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Explanation of Solving a Linear Equation with One Variable - Succeed in Algebra. Also refer to solution, equal sign, combine like terms, distributive law, property, mathematics, math, maths, Ron Kurtus, School for Champions. Copyright © Restrictions Solving a Linear Equation with One Variableby Ron Kurtus (revised 3 April 2008) A linear equation with one variable consists of numbers or constants and multiplies of a variable. The standard form of such an equation is ax + b = 0, where a and b are constants and x is the variable. Often, the equation is in a more complex form. The solution of the equation is found by operating on both sides of the equation to get it into the form similar to x = −b/a. Questions you may have are:
This lesson will answer those questions. There is a mini-quiz near the end of the lesson. Rules for solutionWhen you have a linear equation with one variable, your goal is to manipulate the expressions, such that you you end up with the variable x on the left side of the equal sign and the constants on the right side. That is solving the equation. For example, the solution of the equation 4a = 3 − x is x = 3 − 4a. Basic ruleThe basic rule used in solving equations in Algebra is:
If you add a term on the left side, you must add the same term on the right side. If you multiply a term on the left side, you must multiply the same term on the right side. ExamplesIn the equation 4a = 3 − x, you want to get the x on the left side and the other items on the right side. You perform the following operations: Add x to both sides of the equation.
Subtract 4a from both sides of the equation.
Solving by combining like termsYou can solve an equation like 2x + 3 = −4x − 7 first getting all the x terms on the left side and all the constant terms on the right side. Next, you combine like terms. Then you divide by the multiple of x to get your solution. ExampleConsider the equation:
Add 4x to both sides.
Combine like terms.
Subtract 3 from both sides.
Combine like terms.
Divide both sides by 6.
Simplify the fraction.
Another exampleConsider the equation:
Multiply out to get rid of the parentheses.
Get rid of the fraction by multiplying both sides by 3.
Multiply out to get rid of the parentheses.
Combine like terms.
Subtract 9 from both sides.
Subtract 6x from both sides.
Divide by −7.
Variable in a fractionThere are equations where the x term is part of the denominator in an equation. In such a case, you must multiply both sides of the equation by the x term, so that it does not contain variable fractions. Likewise, you want to remove any fractions in the equation but multiplying by the denominator of the equation. ExampleConsider the equation:
Multiply both sides by (x + 1).
Simplify the fraction (x + 1)/(x + 1) = 1.
Multiply both sides by 12.
Multiply with distributive law or multiply out to get rid of the parentheses.
Subtract 7x from both sides.
Combine like terms.
Divide by 17 to get solution of equation.
Another exampleConsider the equation:
Multiply both sides by (5x − 3).
Multiply both sides by x.
Multiply with distributive law (remove parentheses).
Subtract 15x from both sides.
Divide both sides by −14x.
SummaryA linear equation with one variable consists of numbers or constants and multiplies of a variable. The standard form of such an equation is ax + b = 0, where a and b are constants and x is a variable. Often, the equation is in a more complex form. The solution of the equation is found by operating on the equation to get it into the form similar to x = −b/a. In other words, you want the x alone on the left side and the other items on the right side of the equation. The rule is what you do on the left side, you do on the right side. Go step-by-step ResourcesThe following resources provide information on this subject: WebsitesBooksMiscellaneousMini-quiz to check your understanding1. What is the solution of the equation 2x = x + 3? 2. What is the solution of the equation (x − 2a)/2 = 0? 3. What is the solution of the equation 1/(x + 1) = 5? If you got all three correct, you are on your way to becoming a Champion in Algebra. If you had problems, you had better look over the material again. What do you think?Do you have any questions, comments, or opinions on this subject? If so, send an email with your feedback. We will try to get back to you as soon as possible. Share linkFeel free to establish a link from your website to pages in this site. Or use our form to send this link to yourself or a friend. Students and researchersThe Web address of this page is Please include it as a reference in your report, document, or thesis. Where can you go from here?
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