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1. Evaluate –3 x – 5 y for x = –3 and y = 4 .

–11. ANSWER. (1, 1). ANSWER. Warm Up #4. 1. Evaluate –3 x – 5 y for x = –3 and y = 4. 2. Solve the system by graphing. x + y = 2 2 x + y = 3. Solving a system of equations using Substitution. Step 1: Get one of the 4 variables by itself on one side of the equals sign

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1. Evaluate –3 x – 5 y for x = –3 and y = 4 .

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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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