9.
solve |4x + 3| = 9 + 2x and check for extraneous solutions.
the first answer I got was:
|4x +3| = 9 + 2x
|4x| = 6 + 2x
2x = 6
x = 3
can someone help me find the other answer? thank you!
hello emily,remember the properties of modulus
|4x + 3| = 9 + 2x
case1 4x+3=9+2x =>x=3
case2 -4x-3=9+2x =>x= -2
To find the other answer, let's consider the case where the absolute value is negative:
|4x + 3| = -(9 + 2x)
When the absolute value is negative, it means that the expression inside the absolute value is negative as well. So we can write:
4x + 3 = -(9 + 2x)
Now, let's solve for x:
4x + 3 = -(9 + 2x)
4x + 3 = -9 - 2x
To continue, we can simplify the equation by combining like terms:
4x + 2x + 3 = -9
6x + 3 = -9
Next, isolate the variable by subtracting 3 from both sides:
6x + 3 - 3 = -9 - 3
6x = -12
To solve for x, divide both sides by 6:
(6x)/6 = (-12)/6
x = -2
So the other solution is x = -2.
To check for extraneous solutions, we substitute both solutions into the original equation:
For x = 3:
|4x + 3| = 9 + 2x
|4(3) + 3| = 9 + 2(3)
|12 + 3| = 9 + 6
|15| = 15
15 = 15 (true)
For x = -2:
|4x + 3| = 9 + 2x
|4(-2) + 3| = 9 + 2(-2)
|-8 + 3| = 9 - 4
|-5| = 5
5 = 5 (true)
Both solutions satisfy the original equation, so there are no extraneous solutions. The solutions are x = 3 and x = -2.
To find the other solution, we need to consider the case when the absolute value is negative. Remember that |a| = -a only when a is negative.
Given the equation |4x + 3| = 9 + 2x, let's consider the case when 4x + 3 is negative:
When 4x + 3 < 0, we apply the property that |a| = -a. This gives us:
-(4x + 3) = 9 + 2x
Simplifying this, we get:
-4x - 3 = 9 + 2x
Now, let's solve for x:
-4x - 3 + 4x = 9 + 2x + 4x
-3 = 9 + 6x
-3 - 9 = 6x
-12 = 6x
x = -2
So, the second solution is x = -2.
To check for extraneous solutions, substitute both x = 3 and x = -2 back into the original equation:
For x = 3, |4x + 3| = 9 + 2x becomes:
|4(3) + 3| = 9 + 2(3)
|12 + 3| = 9 + 6
|15| = 15
Since |15| = 15, this solution is valid.
For x = -2, |4x + 3| = 9 + 2x becomes:
|4(-2) + 3| = 9 + 2(-2)
|-8 + 3| = 9 - 4
|-5| = 5
Again, since |-5| = 5, this solution is valid.
Therefore, there are no extraneous solutions for this equation, and the solutions are x = 3 and x = -2.