# Solve Equations Using Elimination

how do you solve by using elimination an example like;
4x-3y=14
5x-4y=18

### Comments for Solve Equations Using Elimination

 Apr 05, 2011 Solve Equations Using Elimination by: Staff The question: how do you solve by using elimination an example like; 4x-3y=14 5x-4y=18 The answer: Using the elimination method to solve simultaneous equations is also called solving by addition. The basic idea is to add the two equations together in order to eliminate one of the variables. 4x - 3y = 14 5x - 4y = 18 But first, multiply each side of the second equation by -¾. This will change the second term in the equation to +3y. 5x - 4y = 18 (-¾)*(5x - 4y) = (-¾)*18 -15x/4 + 3y = -54/4 -15x/4 + 3y = -27/2 The original two equations can now be written as: 4x - 3y = 14 -15x/4 + 3y = -27/2 Add the two equations together. This will eliminate the y variable. 4x - 3y = 14 + (-15x/4 + 3y = -27/2) -------------------------------- 4x - 15x/4 - 3y + 3y = 14 - 27/2 (4x - 15x/4) + (- 3y + 3y) = 14 - 27/2 (4x - 15x/4) + 0 = 14 - 27/2 The y variable has been eliminated. Solve for x 16x/4 - 15x/4 = 14 - 27/2 x/4 = 14 - 27/2 x/4 = 28/2 - 27/2 x/4 = 1/2 4 * (x/4) = 4 * (1/2) x * (4/4) = 4 * (1/2) x * 1 = 4 * (1/2) x = 4 * (1/2) x = 4/2 x = 2 Now that the value of x is known, substitute 2 for x in equation 1 and solve for y. 4x - 3y = 14 4 * 2 - 3y = 14 8 - 3y = 14 8 - 3y + 3y = 14 + 3y 8 + 0 = 14 + 3y 8 = 14 + 3y 8 - 14 = 14 – 14 + 3y 8 - 14 = 0 + 3y 8 - 14 = 3y -6 = 3y -6/3 = 3y/3 -6/3 = y * (3/3) -6/3 = y * 1 -6/3 = y -2 = y The final answer is: x = 2 and y = -2 Check the work. Substitute 2 for x and -2 for y in the original two equations. 4x - 3y = 14 4*2 - 3*(-2) = 14 8 + 6 = 14 14 = 14, correct 5x - 4y = 18 5*2 - 4*(-2) = 18 10 + 8 = 18 18 = 18, correct The values of x = 2 and y = -2 are valid solutions in both of the original equations. This verifies that these values are correct. Thanks for writing. Staff www.solving-math-problems.com

Join in and write your own page! It's easy to do. How? Simply click here to return to Math Questions & Comments - 01.