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How many cases do you have after you remove the absolute value?
w=3 and w=-1/5
Before we begin, let's isolate the absolute value on the left-hand side.
LHS-9=RHS-9
Subtract terms
.LHS /-2.=.RHS /-2.
An absolute value measures an expression's distance from a midpoint on a number line. |5w-7|= 8
lc 5w-7 ≥ 0:5w-7 = 8 & (I) 5w-7 < 0:5w-7 = - 8 & (II)
(I), (II): LHS+7=RHS+7
(I), (II): .LHS /5.=.RHS /5.
Both 3 and - 15 are solutions to the given equation. By substituting w=3 and w=- 15 into the equation and evaluating, we can check if the solutions are correct. First, let's check w=3.
w= 3
Multiply
Subtract terms
|8|=8
Because the left-hand and right-hand sides are equal, we know that the solution is correct. Now we will check the second solution by substituting - 15 for w.
w= -1/5
Put minus sign in numerator
a/5* 5 = a
Subtract terms
|-8|=8
Both sides of the equation are again equal. Therefore, both solutions are correct.