of these coefficients are alike, or add them together if the signs are unlike. In solving the preceding equations, we multiplied both members of each by the coefficient of the quantity to be eliminated in the other equation; but if the coefficients of the letter to be eliminated have any common factor, we may accomplish the same object by the use of smaller multipliers. In such cases, find the least common multiple of the coefficients of the letter to be eliminated, and divide this multiple by each coefficient; the quotients will be the least multipliers which we can employ. 150. Elimination by Substitution.—Take the same equations as before: 5x+4y=35, (1.) 7x-3y=6. (2.) Finding from (1) the value of y in terms of x, we have y 35-50 Y= (3.) 4 Substituting this value of y in (2), we have 105-15% 7ac = 6. 4 Clearing of fractions, 28x—105+15x=24; whence x=3. Substituting this value of x in (3), we have y=5. The method thus exemplified is expressed in the following RULE. Find an expression for the value of one of the unknown quantities in one of the equations; then substitute this value for that quantity in the other equation. 151. Elimination by Comparison.—Take the same equations as before: 5x+4y=35, (1.) 7x-3y=6. (2.) Y= y=1 Derive from each equation an expression for y in terms of x, and we have 35-5x (3.) 4 78-6 and (4.) 3 7.0 -6 35-52 4 Clearing of fractions, 28-24=105-15; whence 43x=129, and x=3. Substituting this value of x in (3), y=5. The method thus exemplified is expressed in the following = RULE. Find an expression for the value of the same unknown quantity in each of the equations, and form a new equation by placing these values equal to each other. In the solution of simultaneous equations, either of the preceding methods can be used, as may be most convenient, and each method has its advantages in particular cases. Generally, however, the last two methods give rise to fractional expressions, which occasion inconvenience in practice, while the first method is not liable to this objection. When the coefficient of one of the unknown quantities in one of the equations is equal to.unity, this inconvenience does not occur, and the method by substitution may be preferable; the first will, however, commonly be found most convenient. EXAMPLES. 1. Given (12+ = 100} to find the values of & 4x7y4 x and y. . Ans. x=8; y=4.. sin =7 2. Given + to find ac and y. y Ans. x=6; y=12. = 3 to find u and y. +10x=192 3. Given y+5 818 nb_mdi y= { } t-em y 5. Given to find u and y. bc-ad bc-ad mc-na Эх — 7y=20 6. Given to find u and y. 9x-11y=44 | 173-13y=144 7. Given to find x and y. 23x+19y=890 1 -mX y 8. Given to find x and y. XC 2 2 y Ans. x= * ; y= mtn m-n 4x+81 6 107-17 9. Given to find u and y. Ans. x=21; y=35. 157–17=4 |