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Solving Systems of Equations: Graphing and Substitution Methods

Learn how to solve systems of equations using graphing and substitution methods step by step. Practice solving equations and find the intersection points. Explore elimination (linear combination) method with examples.

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Solving Systems of Equations: Graphing and Substitution Methods

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  1. –11 ANSWER (1, 1) ANSWER Warm Up #4 1.Evaluate –3x – 5yfor x = –3 and y = 4. 2.Solve the system by graphing. x + y = 2 2x + y = 3

  2. Solving a system of equations using Substitution Step 1: Get one of the 4 variables by itself on one side of the equals sign Step 2: Substitute its value into the other equation and solve Step 3: Substitute the answer back into the equation with the other variable by itself and solve Step 1: Already done Solve: 3y - 2x = 11y = 9 - 2x Step 2: 3(9 – 2x) – 2x = 11 27 – 6x – 2x = 11 Step 3: y = 9 – 2(2) 27 – 8x = 11 - 27 y = 9 – 4 -8x = -16 y = 5 -8 These 2 lines intersect at the point ( 2 , 5 ) x = 2

  3. Step 1: Get one of the 4 variables by itself on one side of the equals sign Step 2: Substitute its value into the other equation and solve Step 3: Substitute the answer back into the equation with the other variable by itself and solve Step 1: Rewrite the first equation as y = 3x + 5 Solve: y – 3x = 5y + x = 3 Step 2: (3x + 5) + x = 3 4x + 5 = 3 - 5 Step 3: y = 3(-½) + 5 4x = -2 4 y = -3/2 + 5 x = -½ y = 3½ These 2 lines intersect at the point ( -½ , 3½ )

  4. Solve Equation 2 for x. STEP 1 EXAMPLE 3 Use the substitution method Solve the system using the substitution method. 2x + 5y = –5 x + 3y = 3 x = –3y + 3 2x+5y = –5 Write Equation 1. 2(–3y + 3) + 5y = –5 Substitute –3y + 3 for x. y = 11 Solve for y. x = –3y+ 3 Solution (-30 , 11) Write revised Equation 2. x = –3(11) + 3 Substitute 11 for y. x = –30 Simplify.

  5. Elimination (Linear Combination) The goal when using Elimination is to add the two equations together to Eliminate one of the variables Ex. 2x – y = 8 -5x + y = 4 You need to have the coefficients of either the x or the y add up to zero (opposites) -3x = 12 Notice the coefficients of the y terms -3 -3 They add up to zero, so we can add the two equations together x = -4 -5(-4) + y = 4 Then solve for the other variable 20 + y = 4 - 20 y = -16 Then substitute it in either equation to find the other variable Solution (-4 , -16)

  6. What if they don’t add up to zero to start with? You can multiply either or both of the equations by something to get them to be opposites 3x + 6y = -6 5x - 2y = 14 3( ) Look at the y coefficients in this problem 3x + 6y = -6 15x - 6y = 42 We can multiply the bottom equation by 3 to make them opposites 18x = 36 18 18 Now add the equations together and solve x = 2 Substitute in one of the equations 3(2) + 6y = -6 6 + 6y = -6 - 6 Solution (2 , -2) 6y = -12 y = -2

  7. Needing to multiply both equations 2( ) 3x + 4y = -25 2x - 3y = 6 We can eliminate either the x or the y terms. Let’s eliminate the x this time. If they were 6x and -6x they would add to zero. -3( ) 6x + 8y = -50 -6x + 9y = -18 Now solve as before 17y = -68 17 17 y = -4 2x – 3(-4) = 6 Solution (-3 , -4) 2x + 12 = 6 - 12 2x = -6 x = -3

  8. Special cases Multiply top equation by -3 -3( ) x + 4y = 1 3x + 12y = 3 -3x - 12y = -3 3x + 12y = 3 0 = 0 Since 0 always equal 0 then this system has infinite solutions.

  9. -6x - 12y = -3 3x + 6y = 4 Multiply the bottom equation by 2 2( ) -6x - 12y = -3 6x + 12y = 8 0 = 5 Since 0 can never equal 5, there are no solutions.

  10. Classwork Assignment WS 3.2 (1-23 odd)

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