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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
Capitolo 6, Problema 33

In Exercises 29–42, solve each system by the method of your choice. {x2+4y2=20x+2y=6\(\begin{cases}\) x^2 + 4y^2 = 20 \\ x + 2y = 6 \(\end{cases}\)

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1
Identify the system of equations: \(x^2 + 4y^2 = 20\) and \(x + 2y = 6\).
From the linear equation \(x + 2y = 6\), solve for one variable in terms of the other. For example, solve for \(x\): \(x = 6 - 2y\).
Substitute the expression for \(x\) into the first equation \(x^2 + 4y^2 = 20\). This gives: \((6 - 2y)^2 + 4y^2 = 20\).
Expand the squared term and simplify the resulting equation to form a quadratic equation in terms of \(y\) only.
Solve the quadratic equation for \(y\), then substitute each \(y\) value back into \(x = 6 - 2y\) to find the corresponding \(x\) values.

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