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Chapter 8: Problem 17

Use the method of substitution to solve the system. $$\left\\{\begin{array}{l} x^{2}+y^{2}=16 \\ 2 y-x=4 \end{array}\right.$$

Short Answer

Expert verified
The solutions are \((-4, 0)\) and \(\left(\frac{12}{5}, \frac{16}{5}\right)\)."

Step by step solution

01

Solve for one variable in terms of the other from the linear equation

Start with the linear equation: \(2y - x = 4\). Solve for \(x\) in terms of \(y\) by adding \(x\) to both sides and subtracting 4: \(x = 2y - 4\).
02

Substitute the expression into the quadratic equation

Substitute \(x = 2y - 4\) into the quadratic equation \(x^2 + y^2 = 16\). This gives:\[(2y - 4)^2 + y^2 = 16.\]
03

Expand and simplify the equation

Expand the equation \((2y - 4)^2\) to get:\[4y^2 - 16y + 16.\]Substitute and simplify:\[4y^2 - 16y + 16 + y^2 = 16.\]Combine like terms:\[5y^2 - 16y + 16 = 16.\]
04

Solve the simplified equation for y

Subtract 16 from both sides:\[5y^2 - 16y = 0.\]Factor out \(y\) from the equation:\[y(5y - 16) = 0.\]Set each factor equal to zero: \(y = 0\) or \(5y - 16 = 0\).
05

Find the corresponding x-values

For \(y = 0\): Substitute back to find \(x\):\[x = 2(0) - 4 = -4.\]For \(5y - 16 = 0\): Solve for \(y\), \(y = \frac{16}{5}\), and substitute back to find \(x\):\[x = 2\left(\frac{16}{5}\right) - 4 = \frac{32}{5} - 4 = \frac{12}{5}.\]
06

Write the solution as ordered pairs

The solutions to the system are the points \( (-4, 0) \) and \( \left( \frac{12}{5}, \frac{16}{5} \right) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Substitution Method
The substitution method is a technique used to solve systems of equations. By solving the system step by step, you will understand how this method fits into the larger picture of algebraic problem-solving. In simple terms, we replace a variable in one equation with an expression derived from another equation.
  • Choose one of the equations: Typically, choose either the simpler equation or solve for a variable with a coefficient of 1.
  • Solve for one variable: Isolate the variable on one side of the equation.
  • Substitute into the other equation: This reduces the system to a single equation in one variable.
  • Solve the remaining equation: This gives the value of one variable.
  • Substitute back to find other variables: Use the known value to find the remaining variables.
The substitution method is particularly useful when one of the equations has a variable that is easy to isolate. It simplifies complex systems by reducing them to a manageable level.
Linear Equations
A linear equation is an equation that forms a straight line when graphed on a coordinate plane. They typically take the form of ax + by = c, where a, b, and c are constants. Linear equations are the foundation of algebra and are used extensively.
Understanding linear equations:
  • They have a constant slope: Changes in one variable lead to proportional changes in another.
  • Simplification: Linear equations often form the simpler part of a system.
  • Comparison: Allows to compare rates and interpret real-life scenarios, like speed versus time.
In our exercise, the equation 2y - x = 4 is linear. We use its linearity to isolate x and perform substitution, showing the interplay between linear forms and more complex types like quadratic equations.
Quadratic Equations
Quadratic equations are polynomial equations of degree 2, generally written in the form ax^2 + bx + c = 0. They can describe various natural phenomena, making them essential in both pure and applied mathematics.
Key attributes of quadratic equations:
  • They have a parabolic shape when graphed: This can open up or down depending on the sign of "a."
  • Zeros or roots: Solutions where the graph intersects the x-axis.
  • Applications: Often involve maximization and minimization problems, such as optimizing the area or cost.
In our exercise, combining the quadratic system x^2 + y^2 = 16 with our simplified expression leads us to solve for one variable. Handling quadratic equations requires techniques like factoring, completing the square, or using the quadratic formula.
Factoring
Factoring involves breaking down an expression into a product of simpler expressions, called "factors." It's a powerful technique in algebra for simplifying expressions and solving quadratic equations.
Steps in factoring:
  • Factor out the greatest common factor (GCF): This reduces the equation into a more workable form.
  • Recognize patterns: Such as difference of squares or perfect square trinomials.
  • Set each factor to zero: This step is crucial in finding solutions to an equation.
In our problem, we use factoring to solve the equation 5y^2 - 16y = 0 by finding common factors. This gives roots that help determine the y-values. Factoring simplifies complexities and makes hitting solutions more feasible.

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