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Elementary Algebra: An introduction to solving linear equations in one variable.

Module overview

Learning how to solve various algebraic equations is one of our main goals in algebra. This module introduces the basic techniques for solving linear equations in one variable. (Prerequisites: Working knowledge of real numbers and their operations.)


  • Define Linear Equations in One Variable
  • Solutions to Linear Equations
  • Solving Linear Equations
  • Combining Like Terms and Simplifying
  • Literal Equations

Define linear equations in one variable

We begin by establishing some definitions.

An equation is a statement indicating that two algebraic expressions are equal.
Linear Equation in One Variable
A linear equation in one variable x size 12{x} {} is an equation that can be written in the form ax + b = 0 size 12{ ital "ax"+b=0} {} where a size 12{a} {} and b size 12{b} {} are real numbers and a 0 size 12{a<>0} {} .

Following are some examples of linear equations in one variable, all of which will be solved in the course of this module.

x + 3 = 5
x 3 + 1 2 = 2 3
5 ( 3 x + 2 ) 2 = 2 ( 1 7 x )

Solutions to linear equations in one variable

The variable in the linear equation 2x + 3 = 13 size 12{2x+3="13"} {} is x size 12{x} {} . Values that can replace the variable to make a true statement compose the solution set. Linear equations have at most one solution. After some thought, you might deduce that x = 5 size 12{x=5} {} is a solution to 2x + 3 = 13 size 12{2x+3="13"} {} . To verify this we substitute the value 5 in for x size 12{x} {} and see that we get a true statement, 2 5 + 3 = 10 + 3 = 13 size 12{2 left (5 right )+3="10"+3="13"} {} .

Is x = 3 size 12{x=3} {} a solution to 2x 3 = 9 size 12{ - 2x - 3= - 9} {} ?

Yes, because 2 3 3 = 6 3 = 9 size 12{ - 2 left (3 right ) - 3= - 6 - 3= - 9} {}

Is a = 1 2 size 12{a= - { { size 8{1} } over { size 8{2} } } } {} a solution to 10 a + 5 = 25 size 12{ - "10"a+5="25"} {} ?

No, because 10 1 2 + 5 = 5 + 5 = 10 25 size 12{ - "10" left ( - { { size 8{1} } over { size 8{2} } } right )+5=5+5="10"<>"25"} {}

When evaluating expressions, it is a good practice to replace all variables with parenthesis first, then substitute in the appropriate values. By making use of parenthesis we could avoid some common errors using the order of operations.

Is y = 3 size 12{y= - 3} {}   a solution to 2y 5 = y 14 size 12{2y - 5= - y - "14"} {} ? 2 (    ) 5 = (    ) 14 Replace variables with parenthesis . 2 ( 3 ) 5 = ( 3 ) 14 Substitute the appropriate value . 6 5 = 3 14 Simplify . 11 = 11 True . Yes because y = 3 produces a true mathematical statement.

Solving linear equations in one variable

When the coefficients of linear equations are numbers other than nice easy integers, guessing at solutions becomes an unreasonable prospect. We begin to develop an algebraic technique for solving by first looking at the properties of equality.

Properties of equality

Given algebraic expressions A and B where c is a real number:

Addition property of equality

If  A = B  then  A + c = B + c

Subtraction property of equality

If  A = B  then  A - c = B - c

Multiplication property of equality

If  A = B  and  c 0  then c A = c B

Division property of equality

If  A = B  and  c 0  then  A c = B c

Multiplying or dividing both sides of an equation by zero is carefully avoided. Dividing by zero is undefined and multiplying both sides by zero will result in an equation 0=0.

To summarize, the equality is retained if we add, subtract, multiply and divide both sides of an equation by any nonzero real number. The central technique for solving linear equations involves applying these properties in order to isolate the variable on one side of the equation.

Use the properties of equality to solve: x + 3 = 5 x + 3 = 5 x + 3 3 = 5 3 Subtract 3 on both sides . x = 8 Simplify The solution set is { 8 } .

Questions & Answers

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Source:  OpenStax, Contemporary math applications. OpenStax CNX. Dec 15, 2014 Download for free at http://legacy.cnx.org/content/col11559/1.6
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