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Published byLorena Paulina Robbins Modified over 9 years ago
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Solve using Calculator Reset your calculator 2 nd, +, 7, 1, 2 Practice A solve by graphing (make sure y is by itself for both equations Enter both in Y1=2x – 1 Y2=3x + 2 Hit graph, adjust window if necessary 2 nd Trace 5 enter enter enter The x and y coordinates at bottom screen 1. (-3, -7)2. (4, 2)3. (2, 7)4. (-3, -7)
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Systems of equations 2 or more equations on the same graph The solution to a system of equations The numbers that make both equations true The coordinate pair where the graphs of the lines intersect
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When do you solve by graphing? When both equations have y by itself When one equation has y by itself and the other can be made that way easily. For example: 1. y = 2x + 4 2. y = -3x +2 3. 2x + 3y = 12 y = 4x – 5 y – 4x = 8 4x + 2y = -6 Yes No
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Solving systems by elimination Is an algebraic method Best done when equations have both variables on the same side, standard form Move the variables around so same variables are stacked, if necessary You may choose to multiply an equation by a number so you can add them
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Recognizing when to add or Multiply by -1, then add Add when coeffecients same, different signs Multiply one equation by -1, then add when Coeffecients same, same signs 5r – 5s = 12 2r – 5s = 6 3r + 5s = 12 2r – 5s = 6 3r – 10s = 2 3r – 5s = 6 add Mult -1, then add 5r – 5s = 12 2r – 5s = 6 3r + 5s = 12 2r – 5s = 6 3r – 10s = 2 3r – 5s = 6 3r – 10s = 2 -3r – 5s = 6 3r – 10s = 2 -3r – 5s = 6
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13x + 5y = -11 13x +11y = 7 -13x – 5y = 11 6y = 18 y = 3 13x + 5(3) = -11 13x + 15 = -11 13x = -26 solution: x = -2 and y = 3 x = -2 ( -2, 3) 1.Eliminate a variable by adding. Add same coefficient and different signs 2. Solve for the variable remaining 3. Substitute in one of the original equations and solve - ( )
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1. Eliminate a variable by adding. Add same coefficient and different signs Sub same coefficient and same signs 2. Solve for the variable remaining 3. Substitute in one of the original equations and solve 6a + 5b = 1 6a – 5b = 11 12a = 12 a = 1 6(1) + 5b = 1 6 + 5b = 1 5b = -5 Solution: (1, -1) is the point where the lines will intersect **Finish the rest of practice B (+)
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What do you do if adding will not eliminate a variable? Coeffecients are not the same Need to change one or both equations by muliplying by a number we choose.
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Equations stay the same value as long as we do the same thing to both sides What are some values that make it true? Put in y =, 2 nd graph y = 2x + 4 (1, 6) (2, 8) (3, 10) 3( ) Now take that equation and multiply everything by 3 y = 2x + 4 3y = 6x + 12 Are the solutions the same? 3y=6x+12 3(6) = 6(1) +12 3(8) = 6(2) + 12 3(10)=6(3) + 12 Yes!!
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What if the coefficients are not the same? 4x – 3y =12 x + 2y = 14 -4x – 8y = -56 4x – 3y =12 Solution: (6,4) 1. Eliminate a variable by adding. Add, same coefficient and different signs Sub, same coefficient and same signs Mult, different coefficients 2. Solve for the variable remaining 3. Substitute in one of the original equations and solve -4( ) c1
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What would you do? 1. 3x – 2y = 11 -x + 6y = 7 2. -2x + 3y = -3 3x + 6y = 15 3 ( ) 3x – 2y = 11 (+)-3x + 18y = 21 -2 ( ) 4x - 6y = 6 (+) 3x + 6y = 15 Solution: (, )Solution: (, 2 )Solution: ( 5, 2 ) These lines intersect at (5,2) Solution: (, )Solution: ( 3, ) Solution: ( 3, 1 ) These lines intersect at (3,1)
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What would you do in this situation? 3x +4y = 37 x = 3 3(3) + 4y = 37 9 + 4y = 37 4y = 28 y = 4 We already know what x equals, all we have to do is substitute it in These two lines intersect at the point (3, 4)
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What would you do in this situation? 2x +3y = 34 y = 5x 2x + 3(5x)= 34 2x + 15y = 34 17x = 34 x = 2 y = 5(2) y = 10 We already know what y equals, all we have to do is substitute it in These two lines intersect at the point (2, 10)
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What would you do in this situation? 2x +3y = 1 y = 3x + 15 2x + 3(3x + 15)= 1 2x + 9x + 45 = 1 11x = -44 x = -4 y = 3(-4)+ 15 y = 3 We already know what y equals, all we have to do is substitute it in These two lines intersect at the point (-4, 3)
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