Textbook Question
Solve each equation in Exercises 15–34 by the square root property.
1
views
Ch. 1 - Equations and Inequalities
Blitzer8th EditionCollege AlgebraISBN: 9780136970514
Not the one you use?Change textbookSelect a different textbook
Table of contents
- Ch. P - Fundamental Concepts of Algebra311
- Ch. 1 - Equations and Inequalities398
- Ch. 2 - Functions and Graphs407
- Ch. 3 - Polynomial and Rational Functions306
- Ch. 4 - Exponential and Logarithmic Functions247
- Ch. 5 - Systems of Equations and Inequalities173
- Ch. 6 - Matrices and Determinants143
- Ch. 7 - Conic Sections124
- Ch. 8 - Sequences, Induction, and Probability251
Chapter 2, Problem 23
In Exercises 1–26, solve and check each linear equation. 16 = 3(x - 1) - (x - 7)
Verified step by step guidance1
Start by expanding the expressions on the right side of the equation: apply the distributive property to both terms. For \(3(x - 1)\), multiply 3 by each term inside the parentheses, and for \(-(x - 7)\), distribute the negative sign to both terms inside the parentheses. This gives you \(3 \times x - 3 \times 1 - 1 \times x + 1 \times 7\).
Simplify the right side by performing the multiplications and combining like terms. This will result in an expression in terms of \(x\) and constants.
Rewrite the equation with the simplified right side, so it looks like \(16 = \text{(simplified expression)}\).
Isolate the variable term by moving all terms involving \(x\) to one side of the equation and constants to the other side. You can do this by adding or subtracting terms on both sides.
Solve for \(x\) by dividing both sides of the equation by the coefficient of \(x\). After finding \(x\), substitute it back into the original equation to check if both sides are equal, confirming your solution.
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
2mWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Distributive Property
The distributive property allows you to multiply a single term by each term inside parentheses. For example, a(b + c) = ab + ac. This is essential for simplifying expressions like 3(x - 1) by multiplying 3 with both x and -1.
Recommended video:
Guided course
04:15
Multiply Polynomials Using the Distributive Property
Combining Like Terms
Combining like terms involves adding or subtracting terms that have the same variable raised to the same power. This simplifies expressions and makes solving equations easier, such as combining terms after distribution in the equation.
Recommended video:
5:22Combinations
Solving Linear Equations
Solving linear equations means finding the value of the variable that makes the equation true. This involves isolating the variable on one side using inverse operations like addition, subtraction, multiplication, or division.
Recommended video:
04:02Solving Linear Equations with Fractions
Related Practice
Textbook Question
In Exercises 21–28, divide and express the result in standard form. 5i/(2 - i)
2
views
Textbook Question
Graph each equation in Exercises 13 - 28. Let x = - 3, - 2, - 1, 0, 1, 2, 3
y = |x| + 1
5
views
Textbook Question
Solve each equation in Exercises 15–34 by the square root property.
Textbook Question
In Exercises 15–26, use graphs to find each set. [3, ∞) ∩ (6, ∞)
3
views
Textbook Question
In Exercises 21–28, divide and express the result in standard form. 2i/(1 + i)
5
views