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Break down the given absolute value equation two separate equations.
k=6 and k=-2/5
When solving an equation involving absolute value expressions, we should consider what would happen if we removed the absolute value symbols. Let's look at an example equation. |ax+b|=|cx+d| Although we can make 4 statements about this equation, there are actually only two possible cases to consider.
| Statement | Result |
|---|---|
| Both absolute values are positive. | ax+b=cx+d |
| Both absolute values are negative. | -(ax+b)=-(cx+d) |
| Only the left-hand side is negative. | -(ax+b)=cx+d |
| Only the right-hand side is negative. | ax+b=-(cx+d) |
lc 3k-2 ≥ 0:3k-2 = 2(k+2) & (I) 3k-2 < 0:3k-2 = - 2(k+2) & (II)
(I):Distribute 2
(II):Distribute -2
(I), (II):LHS+2=RHS+2
(I):LHS-2k=RHS-2k
(II):LHS+2k=RHS+2k
(II):.LHS /5.=.RHS /5.
After solving an absolute value equation, it is necessary to check for extraneous solutions. To do this, we substitute the found solutions into the given equation and determine if a true statement is made.
k= 6
Multiply
Add and subtract terms
|8|=8
Multiply
|16|=16
We ended with a true statement, so k=6 is not extraneous.
k= -2/5
a*b/c= a* b/c
Rewrite 2 as 10/5
Add and subtract fractions
|-16/5|=16/5
|8/5|=8/5
a*b/c= a* b/c
We got another true statement, so k=- 25 is also not extraneous.