HOME David Terr Ph.D. Math, UC Berkeley P.9. Linear Equations A problem of key importance in algebra is the solution of simultaneous linear equations in several variables. (Linear equations in one variable are trivial, so we will not discuss them.) The simplest nontrivial case is two equations in two variables. It is best to illustrate with an example, which we solve using two different methods.   Example 1: Solve the simultaneous equations 3x + 4y = 11 and 2x + 3y = 8 for x and y. Solution 1: One method to solve these equations is substitution. We solve y for x in the first equation and then plug the result into the second equation. Subtracting 3x from both sides of the first equation, we find 4y = 11 - 3x. Dividing both sides by 4, we find y = 11/4 - 3/4 x. Now plugging into the second equation, we find 2x + 3(11/4 - 3/4 x) = 8, which simplifies to 2x + 33/4 - 9/4 x = 8. Upon further simplification, we find 33/4 - 1/4 x = 8 33 - x = 32 -x = -1 x = 1 Once we have obtained x, it is now easy to obtain y by substituting into one of the original equations. Plugging into the second equation, we find (2)(1) + 3y = 8 2 + 3y = 8 3y = 6 y = 2 Thus the solution is x=1; y=2. To check this result, we substitute these values back into the original equations. Thus we verify that 3x + 4y = (3)(1) + (4)(2) = 3 + 8 = 11 and 2x + 3y = (2)(1) + (3)(2) = 2 + 6 = 8. Solution 2: A second method for solving simultaneous equations is cancelling coefficients. This is done by multiplying the first equation by the x-coefficient of the second equation, multiplying the second equation by the x-coefficient of the first equation, and subtracting the two equations, thus eliminating x. Our equations become 6x + 8y = 22 and 6x + 9y = 24. Subtracting the first equation from the second, we find y = 2. Substituting this value of y into the first equation, we find 3x + (4)(2) = 11, which simplifies to 3x + 8 = 11 3x = 3 x = 1 Thus, once again we find x=1 and y=2.   Simultaneous linear equations of three or more variables may be solved similarly, though the math quickly becomes more complicated as the number of equations and variables increases. We will only consider the case of three linear equations in three variables.   Example 2: Solve the following simultaneous linear equations: 2x + 5y - z = 4 x - 3y + 2z = 3 3x - 2y + z = 8 Solution: The trick is to eliminate variables one at a time. Note that if we add the first and third equations, we eliminate z, obtaining the equation 5x + 3y = 12. We can obtain another equation in x and y by adding twice the first equation to the second as follows: 4x + 10y - 2z = 8 x - 3y + 2z = 3 Thus we find 5x + 7y = 11. Subtracting this equation from the previous one in x and y, we find 4y = -1 or y = -1/4. Substituting this value of y into the first equation in x and y, we find 5x + (3)(-1/4) = 12 5x - 3/4 = 12 20x - 3 = 48 20x = 51 x = 51/20 We finally solve for z by substituting the values we found for x and y into any of the three original equations. Upon plugging them into the third equation we find 3(51/20) - 2(-1/4) + z = 8 153/20 + 1/2 + z = 8 153 + 10 + 20z = 160 163 + 20z = 160 20z = -3 z = -3/20 Thus our solution is x = 51/20; y = -1/4; z = -3/20. One should check this solution by substituting these values back into each of the original equations. Copyright © 2007-2009 - MathAmazement.com. All Rights Reserved.