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solving a system of eqauations by substitution

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2 Answers

Since both equations are already solved for y, it should make this a little easier...

y = y                                               Reflexive property

2x2 - 11x + 6 = 3x -5                         Substitution

2x2 - 11x + 6 - 3x + 5 = 0                  Subtract 3x from each side and add 5 to each side

2x2 - 14x + 11 = 0                             Simplify (-11x - 3x = -14x, 6 + 5 = 11)

x = (14 +/-√(142 - 4(2)(11)))/(2*2)    Substitution into quadratic formula

x = (14 +/-√(196 - 88))/4                  Simplify (142 = 196, 4*2*11=88, 2*2=4)

x = (14 +/-√(108))/4                         Simplify (196 - 88 = 108)

x = (14 +/-6√3)/4                             Factor out 36 from 108 and bring the 62 out of the radical

x = (7 +/-3√3)/2                              Reduce the fraction

y = 3 * (7 +3√3)/2 - 5                      Substitute one of the values of x into an original equation

y = (21 +9√3)/2 - 5                          Distribute the 3

y = (21 +9√3)/2 - 10/2                     Find the common denominator

y = (11 +9√3)/2                               Simplify (21/2 - 10/2 = 11/2)

y = 3 * (7 -3√3)/2 - 5                       Substitute the other value of x into an original equation

y = (21 -9√3)/2 - 5                           Distribute the 3

y = (21 -9√3)/2 - 10/2                      Find the common denominator

y = (11 -9√3)/2                                Simplify (21/2 - 10/2 = 11/2)

 

The answers are ([7 + 3√3]/2, [11 + 9√3]/2) and ([7 - 3√3]/2, [11 - 9√3]/2)

 

Substitute y = 3x-5,

2x^2-11x+6 = 3x-5

Collect variables in one side,

2x^2 - 14x = -11

/2, x^2 - 7x = -11/2

Add (7/2)^7 to both sides, and complete the square,

(x - 7/2)^2 = 27/4

x = 7/2 +/- (3/2)sqrt(3)

y = 3x-5 = 11/2 +/- (9/2)sqrt(3)

Answer: [(7/2 + (3/2)sqrt(3), 11/2 + (9/2)sqrt(3)]; [(7/2 - (3/2)sqrt(3), 11/2 - (9/2)sqrt(3)]

 

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