Problem 17 - Algebra and Trigonometry with Analytic Geometry

Question

1-30: Use the method of substitution to solve the system. $$ \left\\{\begin{array}{l} x^{2}+y^{2}=16 \\ 2 y-x=4 \end{array}\right. $$

Step-by-Step Solution

Answer

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

Step 1: Solve for one variable

From the second equation, solve for \(x\) in terms of \(y\):\[2y - x = 4\]\[x = 2y - 4\]

Step 2: Substitute into the first equation

Substitute \(x = 2y - 4\) into the first equation:\[ (2y - 4)^2 + y^2 = 16 \]

Step 3: Simplify the equation

Expand \((2y - 4)^2\):\[ (4y^2 - 16y + 16) + y^2 = 16 \]Combine like terms:\[ 5y^2 - 16y + 16 = 16 \]

Step 4: Solve for \(y\)

Subtract 16 from both sides:\[ 5y^2 - 16y = 0 \]Factor out \(y\):\[ y(5y - 16) = 0 \]Solve for \(y\):\[ y = 0 \text{ or } y = \frac{16}{5} \]

Step 5: Solve for \(x\) using \(y = 0\)

Substitute \(y = 0\) into \(x = 2y - 4\):\[ x = 2(0) - 4 = -4 \]

Step 6: Solve for \(x\) using \(y = \frac{16}{5}\)

Substitute \(y = \frac{16}{5}\) into \(x = 2y - 4\):\[ x = 2\left(\frac{16}{5}\right) - 4 \]Simplify:\[ x = \frac{32}{5} - \frac{20}{5} = \frac{12}{5} \]

Step 7: Verify solutions in original equations

First solution \((x, y) = (-4, 0)\):\[ (-4)^2 + 0^2 = 16 \]Confirmed.Second solution \((x, y) = \, (\frac{12}{5}, \frac{16}{5})\):\[ \left(\frac{12}{5}\right)^2 + \left(\frac{16}{5}\right)^2 = 16 \]\[ \frac{144}{25} + \frac{256}{25} = \frac{400}{25} = 16 \]Confirmed.

Key Concepts

System of Equations

When dealing with mathematics, a system of equations involves two or more equations that have common variables. Solving these systems requires finding values for the variables that satisfy all given equations. In this exercise, we have a system with two equations: a quadratic and a linear. The goal is to find values of \(x\) and \(y\) that make both equations true simultaneously.

• Equation 1: \(x^2 + y^2 = 16\)
• Equation 2: \(2y - x = 4\)

Systems like these can be approached using different methods, such as the substitution method or the elimination method. Here, we use substitution, which is often effective when one of the equations is linear, allowing a simpler expression for substitution into the other equation.

Quadratic Equations

A quadratic equation is an equation of the form \(ax^2 + bx + c = 0\) and involves terms with variables raised to the second power, or squared terms. In our system, the equation \(x^2 + y^2 = 16\) includes squared terms, which makes it a quadratic equation.When solving systems including quadratic equations, substitutions can lead us to quadratic formulas or factored forms, making it easier to solve. Once we substitute \(x = 2y - 4\) into this quadratic equation, we can convert it into a single-variable quadratic equation in terms of \(y\). The solution process often involves simplifying the equation by expanding, combining like terms, and then solving the resulting quadratic equation, often by factoring or using the quadratic formula.

Algebra

Algebra is at the heart of solving systems of equations, particularly when substitution and factoring are involved. Simply put, algebra is a field of mathematics where numbers and quantities are expressed using letters called variables. To solve the given problem using algebra, we mainly rely on these steps:

• Rearranging the linear equation to express one variable in terms of another.
• Substituting the expression into the quadratic equation to eliminate one variable.
• Using algebraic techniques to simplify and solve the resulting equation.
• Eventually, we need to verify all potential solutions by plugging them back into the original equations to ensure they hold true in the system.

The solution involves solving for one variable first and then using that solution to find the corresponding variable using basic algebraic operations.

Problem Solving Strategy

Problem solving in mathematics requires a strategic approach. In the context of solving systems of equations using substitution, we must carefully plan the sequence of steps. Here's a simple breakdown of the substitution method strategy:

• Identify one of the equations (preferably a linear one) to solve for a variable.
• Substitute the expression for that variable into the other equation, simplifying as much as possible.
• Solve the resulting equation for the remaining variable.
• Substitute back to find other variables' values.
• Verify the solutions in the original system of equations to ensure they hold true.

Using this strategy not only helps in solving the given problem but also builds foundational skills for tackling more advanced algebraic problems.