known. The solution is then substituted in one of the original equations, and we solve for the other unknown. Example. 2x + 3y - = 0, 4x - 5y + 6 = 0. Multiply each term by 2 in the upper equation, and we derive the following new equation: 4x + 6y - - 16 = 0. Place the second equation directly below this new equation, and subtract: 4x 6y - 16 = 0, 4x - 5y + 6 = 0. If we subtract the lower equation from the upper, we eliminate the 4x term. Now solve for y. +11y - 22 = 0, 11y=22, Substitute y = 2 in the original upper equation. Then 2x + 6 - 8 = 0, 2x = 2, and X = 1. B.3 INTRODUCTORY ALGEBRA So the solution of these equations is X = 1 and y = 2. Check the solutions by substi- tuting these values into the two original simultaneous equations. Another example: 3x-2y-5=0, - 6x + y + 12 = 0. There are several possibilities to eliminate one of the unknowns. This time lets multiply each term in the lower equation by two and we get 12x + 2y + 24 0. Place this new equation below the original upper equation: 3x - 2y -5=0, - 12x 2y + 24 = 0. Add the two equations and we get 15x + 19 = 0. Solve for X. 15x = -19. X = - 19/5. Substitute this value in the upper equation and solve for y. 3(-19/15) - 2y - = 0 - 57/15 - 75/15 = 2y 2y = - 132/15 and y = -66/15 or - 22/5 The Substitution Method. Select one of the two simultaneous equations and solve for one of the unknowns in terms of the other. Example. (Repeating the first example from above). 3y = 0, 4x - 5y + 6 = 0. We can select either equation. Select the first equation. Then 2x = 8-3y, - X = (8 - 3y)/2. Substitute this value for X in the second equation. 4(8 - 3y)/2 - 5y + 6 : - = 0, 16 - 6y - 5y + = 0, -11y = -22, y ,=2. A REVIEW OF MATHEMATICS FOR TELECOMMUNICATION APPLICATIONS B.4 LOGARITHMS TO THE BASE 10 B.4.1 Definition of Logarithm If b is a positive number different from 1, the logarithm of the number y, written loga y, is defined as follows: If a x = y, then X is a logarithm of y to the base a, and we write loga y = x. This shows, therefore, that a logarithm is an expo