How do you solve #abs(7-2x)=5#?

2 Answers
Aug 26, 2017

Answer:

#x= 1 and 6#

Explanation:

The use of the special brackets of | | means that whatever is inside them is considered as positive.

On the right side of the equals we have +5.

So the left side must end up as #|+-5|# giving:

#|+-5|=+5#

Ok!

Lets consider what will turn #7-2x# into -5

Set #color(white)("d")7-2x=-5#

#2x=7+5#

#x=12/2=+6#

Check #(color(white)(2/2)7-[2xx6]color(white)("d") ) -> 7-12 = -5# as required

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Lets consider what will turn #7-2x# into +5

Set #color(white)("d")7-2x=+5#

#2x=7-5#

#x=2/2=1#

Check #(color(white)(2/2)7-[2xx1]color(white)("d")) -> 5xx1=+5# as required

Aug 26, 2017

Answer:

See a solution process below:

Explanation:

The absolute value function takes any term and transforms it to its non-negative form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.

Solution 1

#7 - 2x = -5#

#-color(red)(7) + 7 - 2x = -color(red)(7) - 5#

#0 - 2x = -12#

#-2x = -12#

#(-2x)/color(red)(-2) = (-12)/color(red)(-2)#

#(color(red)(cancel(color(black)(-2)))x)/cancel(color(red)(-2)) = 6#

#x = 6#

Solution 2

#7 - 2x = 5#

#-color(red)(7) + 7 - 2x = -color(red)(7) + 5#

#0 - 2x = -2#

#-2x = -2#

#(-2x)/color(red)(-2) = (-2)/color(red)(-2)#

#(color(red)(cancel(color(black)(-2)))x)/cancel(color(red)(-2)) = 1#

#x = 1#

The Solutions Are: #x = 6# and #x = 1#