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First, isolate the absolute value on one side of the equation.
x=9 and x=-3
Before we can solve the given equation, we need to isolate the absolute value expression using the Properties of Equality.
LHS+5=RHS+5
Add terms
.LHS /2.=.RHS /2.
An absolute value is always a non-negative number because it measures the expression's distance from a midpoint on a number line. |x-3|= 6
lc x-3 ≥ 0:x-3 = 6 & (I) x-3 < 0:x-3 = - 6 & (II)
(I), (II): LHS+3=RHS+3
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.
x= 9
Subtract terms
|6|=6
Multiply
Subtract terms
Substituting 9 for x in the equation does result in a true statement, so x=9 is not an extraneous solution. Now let's check whether or not x=- 3 is extraneous.
x= - 3
Subtract terms
|-6|=6
Multiply
Subtract terms
Substituting (- 3) for x in the equation also does result in a true statement, so x=- 3 is not an extraneous solution. Therefore, the equation has two solutions.