Solving Systems of Linear Equations By Elimination Unit 2 Lesson 3 Text Topic 3-2 pp.77-82

Warm-Up (SAT Question) 𝑥 2 + 𝑦 2 =153 𝑦=−4𝑥 If (𝑥, 𝑦) is a solution to the system of equations above, what is the value of 𝑥 2 ? Hint: Use substitution to find x first. The second equation gives y in terms of x, so a student can use this to rewrite the first equation in terms of x. Substituting -4x for y in the equation x^2 + y^2 = 153 gives x^2 + (-4x)^2 = 153. This can be simplified to x^2 + 16x^2 =153 or 17x^2 = 153. Since the question asks for the value of x^2 not x, dividing both sides of 17x^2 =153 by 17 gives the answer: x^2 = 153/17 = 9.

Essential Question How is solving by elimination similar to solving by substitution?

What is Elimination? To eliminate means to get rid of or remove. You solve equations by eliminating one of the variables (x or y) using the addition and or subtraction of equivalent equations.

Steps for Solving by Elimination *Line up like terms vertically between the two equations before starting (STANDARD FORM). 1. Choose a variable to eliminate. 2. Eliminate that variable by adding or subtracting one equation from the other. (Sometimes you have to multiply first.) TIP: Use the coefficients of the equations as your multipliers, if necessary. 3. Solve the new equation. 4. Plug in your answer to find the other variable (or do elimination for the other variable). 5. Check your answer & write it as an ordered pair.

Example 1 Solve the following system of linear equations by elimination. (1) (2) 2x – 3y = 15 5x + 3y = 27 Add equation (1) to equation (2) 7x + 0y = 42  7x = 42  By eliminating y, we can now solve for x x = 6

Example 1 Substitute x= 6 into equation (1) to solve for y Check your solution x = 6 and y = -1 in equation (2) 2x – 3y = 15 5x + 3y = 27 2(6) – 3y = 15 5(6) + 3(-1) = 27 30 – 3 = 27 12 – 3y = 15 27 = 27 – 3y = 15 – 12 – 3y = 3 y = -1 Therefore, the solution set = {(6,-1)}

Example 2 (1) 5x + 4y = -28 (2) 3x + 10y = -13  If we were to add these equations we would obtain 8x + 14y = -41  Even though we have only one equation now, we still have 2 variables.  We need to multiply the equations by values that will allow us to eliminate either x or y. (Hint: use the coefficients)

Example 2 5x + 4y = -28 (1) (2) 3x + 10y = -13  If we multiply equation (1) by 5 and equation (2) by -2, we be able to eliminate y using a 20 and -20. (1) x 5 (2) x -2 25x + 20y = -140 -6x – 20y = 26 (3) (4)  When you change the equations you need to renumber them. Add (3) & (4)  19x = -114 x = -6

Example 2 Substitute x = -6 into equation (1) 5x + 4y = -28 Check your answer x = -6 and y = ½ into equation (2) 5(-6) + 4y = -28 -30 + 4y = -28 3(-6) + 10(½) = -13 4y = -28 +30 -18 + 5 = -13 4y = 2 -13 = -13 Therefore, the solution set = {(-6, ½)}

Assignment 19. 20. 21. 22. 23. 24. 25. 26. 27. TB pg. 81 (19-27)