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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 1
Capitolo 6, Problema 1

Solve each system by the substitution method.
{x+y=2y=x2−4\(\begin{cases}\) x + y = 2 \\ y = x^2 - 4 \(\end{cases}\)

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1
Start with the given system of equations: \(x + y = 2\) and \(y = x^{2} - 4\).
From the first equation, solve for \(y\) in terms of \(x\): \(y = 2 - x\).
Substitute this expression for \(y\) into the second equation: \(2 - x = x^{2} - 4\).
Rewrite the equation to set it equal to zero: \(x^{2} - 4 - (2 - x) = 0\), which simplifies to \(x^{2} - 4 - 2 + x = 0\).
Combine like terms to get a quadratic equation: \(x^{2} + x - 6 = 0\). This quadratic can now be solved using factoring, completing the square, or the quadratic formula.

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System of Equations

A system of equations consists of two or more equations with the same set of variables. The goal is to find values for the variables that satisfy all equations simultaneously. In this problem, the system includes a linear and a quadratic equation involving x and y.
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Introduction to Systems of Linear Equations

Substitution Method

The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. This reduces the system to a single equation with one variable, making it easier to solve. Here, y is expressed from the first equation and substituted into the second.
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Choosing a Method to Solve Quadratics

Solving Quadratic Equations

After substitution, the resulting equation is quadratic, which can be solved by factoring, completing the square, or using the quadratic formula. The solutions to the quadratic give possible values for x, which can then be used to find corresponding y values.
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Solving Quadratic Equations by Factoring
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