 Solving Algebraically ELIMINATION The Elimination method is most useful when the equations are both in standard form, or near standard form. It also lends itself well to alternate solving methods, such as using a calculator solver function, deriving a formula, or solving using determinants (Math 31). We'll be mixing the equations together and solving the result in order to find the mixing point, or intersection point, which will be the solution. The elimination method mixes the two equations by adding or subtracting them, term by term, in order to 'eliminate' one variable, allowing us to solve for the other. In order to do this, we'll want to make sure both equations are in the form Example 1: 4x - 2y - 10 = 0 3x + 4y = 13 We need to rearrange the first equation slightly to get: 4x - 2y = 10 3x + 4y = 13 We're going to double the first equation: 8x - 4y = 20 This new equation is completely equivalent to the original 3x + 4y = 13 Why did we do that? Well, watch what happens when we add the two equations, term by term: 8x - 4y = 20 3x + 4y = 13 11x + 0y = 33 The y terms disappear! You can now solve this for x: 11x = 33 x = 3 Just like with the other algebraic methods, we now substitute this into one of the original two equations to find 'y'. We'll use the first original equation: 4x - 2y - 10 = 0 4(3) - 2y - 10 = 0 2 - 2y = 0 -2y = -2 y = 1 Solution is (3, 1) Here's a summary of how the Elimination method works:
Example 2: 5x + 2y = 4 3x - 2y = -20 Both equations are already in the form Ax + By = C The coefficients of the y terms are the same size We can start with step 3: 5x + 2y = 4 3x - 2y = -20 The signs on the y terms are opposite, so add the equations: 8x + 0y = -16 8x = -16 x = -2 Substitute this in the first equation (fewer negative signs) 5x + 2y = 4 5(-2) + 2y = 4 -10 + 2y = 4 2y = 14 y = 7 The solution is (-2, 7) Example 3a: 2x + 3y = 11 5x - 2y = -10.5 Both equations are already in the form Ax + By = C so we can start at step 2: We can't multiply the first equation by any whole number to get its coefficients to be the same size as those of x or y in the second equation. Similarly we can't do it with the second equation We'll have to multiply both equations by something. Let's make the coefficients of x the same; we'll mutliply the first equation by 5 and the second equation by 2. The will make both x coefficients 10: 2x + 3y = 11 x 5 5x - 2y = -10.5 x 2 10x + 15y = 55 10x - 4y = -21 The signs on the x terms are the same, so we subtract: 10x + 15y = 55 10x - 4y = -21 0x + 19y = 76 be careful subtracting negatives y = 4 Substitute in original equation 1: 2x + 3(4) = 11 2x = -1 x = -0.5 solution is (-0.5, 4) Example 3b: 2x + 3y = 11 5x - 2y = -10.5 We're going to do the same example again, only this time we're going to eliminate the y's, just to show you that it doesn't matter what you choose 2x + 3y = 11 5x - 2y = -10.5 We're going to make both y coefficients have size 6; Multiply the first equation by 2 and the second equation by 3: 2x + 3y = 11 x2 5x - 2y = -10.5 x3 4x + 6y = 22 15x - 6y = -31.5 The signs on the y's are opposite, so to get zero we add the equations: 4x + 6y = 22 15x - 6y = -31.5 19x + 0y = -9.5 x = -0.5 Now substitute and solve; again choose equation one: 2(-0.5) + 3y = 11 -1 + 3y = 11 3y = 12 y = 4 solution is (-0.5, 4) The elimination method is the most important of the three methods to master, particularly because it lends itself well to using a formula, and is often the best choice to solve real-world problems. |