Answer :
1.
[tex]a|x+b|+c=d \ \ \ |-c \\ a|x+b|=d-c \\ \\ c=d \hbox{ so } d-c=0 \\ \\ a|x+b|=0 \ \ \ |\div a, \ a>0 \\ |x+b|=\frac{0}{a} \\ |x+b|=0 \\ \\ \hbox{if } |x|=0, \hbox{ then } x=0 \\ \\ x+b=0 \\ x=-b[/tex]
If a>0 and c=d, the equation has one solution.
2.
[tex]a|x+b|+c=d \ \ \ |-c \\ a|x+b|=d-c \ \ \ |\div a, a <0 \\ |x+b|=\frac{d-c}{a} \\ \\ c>d \hbox{ so } d-c<0 \\ a<0 \\ \hbox{a negative number divided by a negative number is positive number so} \\ \frac{d-c}{a} > 0 \\ \hbox{if } |x|=a, \ a>0, \hbox{ then } x=a \ \lor \ x=-a \\ \\ |x+b|=\frac{d-c}{a} \\ x+b=\frac{d-c}{a} \ \lor \ x+b=-\frac{d-c}{a} \\ x=\frac{d-c}{a}-b \ \lor \ x=-\frac{d-c}{a}-b[/tex]
If a<0 and c>d, the equation has two solutions.
[tex]a|x+b|+c=d \ \ \ |-c \\ a|x+b|=d-c \\ \\ c=d \hbox{ so } d-c=0 \\ \\ a|x+b|=0 \ \ \ |\div a, \ a>0 \\ |x+b|=\frac{0}{a} \\ |x+b|=0 \\ \\ \hbox{if } |x|=0, \hbox{ then } x=0 \\ \\ x+b=0 \\ x=-b[/tex]
If a>0 and c=d, the equation has one solution.
2.
[tex]a|x+b|+c=d \ \ \ |-c \\ a|x+b|=d-c \ \ \ |\div a, a <0 \\ |x+b|=\frac{d-c}{a} \\ \\ c>d \hbox{ so } d-c<0 \\ a<0 \\ \hbox{a negative number divided by a negative number is positive number so} \\ \frac{d-c}{a} > 0 \\ \hbox{if } |x|=a, \ a>0, \hbox{ then } x=a \ \lor \ x=-a \\ \\ |x+b|=\frac{d-c}{a} \\ x+b=\frac{d-c}{a} \ \lor \ x+b=-\frac{d-c}{a} \\ x=\frac{d-c}{a}-b \ \lor \ x=-\frac{d-c}{a}-b[/tex]
If a<0 and c>d, the equation has two solutions.
