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Linear Equations: Definition and Examples | EDU.COM

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Linear Equations: Definition and Examples | EDU.COMEDU.COMResourcesBlogGuidePodcastPlanBackHomesvg]:size-3.5">Math Glossarysvg]:size-3.5">Linear EquationsLinear Equations: Definition and ExamplesTable of ContentsLinear Equations Definition of Linear Equations

A linear equation is an algebraic equation in which each variable is raised to the power of 1. The degree of a linear equation is 1, which means no variable has an exponent greater than 1. When graphed, linear equations in one or two variables always form a straight line. These equations can be written in different ways based on the number of variables present. For example, a linear equation in one variable can be written as Ax+B=0Ax + B = 0Ax+B=0, where A and B are real numbers and x is a variable.

Linear equations come in different standard forms depending on the number of variables. For a linear equation in one variable, the standard form is Ax+B=0Ax + B = 0Ax+B=0 (where A≠0A ≠ 0A=0), and it has only one solution. For a linear equation in two variables, the standard form is Ax+By+C=0Ax + By + C = 0Ax+By+C=0 (where A≠0A ≠ 0A=0, B≠0B ≠ 0B=0), and it has infinitely many solutions. Linear equations can also be expressed in other forms like slope-intercept form (y=mx+by = mx + by=mx+b) and slope-point form (y−y1=m(x−x1)y - y_1 = m(x - x_1)y−y1​=m(x−x1​)).

Examples of Linear Equations Example 1: Solving a Basic Linear Equation Problem:

Solve the linear equation 5x−12=185x - 12 = 185x−12=18.

Step-by-step solution:

Step 1, Add 121212 to both sides of the equation.

5x−12+12=18+125x - 12 + 12 = 18 + 125x−12+12=18+12 5x=305x = 305x=30

Step 2, Divide both sides by 555 to find the value of xxx.

5x5=305\frac{5x}{5} = \frac{30}{5}55x​=530​ x=6x = 6x=6

So, the answer is x=6x = 6x=6.

Example 2: Solving a Linear Equation with Fractions Problem:

Solve for xxx: 2x+53=x−5\frac{2x + 5}{3} = x - 532x+5​=x−5.

Step-by-step solution:

Step 1, Multiply both sides of the equation by 333 to eliminate the fraction.

2x+53×3=(x−5)×3\frac{2x + 5}{3} \times 3 = (x - 5) \times 332x+5​×3=(x−5)×3 2x+5=3x−152x + 5 = 3x - 152x+5=3x−15

Step 2, Subtract 555 from both sides of the equation.

2x+5−5=3x−15−52x + 5 - 5 = 3x - 15 - 52x+5−5=3x−15−5 2x=3x−202x = 3x - 202x=3x−20

Step 3, Subtract 3x3x3x from both sides.

2x−3x=3x−20−3x2x - 3x = 3x - 20 - 3x2x−3x=3x−20−3x −x=−20-x = -20−x=−20

Step 4, Multiply both sides by −1-1−1.

−x×(−1)=−20×(−1)-x \times (-1) = -20 \times (-1)−x×(−1)=−20×(−1) x=20x = 20x=20

So, the answer is x=20x = 20x=20.

Example 3: Solving a Word Problem Problem:

The sum of two numbers is 555555. If one number is 111111 less than the other, find the numbers by framing a linear equation.

Step-by-step solution:

Step 1, Let's call the first number xxx.

Step 2, Since the second number is 111111 less than the first, we can write it as x−11x - 11x−11.

Step 3, According to the problem, the sum of the two numbers is 555555.

x+(x−11)=55x + (x - 11) = 55x+(x−11)=55

Step 4, Simplify the left side of the equation.

2x−11=552x - 11 = 552x−11=55

Step 5, Add 111111 to both sides of the equation.

2x−11+11=55+112x - 11 + 11 = 55 + 112x−11+11=55+11 2x=662x = 662x=66

Step 6, Divide both sides by 222.

2x2=662\frac{2x}{2} = \frac{66}{2}22x​=266​ x=33x = 33x=33

Step 7, Find the second number by using the relationship x−11x - 11x−11.

Second number =33−11=22= 33 - 11 = 22=33−11=22

So, the two numbers are 333333 and 222222.

Comments(4)SSoftballDevoteeTheoNovember 6, 2025This glossary page on linear equations is a lifesaver! I've used it to help my students grasp the concept. Thanks for the clear examples!

WWindsurferZoeNovember 6, 2025I've used this linear equations glossary with my students. It's super helpful for explaining concepts clearly and making learning fun!

CConsultantNoraNovember 5, 2025This glossary page on linear equations is great! It's helped my students grasp the concept. Clear defs and examples are super useful.

SStylistIvyNovember 4, 2025I've used this linear equations glossary page with my students. It's a great resource for making the concept clear and helping them with problem-solving.

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