Exercise 36 Page 598

Practice makes perfect

a Let's start by isolating the absolute value expression on the left-hand side.

∣ x - 3 ∣ + 3 = 4

SubEqn

$LHS - 3 = RHS - 3$

∣ x - 3 ∣ = 1

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.

∣ x - 3 ∣ = 1 ⇓ x - 3 = 1 x - 3 = - 1

Let's solve these one at the time.

x - 3 = 1

AddEqn

$LHS + 3 = RHS + 3$

x = 4

Let's also solve the last equation.

x - 3 = - 1

AddEqn

$LHS + 3 = RHS + 3$

x = 2

To test the solutions we have to substitute them into the original equation and simplify. If the left-hand side and right-hand side are equal, we have the correct solutions.

∣ x - 3 ∣ + 3 = 4

Substitute

$x = 4$

∣ 4 - 3 ∣ + 3 = ? 4

▼

Simplify left-hand side

SubTerm

Subtract term

∣ 1 ∣ + 3 = ? 4

AbsPos

$∣ 1 ∣ = 1$

1 + 3 = ? 4

AddTerms

Add terms

4 = 4

The first solution is correct. Let's also check the second solution.

∣ x - 3 ∣ + 3 = 4

Substitute

$x = 2$

∣ 2 - 3 ∣ + 3 = ? 4

▼

Simplify left-hand side

SubTerm

Subtract term

∣ - 1 ∣ + 3 = ? 4

AbsNeg

$∣ - 1 ∣ = 1$

1 + 3 = ? 4

AddTerms

Add terms

4 = 4

The second solution is also correct.

b Again, let's perform inverse operations until

x

is solved for.

(x - 5)^{3} = 8

RadicalEqn

$3 LHS = 3 RHS$

x - 5 = 2

AddEqn

$LHS + 5 = RHS + 5$

x = 7

To test the solution, we have to substitute it into the original equation and simplify. If the left-hand side and right-hand side are equal, we have the correct solution.