/*! 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 73 Solve for \(x,\) given that \(a,... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve for \(x,\) given that \(a, b,\) and \(c\) are real numbers and \(a \neq 0\) $$a x+b=c$$

Short Answer

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

Step by step solution

01

Start with the Given Equation

The problem gives us the linear equation \(a x + b = c\). Our goal is to solve this equation for \(x\).
02

Isolate the Term with x

First, we need to get the term involving \(x\) by itself on one side. To do this, subtract \(b\) from both sides of the equation: \[a x = c - b\]
03

Solve for x

Since we need to find \(x\), divide both sides of the equation by \(a\) to isolate \(x\): \[x = \frac{c-b}{a}\].
04

Confirm the Solution

Verify this solution by substituting \(x = \frac{c-b}{a}\) back into the original equation. Calculating \(a \left( \frac{c-b}{a} \right) + b = c\), we simplify to find \(c\). Hence, the solution is correct.

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 91Ó°ÊÓ!

Key Concepts

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

Isolation of Variables
The key to solving linear equations like \( ax + b = c \) is to skillfully isolate the variable of interest, which in this case is \( x \). This ensures that we focus exclusively on solving for \( x \) by getting rid of any other terms cluttering its path. Here's how you can achieve that:

  • Subtracting Terms: Start by moving constants to the other side of the equation. For example, subtract \( b \) from both sides to clear it from the left side, giving us \( ax = c - b \).
  • Dividing by Coefficients: Once \( x \) is free from additions or subtractions, manage any multiplication. Since \( a \) is multiplied by \( x \), divide both sides by \( a \) to isolate \( x \), resulting in the expression \( x = \frac{c-b}{a} \).
Remember, when performing these operations, ensure you apply the action uniformly to both sides of the equation to maintain equality. A common pitfall is forgetting this, which leads to incorrect results.
Verification of Solutions
Verification is a crucial step to ensure the solution to a linear equation is accurate and valid. Neglecting this step may yield incorrect answers that could act as misleading conclusions.

Here's how to verify your solutions effectively:
  • Substitution: Once you determine \( x \), plug it back into the original equation. In our example, substitute \( x = \frac{c-b}{a} \) into \( ax + b = c \).
  • Simplification: Simplify the equation to check if both sides are equal. Calculate \( a\left( \frac{c-b}{a} \right) + b \). The result should simplify back to \( c \).
If both sides of the equation balance out after substitution and simplification, the solution is confirmed correct. This affirmation reassures that the process was executed correctly.
Linear Equation Steps
Breaking down the solution process into clear, manageable steps is essential when tackling linear equations. This structured approach makes solving equations systematic and less daunting.

Here is a simplified step-by-step plan:
  • Identify the Equation: Start by precisely understanding the given equation, just like \( a x + b = c \) in our example.
  • Isolate the Variable: As described previously, move terms and divide coefficients to single out the variable. After isolating, we have \( x = \frac{c-b}{a} \).
  • Verification: Perform a quick check by substituting \( x \) back into the starting equation to ensure consistency and correctness.
Following these steps guarantees a coherent and structured solution, reducing mistakes and bolstering confidence while solving linear equations.

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

Explain the mistake that is made. Find the \(x\) - and \(y\) -intercepts of the line with equation \(2 x-3 y=6\) Solution: \(x\) -intercept: set \(x=0\) and solve for \(y\) \(-3 y=6\) The \(x\) -intercept is (0,-2) \(y=-2\) \(y\) -intercept: set \(y=0\) and solve for \(x . \quad 2 x=6\) The \(y\) -intercept is (3,0) \(x=3\)

Mike's home phone plan charges a flat monthly fee plus a charge of \(\$ .05\) per minute for long-distance calls. The total monthly charge is represented by \(y=0.05 x+35\) \(x \geq 0,\) where \(y\) is the total monthly charge and \(x\) is the number of long-distance minutes used. Interpret the meaning of the \(y\) -intercept.

A golf club membership has two options. Option A is a \(\$ 300\) monthly fee plus \(\$ 15\) cart fee every time you play. Option B has a \(\$ 150\) monthly fee and a \(\$ 42\) fee every time you play. Write a mathematical model for monthly costs for each plan and graph both in the same viewing rectangle using a graphing utility. Explain when Option A is the better deal and when Option \(\mathrm{B}\) is the better deal.

Show that two lines with equal slopes and different \(y\) -intercepts have no point in common. Hint: Let \(y_{1}=m x+b_{1}\) and \(y_{2}=m x+b_{2}\) with \(b_{1} \neq b_{2} .\) What equation must be true for there to be a point of intersection? Show that this leads to a contradiction.

Weight. The average weight of a baby born in 1900 was 6 pounds 4 ounces. In 2000 the average weight of a newborn was 6 pounds 10 ounces. What is the rate of change of birth weight in ounces per year? What do we expect babies to weigh at birth in \(2040 ?\)

See all solutions

Recommended explanations on Math Textbooks

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.