Ch. 1 - Equations and Inequalities

Lial13th EditionCollege AlgebraISBN: 9780136881063Not the one you use?Change textbook

Lial 13th Edition

Ch. 1 - Equations and Inequalities

Problem 45

Chapter 2, Problem 45

Solve each equation or inequality. |6 - 2x | + 1 = 3

Verified step by step guidance

Start by isolating the absolute value expression. Subtract 1 from both sides of the equation to get: \(|6 - 2x| = 3 - 1\).

Simplify the right side to have: \(|6 - 2x| = 2\).

Recall that if \(|A| = B\), then \(A = B\) or \(A = -B\). Apply this to get two separate equations: \(6 - 2x = 2\) and \(6 - 2x = -2\).

Solve each equation separately. For \(6 - 2x = 2\), subtract 6 from both sides and then divide by -2 to isolate \(x\). For \(6 - 2x = -2\), do the same: subtract 6 and divide by -2.

Write the solutions for \(x\) from both equations. These values are the solutions to the original absolute value equation.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Absolute Value Definition

The absolute value of a number represents its distance from zero on the number line, always as a non-negative value. For an expression |A| = B, it means A = B or A = -B, provided B ≥ 0. Understanding this helps in splitting the equation into two separate cases to solve.

Solving Linear Equations

Linear equations involve variables raised only to the first power and can be solved by isolating the variable using inverse operations like addition, subtraction, multiplication, and division. After splitting the absolute value equation, each resulting linear equation is solved separately.

Checking for Extraneous Solutions

When solving absolute value equations, some solutions may not satisfy the original equation due to the definition of absolute value. It is important to substitute solutions back into the original equation to verify their validity and discard any extraneous solutions.

Recommended video:

05:21

Restrictions on Rational Equations

Related Practice

Textbook Question

Solve each equation using completing the square. -2x² + 4x + 3 = 0

views

Textbook Question

Solve each formula for the specified variable. Assume that the denominator is not 0 if variables appear in the denominator. F = GMm/r², for m (force of gravity)

Textbook Question

Solve each equation. x - √(2x+3) = 0

Textbook Question

Height of a Projectile A projectile is launched from ground level with an initial velocity of v₀ feet per second. Neglecting air resistance, its height in feet t seconds after launch is given by s=-16t²+v₀t. In each exercise, find the time(s) that the projectile will (a) reach a height of 80 ft and (b) return to the ground for the given value of v₀. Round answers to the nearest hundredth if necessary. v₀=32

Textbook Question

Write each number in standard form a+bi. -3+ √-18 / 24

views

Textbook Question

Solve each equation. x²- √(5x) -1 = 0

views