Solve each of the absolute value equations algebraically.
|(x/2)+3|=3x-3
Now the work isn't an issue, I did all that & I have my answer. My problem is, I double checked my answer on mathways & it said only my first answer: x=(12/5) but I also got x=0 & when double checked that checks out... Is there a reason that 0 can't be a suitable answer?
Thanks a bunch! :)
Well, first of all let's look at
|(x/2)+3|=3x-3 , and just follow the definition of |....|, it has to be zero or positive, so
3x-3 ≥ 0
x ≥ 1
I will trust your algebra, and that you got x = 0
but x≥ 1, so x cannot be zero
you could have also checked that answer:
if x = 0
LS = |0/2+3| = 3
RS = 3(0) - 3 = -3
so it did not work!!
Thank you so much, that makes total sense, I wasn't thinking about that & now I am!
To solve the absolute value equation algebraically, you correctly started by isolating the expression inside the absolute value by subtracting 3x-3 from both sides:
|(x/2) + 3| = 3x - 3
(x/2) + 3 = 3x - 3 or (x/2) + 3 = -(3x - 3)
Next, you can solve each equation separately.
For the first equation:
(x/2) + 3 = 3x - 3
Simplify this equation:
x/2 + 3 = 3x - 3
x/2 - 3x = -3 - 3
x - 6x = -6
-5x = -6
x = -6 / -5
x = 6/5
So, x = 6/5 or x = 1.2.
For the second equation:
(x/2) + 3 = -(3x - 3)
Simplify this equation:
x/2 + 3 = -3x + 3
x/2 + 3x = 3 - 3
x + 6x = 6
7x = 6
x = 6 / 7
So, x = 6/7 or approximately 0.857.
Now, let's address your concern about the solution x = 0. You are correct in including this as a possible solution. When solving an absolute value equation, it is essential to consider all possible cases.
In the original equation, |(x/2) + 3| = 3x - 3, if we substitute x = 0, we have:
|(0/2) + 3| = 3(0) - 3
|3| = -3
Since the absolute value of any number is always positive or zero, it cannot be equal to a negative value like -3. Hence, x = 0 is not a suitable solution in this case.
Therefore, the only solution is x = 6/5 or x = 1.2.
It's always essential to double-check solutions by substituting them back into the original equation to ensure they satisfy all conditions.