College Algebra (Abridged)

Case II: System of Equations with Indefinitely Many Solutions 2 – = 1 …………………………………………………..(1) 8 – 4 = 4 ………………………………………………….(2) Solving equation (1) for the value of = 2 – 1 Substituting the value of in equation 2 8 – 4(2 − 1) = 4 Solving the equation for the value of 8 – 8 + 4 = 4 0 = 0 In this case, the unique value(s) of cannot be obtained since the values of variables, as well as constants, cancel out. This is primarily because both the given equations are actually the same. If we multiply equation (1) by 4, we get equation (2). The solution for this set of linear equations is any set of ordered pairs ( , ) which satisfies the equation 2 − = 1 . In order to find the solution to such problems, let us consider = 0 , so we get the values of and as (0, − 1) which becomes a possible solution. Now, consider the value of = 1 . The values of ordered pair ( , ) would be (1,1) . This is another possible solution.

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