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Ch. 5 - Systems of Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBlitzer 8th EditionCh. 5 - Systems of Equations and InequalitiesProblema 47
Capitolo 6, Problema 47

In Exercises 47–52, solve each system by the method of your choice. {2x2+xy=6x2+2xy=0\(\begin{cases}\) 2x^2 + xy = 6 \\ x^2 + 2xy = 0 \(\end{cases}\)

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Start by writing down the system of equations clearly: \$2x^{2} + xy = 6$ and $x^{2} + 2xy = 0$.
From the second equation \(x^{2} + 2xy = 0\), factor out the common term \(x\): \(x(x + 2y) = 0\).
Set each factor equal to zero to find possible cases: either \(x = 0\) or \(x + 2y = 0\).
For the case \(x = 0\), substitute into the first equation \$2(0)^{2} + 0 imes y = 6$ and check if it holds true.
For the case \(x + 2y = 0\), solve for \(y\) as \(y = -\frac{x}{2}\), then substitute this expression for \(y\) into the first equation \(2x^{2} + x\left(-\frac{x}{2}\right) = 6\) to find values of \(x\).

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In Exercises 47–48, solve each system by the method of your choice. (x + 2)/2 - (y + 4)/3 = 3 (x + y)/5 = (x - y)/2 - 5/2

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Graph the solution set of each system of inequalities or indicate that the system has no solution.

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In Exercises 47–48, solve each system by the method of your choice. (x - y)/3 = (x + y)/2 - 1/2 (x + 2)/2 - 4 = (y + 4)/3

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In Exercises 47–52, solve each system by the method of your choice. {−4x+y=12y=x3+3x2\(\begin{cases}\) -4x + y = 12 \\ y = x^3 + 3x^2 \(\end{cases}\)

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Graph the solution set of each system of inequalities or indicate that the system has no solution.

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In Exercises 46–55, graph the solution set of each system of inequalities or indicate that the system has no solution.


This

is a piecewise function. Refer to the textbook.

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