Exercise 133 Page 628

Practice makes perfect

a When we remove the absolute value from the left-hand side we get two equations — one equation where the right-hand side is positive, and one where the right-hand side is negative.

0 = ∣ 3 x - 3 ∣ ⇓ 0 = 3 x - 3 - 0 = 3 x - 3

However, since

0 = - 0

we only have to solve the first equation.

0 = 3 x - 3

AddEqn

$LHS + 3 = RHS + 3$

3 = 3 x

RearrangeEqn

Rearrange equation

3 x = 3

DivEqn

$LHS / 3 = RHS / 3$

x = 1

b By simplifying the left-hand side and then performing inverse operations we can solve for

x .

x^{2} - 3 (2 x + 3) = x^{2} - 3

SubEqn

$LHS - x^{2} = RHS - x^{2}$

- 3 (2 x + 3) = - 3

MultEqn

$LHS \cdot (- 1) = RHS \cdot (- 1)$

3 (2 x + 3) = 3

DivEqn

$LHS / 3 = RHS / 3$

2 x + 3 = 1

SubEqn

$LHS - 3 = RHS - 3$

2 x = - 2

DivEqn

$LHS / 2 = RHS / 2$

x = - 1

c By simplifying the right-hand side and then performing inverse operations we can solve for

x .

0 = (p - 3)^{2} - p^{2}

ExpandNegPerfectSquare

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$

0 = p^{2} - 6 p + 9 - p^{2}

SubTerm

Subtract term

0 = - 6 p + 9

AddEqn

$LHS + 6 p = RHS + 6 p$

6 p = 9

DivEqn

$LHS / 6 = RHS / 6$

p = \frac{9}{6}

ReduceFrac

$\frac{a}{b} = \frac{a / 3}{b / 3}$

p = \frac{3}{2}

d Let's perform inverse operations until

w

is isolated on one side.

u = \frac{v + 2}{w}

MultEqn

$LHS \cdot w = RHS \cdot w$

u w = v + 2

DivEqn

$LHS / u = RHS / u$

w = \frac{v + 2}{u}

Subchapter links

11.2.1 Solving Work and Mixture Problems p.603-604

11.2.2 Solving Equations and Systems Graphically p.607-608

11.2.3 Using a Best-Fit Line to Make a Prediction p.610-614

11.2.4 Treasure Hunt p.616-617

11.2.5 Using Coordinate Geometry and Constructions to Explore Shapes p.620-621

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11.2.6 Modeling with Exponential Functions and Linear Inequalities p.625-628

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Exercise 133 Page 628

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