/*! 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 75 Solve for \(x.\) Simple Literal ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 75

Solve for \(x.\) Simple Literal Equations. $$a x+4=b-3$$

Short Answer

Expert verified
The solution for \(x\) is \(x = \frac{b-7}{a}\).

Step by step solution

01

Isolate the Variable Term

Start by moving the constant on the same side as the variable term to the other side of the equation. This can be done by subtracting 4 from both sides of the equation to get the variable term by itself on one side. The equation becomes: \(ax = b - 3 - 4\).
02

Combine Like Terms

Combine the constants on the right side of the equation. Subtract 3 from 4 to simplify the equation to \(ax = b - 7\).
03

Divide by the Coefficient

To solve for the variable \(x\), divide both sides of the equation by the coefficient \(a\). This will isolate \(x\) on one side of the equation. The final step gives you \(x = \frac{b-7}{a}\).

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.

Isolate the Variable
One of the foundational techniques in algebra is learning how to isolate the variable. This means rearranging an equation so that the variable we are solving for is alone on one side of the equation. Isolation is the first critical step to solving literal equations. For instance, if we have an equation such as \( ax+4=b-3 \), our goal is to get \( x \) by itself.

To do this, we need to perform the same operation on both sides of the equation to maintain the balance鈥攊t's like keeping a seesaw level. We can subtract 4 from each side to move the constant away from the variable term, leading to \( ax = b - 3 - 4 \). This now readies us for the next important step: combining like terms.
Combine Like Terms
Once we've started the process of isolating the variable, we'll often encounter terms that can be simplified. This is where we combine like terms. In algebra, like terms are terms that have the same variable raised to the same power. Numbers without variables, known as constants, can also be like terms.

In our example with \( ax = b - 3 - 4 \), the numbers -3 and -4 are like terms because they are both constants. We combine them by simple addition or subtraction: -3 and -4 add together to give -7, which simplifies our equation to \( ax = b - 7 \). This act not only tidies up equations but also brings us one step closer to finding the value of our variable.
Coefficients in Algebra
In any algebraic expression, numbers that multiply a variable are called coefficients. For example, in the term \( ax \), \( a \) is the coefficient of \( x \). Understanding coefficients is crucial because it allows us to manipulate equations to solve for variables.

To solve for \( x \) in \( ax = b - 7 \), we need to divide both sides of the equation by the coefficient \( a \). This is how we 'cancel' the effect of the coefficient, leaving us with the isolated variable: \[ x = \frac{b-7}{a} \]. Don't forget that dividing by a coefficient is the same as multiplying by its inverse, enabling us to isolate the variable cleanly and solve the equation.

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

Solve and check each equation. Treat the constants in these equations as exact numbers. Leave your answers in fractional, rather than decimal, form. Equations Having Symbols of Grouping. $$5 x+6=6(x-3)+2$$

Treat the percents given in this exercise as exact numbers, and work to three significant digits. How many kilograms of brass containing \(63 \%\) copper must be melted with \(1120 \mathrm{kg}\) of brass containing \(72 \%\) copper to produce a new brass containing \(67 \%\) copper?

Solve and check each equation. Treat the constants in these equations as exact numbers. Leave your answers in fractional, rather than decimal, form. Simple Fractional Equations. $$\frac{2 x-4}{7}=\frac{2-2 x}{4}$$

Solve and check each equation. Treat the constants in these equations as exact numbers. Leave your answers in fractional, rather than decimal, form. $$10-5 y=1-y$$

Solve and check each equation. Treat the constants in these equations as exact numbers. Leave your answers in fractional, rather than decimal, form. Simple Fractional Equations. $$5+\frac{7 x}{3}=\frac{x}{2}$$
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Logic and Functions

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.