Chapter Closure
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Start with the Rewriting strategy, and simplify the given equation by applying the Properties of Equality.
You can use the Undoing strategy and raise both sides of the equation to the power of 2.
Use the Rewriting strategy, and simplify the given equation by applying the Properties of Equality.
You can use the Looking Inside strategy to find the solutions.
y=7 and y=-5
x=-99/2
y=1
x=2 and x=-3
To solve the given equation, we will first rewrite our equation using the Properties of Equality. Our goal is to isolate the expression raised to the power of 2.
Now, using the undoing strategy, we will take the square root to eliminate squaring.
sqrt(LHS)=sqrt(RHS)
sqrt(a^2)=|a|
Calculate root
An absolute value measures an expression's distance from a midpoint on a number line. |y-1|= 6 This equation means that the distance is 6, either in the positive direction or the negative direction. |y-1|= 6 ⇒ ly-1= 6 y-1= -6 To find the solutions to this absolute value equation, we need to solve both of these cases for y.
lc y-1 ≥ 0:y-1 = 6 & (I) y-1 < 0:y-1 = - 6 & (II)
(I), (II): LHS+1=RHS+1
Both 7 and -5 are solutions to the given equation.
According to the Undoing strategy, we should first raise both sides of the equation to the power of 2 to eliminate the square root.
LHS^2=RHS^2
( sqrt(a) )^2 = a
Calculate power
Now we can continue to solve using the Properties of Equality.
LHS-1=RHS-1
.LHS /-2.=.RHS /-2.
Put minus sign in front of fraction
To find the solution, we should rewrite our equation using the Properties of Equality to make it easier to solve. Notice that y cannot equal 0, because it is in the denominator of a fraction.
Now, let's fully isolate the variable on the left-hand side of the equation.
An absolute value measures an expression's distance from a midpoint on a number line.
|2x+1|= 5
lc 2x+1 ≥ 0:2x+1 = 5 & (I) 2x+1 < 0:2x+1 = - 5 & (II)
(I), (II): LHS-1=RHS-1
(I), (II): .LHS /2.=.RHS /2.
Both 2 and -3 are solutions to the absolute value equation.