The legitimacy of the successive steps made in arriving at the solution x= = 10, y = 6, is shown by exhibiting the successive systems of equivalent equations. In practice this work may be abbreviated as follows: To eliminate y, multiply both members of equation (2) by 2. Then the coefficients of y in the resulting equation and in equation (1) are the same with different signs, and the work may be arranged as follows: (3) (4) (5) (6) (7) (8) Multiply (6) by 3 12x16y=240. 12x-9y=90. Observe that (1) and (2), (3) and (4), (5) and (6), (7) and (8), (10) and (11) form systems of equivalent equations. With (10) is associated the simplest of the equations, number (5), which is equivalent to (11). Clear (1) of fractions; 3.124 y 12 36. Transpose and unite terms Multiply (5) by 4 3x+4y60. 4x-3y=30. 215. The examples discussed in the preceding section illustrate the following method of elimination by addition and subtraction for two equations in two unknown quantities. Simplify the equations by removing parentheses or clearing fractions, and transpose the x and y terms to the first members and the terms free from x and y to the second members, multiply both members of each equation by such a number as will make the absolute values of the coefficients of one of the unknown quantities the same in the two resulting equations. The unknown quantity whose coefficients are now equal, with the same or opposite signs, can be eliminated in the first case by subtracting and in the second case by adding corresponding members, and equating the results. The solution of the given system of equations can now be found by solving this derived equation, and substituting the value of the unknown quantity thus found in the simpler of the preceding equations. EXERCISE XL Solve the following systems of equations by the method of addition and subtraction: 216. Particular Case.-Consider first the case in which one of the equations involves but one of the unknown quantities. and the system of two equations is equivalent to the following, or further to the system, x = 7 3.7-4y = 9 Sx = 7 y = 3. 217. General Case. Reduce the solution of the general to this particular case, by replacing the proposed system by another system. equivalent to it, in which one of the equations involves but one unknown quantity. Let, for example, the system of two equations be Derive the value of x from equation (1) as though y were known, thus, (3) x = 3 +12 y; substitute 3+ 12 y instead of x in equation (2); thus, and, according to the theorem 3, 212, the given system is equiv alent to the system, II ƒ (3) x = 3 +12 y { @ +48 (4) 3+12y + 4y = 19. But the equation (4) in system II involves y only, and by solving Then substitute 1 for y in equation (3), thus, System II is therefore equivalent to x = 3 +12 = 15. x = 15, and y = 1. Consequently system II has the solution x = 15 and y = no other solutions. 1, and 218. RULE. To solve a system of two equations of the first degree in two unknown quantities, the following rule can be stated: Derive from one of the equations the value of one of the unknown quantities, as if the other were known, and substitute this value in the other equation; and thus obtain an equation of the first degree in one unknown quantity. Solve this equation. Substitute the value found in the first equation, and solve the resulting equation for the first unknown quantity. 219. Applications. In practice circumstances arise which tend to simplify calculation. EXAMPLE 1. Solve the two equations, ((1) 4x+3y=61 (2) 7x - y = 38. Deduce the value of y from equation (2); then, an equation all of whose coefficients are integers. From simplifying, it follows, that (5) 4 x + 21 x = 61 +114 x = 7. Substitute 7 for x in (3). Then y = 7.7-38 = 11. The given equations have, therefore, the solution, Since the coefficients of x are equal, solve (1) for 4x; thus, By substituting 3 for y in (3) the equation follows: Since the coefficient 5 of y in (2) is a divisor of the coefficient 10 of y in (1), solve (2) for y; thus |