Algebra is the division of math concerned with operations and relations. Its concentrations range from solving equations and inequalities to graphing functions and polynomials. Algebra’s complexity grows with increasing variables and operations, but it starts its foundation in linear equations and inequalities.
A linear equation is any equation involving one or two variables whose exponents are one. In the case of one variable, one solution exists for the equation. For example, with 2x = 6, x can only be 3.
A linear inequality is any statement involving one or two variables whose exponents are one, where inequality rather than equality is the center of focus. For instance, with 3y < 2, the “<” represents less than and the solution set includes all numbers y < 2/3.
One obvious difference between linear equations and inequalities is the solution set. A linear equation of two variables can have more than one solution. For instance, with x = 2y + 3, (5, 1), (3, 0) and (1, -1) are all solutions to the equation. In each pair, x is the first value and y is the second value. However, these solutions fall on the exact line described by y = ½ x – 3/2.
If the inequality were x ? 2y + 3, these same linear solutions would exist in addition to (3, -1), (3, -2) and (3, -3), where multiple solutions can exist for the same value of x or the same value of y only for inequalities. The "?" means that it is unknown whether x is greater than or less than 2y + 3. The first number in each pair is the x value and the second is the y value.
The graph of linear inequalities include a dashed line if they are greater than or less than but not equal to. Linear equations, on the other hand, include a solid line in every situation. Moreover, linear inequalities include shaded regions whereas linear equations do not.
The complexity of linear inequalities outweighs the complexity of linear equations. While the latter involves simple slope and intercept analysis, the former (linear inequalities) involves the assessment of where to shade when resolving how to account for the additional set of solutions.