Exercise 106 Page 205

a 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.

2(y-1)^2+8=80

SubEqn

LHS-8=RHS-8

2(y-1)^2=72

DivEqn

.LHS /2.=.RHS /2.

(y-1)^2=36

Now, using the undoing strategy, we will take the square root to eliminate squaring.

(y-1)^2=36

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt((y-1)^2)=sqrt(36)

SqrtToAbs

sqrt(a^2)=|a|

|y-1|=sqrt(36)

CalcRoot

Calculate root

|y-1|=6

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.

| y-1|=6

lc y-1 ≥ 0:y-1 = 6 & (I) y-1 < 0:y-1 = - 6 & (II)

lcy-1=6 & (I) y-1=-6 & (II)

(I), (II): LHS+1=RHS+1

ly_1=7 y_2=-5

Both 7 and -5 are solutions to the given equation.

b According to the Undoing strategy, we should first raise both sides of the equation to the power of 2 to eliminate the square root.

sqrt(1-2x)=10

RaiseEqn

LHS^2=RHS^2

(sqrt(1-2x))^2=10^2

PowSqrt

( sqrt(a) )^2 = a

1-2x=10^2

CalcPow

Calculate power

1-2x=100

Now we can continue to solve using the Properties of Equality.

1-2x=100

SubEqn

LHS-1=RHS-1

-2x=99

DivEqn

.LHS /-2.=.RHS /-2.

x=99/-2

MoveNegDenomToFrac

Put minus sign in front of fraction

x=-99/2

c 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.

6y-1/y-2=3

AddEqn

LHS+2=RHS+2

6y-1/y=5

MultEqn

LHS * y=RHS* y

6y-1=5y

Now, let's fully isolate the variable on the left-hand side of the equation.

6y-1=5y

AddEqn

LHS+1=RHS+1

6y=5y+1

SubEqn

LHS-5y=RHS-5y

y=1

d An absolute value measures an expression's distance from a midpoint on a number line.

|2x+1|= 5This equation means that the distance is 5, either in the positive direction or the negative direction. |2x+1|= 5 ⇒ l2x+1= 5 2x+1= - 5 To find the solutions, we need to look inside the absolute value expression and consider both of these cases.

|2x+1|=5

lc 2x+1 ≥ 0:2x+1 = 5 & (I) 2x+1 < 0:2x+1 = - 5 & (II)

lc2x+1=5 & (I) 2x+1=- 5 & (II)

(I), (II): LHS-1=RHS-1

lc2x=4 & (I) 2x=-6 & (II)

(I), (II): .LHS /2.=.RHS /2.

lx_1=2 x_2=-3

Both 2 and -3 are solutions to the absolute value equation.