How do you solve #4x-9=6x+19#?

1 Answer
Nov 20, 2015

Method in details showing the basis of manipulation shortcuts.

#color(green)(x=-14)#

Explanation:

#color(green)(Given: color(white)(xx) 4x-9=6x+19)#

Objective: Have a single #x# on one side of the = and everything else on the other.

As #6x# is bigger than #4x# I am choosing to move the left #4x# to the right.

#color(blue)(Step 1)#

Subtract #color(blue)(4x)# from both sides; this will bring all the x-terms together

#color(brown)((4x-9) color(blue)(-4x) = color(brown)((6x+19) color(blue)(-4x)#
The purpose of the brackets is to show you what is being changed. They serve no other purpose than that or of grouping to make things clearer.

#color(brown)((4xcolor(blue)(-4x)) -9 =(6xcolor(blue)(-4x))+19)#

#0 -9 =2x+19#

#-9=2x+19#

~~~~~~~~~~This process explains the short cut ~~~~~~~~~~~~~~~~
By subtracting #4x# from both sides I have changed the one on the left to the value of 0. The consequence of this is that there is now a #4x# on the other side of the = but its sign has changed.

For addition and subtraction
#color(brown)("The shortcut is:" ) color(white)(x)color(blue)("move it to the other side of = and change its sign from + to -")#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)(Step 2)#

Subtract #color(blue)(19)# from both sides; this will isolate the x-terms.

#color(brown)((-9 )color(blue)( -19) = (2x+19))color(blue)(-19)#

#-28 = 2x +0#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)(Step 3)#
-
Divide both sides by 2 ( #divide 2 " is the same as" times 1/2#)

#color(brown)((-28))/(color(blue)(2)) = color(brown)((2x))/(color(blue)(2))#

#-14 = 2/2 x#

But #2/2 =1# giving

#-14=x#

Convention is that the #x# be written on the left so we have

#color(green)(x=-14)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

For multiplication and division
#color(brown)("The shortcut is:" ) color(white)(x)color(blue)("move it to the other side of = and multiply by its inverse")#