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

2 Answers
Nov 23, 2016

Answer:

#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

Answer:

#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.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#