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Break down the given absolute value equation two separate equations.
h=0.25
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 3h+1 ≥ 0:3h+1 = 7h & (I) 3h+1 < 0:3h+1 = - 7h & (II)
LHS-3h=RHS-3h
(I):.LHS /4.=.RHS /4.
(II):.LHS /(-10).=.RHS /(-10).
(I), (II): Rearrange equation
(I), (II): Calculate quotient
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.
h= 0.25
Multiply
Add terms
|1.75|=1.75
We ended with a true statement, so h=0.25 is not extraneous. Now let's check h=-0.1.
h= -0.1
Multiply
Add terms
|0.7|=0.7
We ended with a false statement, so h=-0.1 is extraneous. Therefore, it is not a valid solution to the equation.