How do you solve (2x - 7) ( x + 1) = 6x - 19?

2 Answers
Nov 23, 2016

x=3/2 or 4

Explanation:

Multiply out the Left hand side of the equation
2x^2+2x-7x-7=2x^2-5x-7
Therefore 2x^2-5x-7=6x-19
Collecting like terms on the left
2x^2-11x+12=0
Factorise
(2x-3)(x-4)=0
x=3/2 or x=4

Nov 24, 2016

x=4, 3/2

Explanation:

NOTE: This is a long answer.

Solve (2x-7)(x+1)=6x-19

FOIL the left-hand side.
http://www.mesacc.edu/~scotz47781/mat120/notes/polynomials/foil_method/foil_method.htmlhttp://www.mesacc.edu/~scotz47781/mat120/notes/polynomials/foil_method/foil_method.html

(2x-7)(x+1)=2x*x+2x*1+(-7)*x+(-7*1)

(2x-7)(x+1)=2x^2+2x-7x-7

(2x-7)(x+1)=2x^2-5x-7

Put the equation back together.

2x^2-5x-7=6x-19

Subtract 5x from both sides.

2x^2-6x-5x-7=-19

x^2-11x-7=-19

Add 19 to both sides.

2x^2-11x-7+19=0

2x^2-11x+12=0

This is a quadratic equation, ax+bx+c, where a=2, b=-11, and c=12.

Use the quadratic formula to solve for x.

Quadratic Formula

x=(-b+-sqrt(b^2-4ac))/(2a)

x=(-(-11)+-sqrt(-11^2-4*2*12))/(2*2)

x=(11+-sqrt(121-96))/4

x=(11+-sqrt25)/2

x=(11+-5)/4

x=16/4=4

x=6/4=3/2

x=4, 3/2