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}{x+y=2y=x2−4x+4

Ch. 5 - Systems of Equations and Inequalities

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Blitzer 8th Edition

Ch. 5 - Systems of Equations and Inequalities

문제 3

6장, 문제 3

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

검증된 단계별 안내

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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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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.

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.

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.

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 \(\end{cases}\){x+y=2y=x2−4x+4