Linear Equation Lab: Solve Equations and Inequalities

Change both sides of a linear equation or inequality. Follow the steps, check a value, and see the solution set on a number line.

Free, with no sign-up.

For High school students and adults reviewing the topic.

About 10 minutes.

How does it work?

  1. Choose a comparison and change the coefficients to update the steps, table, and number line.
  2. Open a preset to compare one solution, an interval, no solution, and every number.

What will you get?

Compare solution sets, check substitutions, and solve two coefficient challenges.

What does the model assume?

  • The variable ranges over all real numbers. Coefficients range from −100 to 100 in steps of 0.001.
  • The model solves ax + b compared with cx + d. Collecting terms gives (a − c)x compared with d − b. A negative divisor reverses an inequality.
  • When a = c, the comparison between b and d decides whether every real number or no real number satisfies the relation.
  • Integer arithmetic preserves the entered coefficient values. Reduced fractions with denominators up to 100 are displayed exactly; other values are shown to four decimal places. Decimal display rounding does not affect the solution set.
  • The substitution checks one example. It illustrates the algebraic result rather than proving the entire solution set. Strict inequalities use a sample one unit inside the permitted interval.
  • The number line centers on the boundary and rescales its ticks. Its arrows show continuation, not a finite end to the interval.

Which setup would you like to try?

What can you change?

Range: -100 to 100 .

Range: -100 to 100 .

Range: -100 to 100 .

Range: -100 to 100 .

What does the model predict?

QuantityValueUnit
Solution typeA boundary or one solution
Collected coefficient2
Collected constant1
Boundary or equation solution1/2
Comparison after division, if needed=
Inequality direction reversedNo
Value checked by substitution1/2
Left side at the checked value2
Right side at the checked value2
Does the checked value satisfy the relation?Yes

Follow the transformations

Original relation: (2)x + (1) = (0)x + (2).

  1. Collect x terms: (2)x + (1) = 2.
  2. Collect constants: (2)x = 1.
  3. Divide by 2: x = 1/2.

x = 1/2

At x = 1/2, the left side is 2 and the right side is 2. The comparison = is true.

For an interval, the checked value is an example inside it. For a constant comparison, the example is x = 0.

Solution set: x = 1/2.-1.5-0.51/21.52.5x

Solution set: x = 1/2.

A filled point includes the boundary; an open circle excludes it. A solid ray and arrow show the permitted direction. A dashed line provides the reference axis.

Can you reach these targets?

Set a = 2, b = 1, c = 0 and relation =. Choose d so the solution is x = 3.

  • Target: Coefficient a = 2 ; tolerance 0 .
  • Target: Constant b = 1 ; tolerance 0 .
  • Target: Coefficient c = 0 ; tolerance 0 .
  • Target: Boundary or equation solution = 3 ; tolerance 0 .
  • Solution type (A boundary or one solution)
  • Relation (=)

Set b = 2, c = 4, d = 11 and relation <. Choose a so dividing reverses the inequality.

  • Target: Constant b = 2 ; tolerance 0 .
  • Target: Coefficient c = 4 ; tolerance 0 .
  • Target: Constant d = 11 ; tolerance 0 .
  • Inequality direction reversed (Yes)
  • Solution type (A boundary or one solution)
  • Relation (<)
Read the topic guide