How do you solve #\frac { 5- b } { 3} = \frac { 2b + 2} { - 4}#?

3 Answers
May 2, 2018

First, let's clear the denominators:

#-4 xx (5-b) = (2b+2) xx 3#

#-20 + 4b = 6b + 6#

#4b = 6b = 26#

#-2b = 26#

#b =-13#

Let's check our work:

#(5+13)/3 = 6#

#(2 xx -13 + 2)/(-4) = 6 #

Yep! #b = -13#

May 2, 2018

Answer:

#b=-13#

Explanation:

There is ONE fraction on each side of the equal sign, so you may cross-multiply.

#color(blue)((5-b))/color(red)(3) = color(red)((2b+2))/color(blue)(-4)#

#color(red)(6b +6)=color(blue)(-20+4b)#

#6b-4b=-20-6#

#2b = -26#

#b =-13#

May 2, 2018

Answer:

multiply both sides by each fraction's denominator, then solve for #b#, which would give you #b=-13#

Explanation:

What we will do is multiply both sides by the denominator of each side, thereby eliminating the need for fractions.

Starting with the left hand side:

#(5-b)/cancel(3)cancel(color(red)(xx3))=(2b+2)/(-4)color(red)(xx3)#

#5-b=((2b+2)color(red)(xx3))/(-4)#

#5-b=(6b+6)/(-4)#

Now, we multiply both sides by the denominator of the right-hand side:

#(5-b)color(red)(xx-4)=(6b+6)/cancel(-4)cancel(color(red)(xx-4))#

#-20+4b=6b+6#

Now, we'll re-arrange the equation so all of the terms related to #b# are on one side, and the constants are on the other:

#-20+cancel(4b)color(red)(-cancel(4b))=6b+6color(red)(-4b)#

#-20=2b+6#

#-20color(red)(-6)=2b+cancel(6)color(red)(-cancel(6))#

#-26=2b#

Finally, divide both sides by the constant #b# is multiplied by:

#-26/color(red)(2)=(cancel(2)b)/cancel(color(red)(2))#

#color(green)(b=-13)#