In Exercises 29–42, solve each system by the method of your choice. {x2+(y−2)2=4x2−2y=0\begin{cases} $$x^2 + (y$$ - $$2)^2 = 4$$ \\ $$x^2 - 2y = 0$$ \end{cases}{x2+(y−2)2=4x2−2y=0

Ch. 5 - Systems of Equations and Inequalities

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Blitzer 8th Edition

Ch. 5 - Systems of Equations and Inequalities

문제 37

6장, 문제 37

In Exercises 29–42, solve each system by the method of your choice. $\begin{cases}\) x^2 + (y - 2)^2 = 4 \\ x^2 - 2y = 0 \(\end{cases}$

검증된 단계별 안내

Identify the system of equations to solve: $x^2 + (y - 2)^2 = 4$ and $x^2 - 2y = 0$.

From the second equation, isolate $y$ in terms of $x$: $x^2 - 2y = 0 \implies 2y = x^2 \implies y = \frac{x^2}{2}$.

Substitute the expression for $y$ from step 2 into the first equation: $x^2 + \left( \frac{x^2}{2} - 2 \right)^2 = 4$.

Expand and simplify the equation from step 3 to form a polynomial equation in terms of $x$ only. This will involve squaring the binomial and combining like terms.

Solve the resulting polynomial equation for $x$. Then, substitute each $x$ value back into $y = \frac{x^2}{2}$ to find the corresponding $y$ values, giving the solution points $(x, y)$ for the system.

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주요 개념

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

Solving Systems of Equations

A system of equations consists of two or more equations with the same variables. Solving the system means finding all variable values that satisfy all equations simultaneously. Methods include substitution, elimination, and graphing, chosen based on equation types and complexity.

Equations of Circles

The equation x² + (y - k)² = r² represents a circle with center at (0, k) and radius r. Understanding this form helps identify geometric constraints and possible solution points when combined with other equations in a system.

Substitution Method

The substitution method involves solving one equation for one variable and substituting that expression into the other equation. This reduces the system to a single equation with one variable, simplifying the process of finding solutions.

In Exercises 29–42, solve each system by the method of your choice. {x2+(y−2)2=4x2−2y=0\(\begin{cases}\) x^2 + (y - 2)^2 = 4 \\ x^2 - 2y = 0 \(\end{cases}\){x2+(y−2)2=4x2−2y=0