/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 Solve each of the given equation... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 43

Solve each of the given equations for \(\mathrm{x}\). $$45 x+12=0$$

Short Answer

Expert verified
x = -\frac{4}{15}

Step by step solution

01

Isolate the term with x

To solve for \( x \), we first need to isolate the term with \( x \). The equation is \( 45x + 12 = 0 \). We subtract \( 12 \) from both sides to remove the constant term from the left side of the equation. This gives us: \( 45x + 12 - 12 = 0 - 12 \), which simplifies to \( 45x = -12 \).
02

Solve for x

Now that we have isolated the \( x \) term, we need to solve for \( x \) by dividing both sides by \( 45 \), the coefficient of \( x \). This gives us: \( x = \frac{-12}{45} \). We simplify \( \frac{-12}{45} \) by finding the greatest common divisor of 12 and 45, which is 3, and dividing the numerator and the denominator by 3. So, \( x = \frac{-12 \div 3}{45 \div 3} = \frac{-4}{15} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solving Equations
Solving equations involves finding the value of the unknown variable that makes the equation true. This is a fundamental concept in algebra. Let's break it down with an example equation: \(45x + 12 = 0\).
  • Begin by identifying the variable you want to solve for, which is \(x\) in this case.
  • Next, you'll isolate and solve for \(x\) using inverse operations, which let us "undo" addition, subtraction, multiplication, or division.

The strategy generally involves rearranging the equation to get the variable on one side and the numbers on the other. This requires a clear understanding of how to manipulate equations, such as adding or subtracting the same value from both sides to maintain equality. Persistently practicing this process will help you become proficient in solving equations efficiently.
Isolating Variables
Isolating variables means getting the variable alone on one side of the equation. This is a pivotal step in solving equations. Take our initial problem \(45x + 12 = 0\) for example:
  • To isolate \(x\), first remove the constant term. In this case, subtract \(12\) from both sides to eliminate it from the left side: \(45x = -12\).
  • Now \(x\) is attached to its coefficient \(45\). To isolate \(x\), you'll divide both sides by \(45\).

Through this isolation, the equation becomes simpler, moving closer to finding the exact value of \(x\). This process underscores the algebraic principle of maintaining balance in an equation, ensuring what you do to one side, you do to the other.
Fraction Simplification
Fraction simplification is an essential skill when you end up with a fraction as your solution. It makes the result easier to understand and use. Let's see it in action with the fraction \(\frac{-12}{45}\):
  • Identify the greatest common divisor (GCD) of the numerator and the denominator. Here, the GCD of \(12\) and \(45\) is \(3\).
  • Divide both the numerator and the denominator by this GCD to simplify the fraction. So, \(\frac{-12}{45}\) becomes \(\frac{-4}{15}\) after division by \(3\).

By simplifying fractions, you achieve a cleaner, more manageable expression. This process not only aids in clearer communication but also helps in comparing fractions more easily in further mathematical operations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A particle moves along a line with constant acceleration. It is known the velocity of the particle, as a function of the amount of time that has passed, is given by the equation $$ v=v_{0}+a t $$ where \(v\) is the velocity at time t, v0 is the initial velocity of the particle (at time \(t=0\) ), and a is the acceleration of the particle. i. Solve formula (2) for t. ii. You know that the current velocity of the particle is \(120 \mathrm{~m} / \mathrm{s}\). You also know that the initial velocity was \(40 \mathrm{~m} / \mathrm{s}\) and the acceleration has been a constant \(a=2 \mathrm{~m} / \mathrm{s}^{2}\). How long did it take the particle to reach its current velocity?

Solve the inequality. Express your answer in both interval and set notations, and shade the solution on a number line. $$-14 x+4>-6 x+8$$

Perform each of the following tasks in Exercises \(1.5 .96-1.5 .99\) i. Write out in words the meaning of the symbols which are written in set- builder notation. ii. Write some of the elements of this set. iii. Draw a real line and plot some of the points that are in this set. $$C=\\{x \in \mathbb{Z}: x \leq 2\\}$$

Solve each of the given equations for \(x\). Check your solutions using your calculator. $$-\frac{3}{2} x+9=\frac{1}{4} x+7$$

Solve each of the given equations for the indicated variable. \(\frac{V}{t}=k\) for \(t\)
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.