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Ch. 5 - Systems of Equations and Inequalities
Blitzer - College Algebra 8th Edition
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Tutti i libri di testoBlitzer 8th EditionCh. 5 - Systems of Equations and InequalitiesProblema 3
Capitolo 6, Problema 3

In Exercises 1–18, solve each system by the substitution method. {x+y=2y=x2−4x+4\(\begin{cases}\) x + y = 2 \\ y = x^2 - 4x + 4 \(\end{cases}\)

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Start with the given system of equations: \(x + y = 2\) and \(y = x^{2} - 4x + 4\).
From the first equation, solve for \(y\) in terms of \(x\): \(y = 2 - x\).
Substitute this expression for \(y\) into the second equation: \(2 - x = x^{2} - 4x + 4\).
Rewrite the equation to set it equal to zero by moving all terms to one side: \(0 = x^{2} - 4x + 4 - (2 - x)\).
Simplify the equation and solve the resulting quadratic equation for \(x\). Once you find the values of \(x\), substitute them back into \(y = 2 - x\) to find the corresponding \(y\) values.

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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.
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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.
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Choosing a Method to Solve Quadratics

Quadratic Equations

A quadratic equation is a polynomial equation of degree two, typically in the form ax² + bx + c = 0. Solving quadratic equations often involves factoring, completing the square, or using the quadratic formula to find the variable's values.
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Introduction to Quadratic Equations
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