Exercise 41 Page 180

Practice makes perfect

a Let's first isolate the quadratic expression in the given equation.

3(y+1)^2-5=43

AddEqn

LHS+5=RHS+5

3(y+1)^2=48

DivEqn

.LHS /3.=.RHS /3.

(y+1)^2=16

Now we can solve for y by taking the square root of an equation.

(y+1)^2=16

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt((y+1)^2)=sqrt(16)

SqrtToAbs

sqrt(a^2)=|a|

|y+1|=sqrt(16)

CalcRoot

Calculate root

|y+1|=4

An absolute value measures an expression's distance from a midpoint on a number line. |y+1|= 4 This equation means that the distance is 4, either in the positive direction or the negative direction. |y+1|= 4 ⇒ ly+1= 4 y+1= -4 To find the solutions to the absolute value equation, we need to solve both of these cases for y.

|y+1|=4

lc y+1 ≥ 0:y+1 = 4 & (I) y+1 < 0:y+1 = - 4 & (II)

lcy+1=4 & (I) y+1=-4 & (II)

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

ly_1=3 y_2=-5

Both 3 and -5 are solutions to the absolute value equation.

b Let's first raise each side of the equation to the power of 2.

sqrt(1-4x)=10

RaiseEqn

LHS^2=RHS^2

(sqrt(1-4x))^2=10^2

PowSqrt

( sqrt(a) )^2 = a

1-4x=10^2

CalcPow

Calculate power

1-4x=100

SubEqn

LHS-1=RHS-1

-4x=99

DivEqn

.LHS /(-4).=.RHS /(-4).

x=99/-4

MoveNegDenomToFrac

Put minus sign in front of fraction

x=-99/4

c To solve an equation, we should first gather all of the variable terms on one side of the equation and all of the constant terms on the other side, using the Properties of Equality.

In this case, we need to start by using the Addition Property of Equality to isolate the fraction on the left-hand side of the equation.

6y-1/y-3=2

AddEqn

LHS+3=RHS+3

6y-1/y=5

Now we will multiply both sides by y to eliminate the fraction.

6y-1/y=5

MultEqn

LHS * y=RHS* y

6y-1/y* y=5y

FracMultDenomToNumber

a/y* y = a