 # Section 3-2: Solving Systems Algebraically (Pg.125) By Ms. Beydoun.

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Section 3-2: Solving Systems Algebraically (Pg.125) By Ms. Beydoun

Section 3-2: Solving Systems Algebraically (Pg.125) To solve a system by substitution and elimination Not every system can be solved easily by graphing Substitution allows you to find exact solutions without using graphing

EX. 1. Solve by Substitution. Step 1: 2X – y = 7 ---> y = 2x – 7 Step 2: Substitute the expression for y into other equation. Then, solve for x. 4x + 3(2x – 7) = 4 4x + 6x – 21 = 4 10x – 21 = 4 10x = 25 x = 2.5 Step 3: Substitute value of x into either equation. Then, solve for y. Y = 2(2.5) - 7 y=5-7 y=-2 The solution is (2.5, -2). 4x + 3y = 4 2x – y = 7

 You can solve a system of equations using the Additive Property of Equality. If the quantities you add contain a pair of additive inverses, you can eliminate a variable. You can also eliminate a variable by subtracting like terms. EX. 2. Solving by Elimination. 4x – 2y = 7 x + 2y = 3 Step 1: Two terms are additive inverses; so add. 4x – 2y = 7 + x + 2y = 3 ________________ 5x + 0 = 10 5x = 10 x = 2 Step 2Choose one of the original equations then substitute for x. x + 2y = 3 2 + 2y = 3 2y = 1 y= 1/2 The solution is (2, 1/2).

Solving an Equivalent System To make 2 terms additive inverses, you may need to multiply one or both equations in a system by a nonzero number. In doing so, you create a system equivalent to the original one. Equivalent Systems are systems that have same solution(s). EX. Solving an Equivalent System. Solve system below by Elimination. 3x + 7y = 15 5x + 2y = -4 To eliminate the y terms, make them additive inverses by multiplying. Step 1: ( 3x + 7y = 15)*2  6x +14y = 30 Step 2: (5x + 2y = -4)*-7  -35x -14y = 28 6x +14y = 30 + -35x -14y = 28 ________________ -29x + 0 = 58 x = -2 Choose an original equation and solve for y. 3x + 7y = 15 3(-2) + 7y = 15 -6 + 7y = 15 7y = 21 y = 3 The solution is (-2, 3).

Solving a system algebraically does not always result in a unique solution as in EX. 3 & 4 on Pg. 126-127. You may get an equation that’s always true or one that’s never true. Ex. Solving a system without a unique solution. 2x – y = 3 -2x + y = -3 2x – y = 3 + -2x + y = -3 ________________ 0 = 0 True Statement! Infinite Number of Solutions

Ex. Solving a system without a unique solution. 2x – 3y = 18 -2x + 3y = -6 2x – 3y = 18 + -2x + 3y = -6 _____________ 0 = 12 False Statement! No Solutions

Practice Solve each system by Substitution. 2x – 3y = 6 x + y = -12 3x – y = 0 4x + 3y = 26

Practice Solve each system by Elimination. X + y = 12 X – y = 2 2w + 5y = -24 3w – 5y = 14

Exit Slip Solve each system by Substitution or Elimination -3X = 5Y = 7 6X – 10Y = -14 -2x + 4y = 6 -3x + 6y = 8 H.W. Pg. 128 – 129 #’s 1 – 11, 19 – 41, 45 – 59 (ODDs)