solve this system of equations by substitution
2x^2-11x+6=y
y=3x-5
solve this system of equations by substitution
2x^2-11x+6=y
y=3x-5
Since both equations are already solved for y, it should make this a little easier...
y = y Reflexive property
2x^{2} - 11x + 6 = 3x -5 Substitution
2x^{2} - 11x + 6 - 3x + 5 = 0 Subtract 3x from each side and add 5 to each side
2x^{2} - 14x + 11 = 0 Simplify (-11x - 3x = -14x, 6 + 5 = 11)
x = (14 +/-√(14^{2} - 4(2)(11)))/(2*2) Substitution into quadratic formula
x = (14 +/-√(196 - 88))/4 Simplify (14^{2} = 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 6^{2} 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)]