# Solve this modulus equation

Gold Member
Homework Statement:
##|2x+3|-x=1##
Relevant Equations:
modulus
##|2x+3|-x=1##
i am getting ##x=-2## and ##x=\frac {-4}{3}## of which none satisfies the original equation, therefore we do not have a solution, right?

Mentor
2022 Award
What do you mean by getting? If there are no solutions, how can you get values for ##x?## You could draw the graph for ##y=|2x+3|## and the graph for ##y=1+x## and see if they intersect or not.

Mentor
Homework Statement:: ##|2x+3|-x=1##
Relevant Equations:: modulus

##|2x+3|-x=1##
i am getting ##x=-2## and ##x=\frac {-4}{3}## of which none satisfies the original equation, therefore we do not have a solution, right?
The equation is equivalent to ##|2x + 3| = x + 1##
Because of the absolute value on the left, it must be true that ##x + 1 \ge 0##, or ##x \ge -1##. Neither of the solutions you found satisfies this additional requirement, so there is no solution.

Gold Member there...i just confirmed with problem owner, it was a typo on textbook part. No solution exists. cheers

Gold Member
What do you mean by getting? If there are no solutions, how can you get values for ##x?## You could draw the graph for ##y=|2x+3|## and the graph for ##y=1+x## and see if they intersect or not.

when we solve modulus equations, we first try getting values for ##x## right? then proceed on checking if they satisfy the equation...

Homework Helper
Gold Member
2022 Award
when we solve modulus equations, we first try getting values for ##x## right?
No, but you can get candidate values by ignoring some constraints (which should be specified) and check which resulting solutions satisfy them.
In the present case, you relaxed the given condition ##|2x+3|-x=1## to be ##±(2x+3)-x=1##.

Gold Member
No, but you can get candidate values by ignoring some constraints (which should be specified) and check which resulting solutions satisfy them.
In the present case, you relaxed the given condition ##|2x+3|-x=1## to be ##±(2x+3)-x=1##.
Either case, by considering those constraints, you will end up with the values that I found...