Algebra 3rd Set of Problems

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Please see attached for the solutions to the given problems.

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Exercises
1. The solution is wrong. Although one of the values is correct, in this case x = 3, this is not enough to be considered a complete solution. This is because a system on linear equations with 2 unknowns needs a solution for each variable, in this case x and y. Since x = 3 is correct value for x, we need to get y. Substituting the value of x in either equation will yield y = 0. So the solution is {3, 0}
2. The solution is wrong. When solving the system and the solution arrives at 0 = 0, the solution to the system is infinitely many. For example, one solution to this system is {0, 4}. Another is {1, 3}. And there is {2, 2}. And there are infinitely many more solutions to this system. To put it simply, any solution pair to equation 1 will fit perfectly as a solution to equation 2 because the system is consistent and dependent.
3. The one who solves for x first will arrive at the solution the faster and less work. Looking at equation 1, the student can simply transpose y and 13 to get y = 3x - 13 and substitute y to 2x - 5y = 20. That is a lot faster than isolating x in either equation because the student will also need to divide the numerical coefficient of x then substitute it to the other equation. The student who solves at y first will most likely deal with fractional terms, which adds to the complexity of solving the system.
4. A system with no solution will arrive at an inequality at the end. For example:

When solved using elimination method will either arrive at or . Therefore, the system has no solution. ...