How do you graph and solve #|3x+1| + |3-2x| =11#?

Question

Avery Alexander · Accepted Answer

#x = -9/5# and #x = 13/5#

#abs(3x+1)+abs(2x-3)=11=3x+1-3x+10# so
#abs(3x+1)/(3x+1)+abs(2x-3)/(3x+1)=1+(10-3x)/(3x+1)# or
#pm1+abs(2x-3)/(3x+1)=1+(10-3x)/(3x+1)#
Now we have two options:

a)
#1+abs(2x-3)/(3x+1)=1+(10-3x)/(3x+1)->abs(2x-3)/(3x+1)=(10-3x)/(3x+1)#

b)
#-1+abs(2x-3)/(3x+1)=1+(10-3x)/(3x+1)->abs(2x-3)/(3x+1)=2+(10-3x)/(3x+1)#

Following with a) we have

a-1)
#abs(2x-3)=10-3x=2x-3-5x+13# or
#abs(2x-3)/(2x-3)=1+(13-5x)/(2x-3)# or
#pm1=1+(13-5x)/(2x-3)# with two options:

a-1-1)
#1=1+(13-5x)/(2x-3)->13-5x=0->color(red)(x=13/5)#

a-1-2)
#-1=1+(13-5x)/(2x-3)->0=2+(13-5x)/(2x-3)->color(red)(x=7)#

Now following with b)

#abs(2x-3)/(3x+1)=2+(10-3x)/(3x+1)->abs(2x-3)=6x+2+10-3x# so
#abs(2x-3)=3x+12# then

b-1)
#abs(2x-3)=2x-3+x+15->abs(2x-3)/(2x-3)=1+(x+15)/(2x-3)# or
#pm1=1+(x+15)/(2x-3)# with two options

b-1-1)
#1=1+(x+15)/(2x-3)->color(red)(x=-15)#

b-1-2)
#-1=1+(x+15)/(2x-3)->0=2+(x+15)/(2x-3)->color(red)(x=-9/5)#

After checking the found values, we pick the feasible values #x = -9/5# and #x = 13/5#

Savannah Anderson · Answer

undefined To graph and solve the equation |3x+1| + |3-2x| = 11:

1. Identify critical points by setting each absolute value expression equal to zero and solving for x.
2. Divide the number line into intervals using these critical points.
3. Test a value within each interval in the original equation to determine the sign of the expression.
4. Graph the solution set on the number line.
5. Solve for x in each interval to find the specific solutions.