Problem 14 - Precalculus: Functions and Graphs

Question

Use the method of substitution to solve the system. $$\left\\{\begin{array}{l} x^{2}+y^{2}=25 \\ 3 x+4 y=-25 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution is \(x = -3\) and \(y = -4\), or the point \((-3, -4)\).

Step 1: Solve for x in terms of y

To begin using the substitution method, solve for one variable in terms of the other from the linear equation. Here, solve for \(x\) in terms of \(y\) from the equation \(3x + 4y = -25\). This gives:\[ x = \frac{-25 - 4y}{3} \]

Step 2: Substitute x in the second equation

Substitute the expression for \(x\) obtained in Step 1 into the circle equation \(x^2 + y^2 = 25\). This results in:\[ \left(\frac{-25 - 4y}{3}\right)^2 + y^2 = 25 \]

Step 3: Simplify and solve the quadratic equation

Expand the equation and simplify:\[ \frac{(-25 - 4y)^2}{9} + y^2 = 25 \]Calculate \((-25 - 4y)^2\) which gives \(625 + 200y + 16y^2\):\[ \frac{625 + 200y + 16y^2}{9} + y^2 = 25 \]Multiply through by 9 to clear the fraction:\[ 625 + 200y + 16y^2 + 9y^2 = 225 \]Combine the \( y^2 \) terms:\[ 16y^2 + 9y^2 + 200y + 625 = 225 \]This simplifies to:\[ 25y^2 + 200y + 400 = 0 \]

Step 4: Further simplify the quadratic equation

Divide the entire quadratic equation by 25:\[ y^2 + 8y + 16 = 0 \]Notice this can be factored as:\[ (y + 4)^2 = 0 \]

Step 5: Solve for y

Since \((y + 4)^2 = 0\), solve for \(y\) to get:\[ y + 4 = 0 \]Thus, \( y = -4 \).

Step 6: Solve for x

Substitute \(y = -4\) back into the expression for \(x\) from Step 1:\[ x = \frac{-25 - 4(-4)}{3} \]This simplifies to:\[ x = \frac{-25 + 16}{3} \]\[ x = \frac{-9}{3} \]\[ x = -3 \]

Step 7: Write the solution

The solution to the system is the point where the two equations intersect. Therefore, \(x = -3\) and \(y = -4\). The point of intersection is \((-3, -4)\).

Key Concepts

Substitution Method

The substitution method is a powerful technique for solving systems of equations. It involves expressing one variable in terms of the other using one of the equations and then substituting this expression into the second equation. This helps reduce the problem to a single equation with one variable, making it simpler to solve.

• Start by isolating one variable in terms of the other.
• Substitute the isolated variable into the other equation.
• Simplify and solve the resulting equation.
• Finally, back-substitute to find the other variable's value.

In the given exercise, we first solved for \(x\) in terms of \(y\) from the linear equation \(3x + 4y = -25\). With \(x\) expressed as \(x = \frac{-25 - 4y}{3}\), we replaced \(x\) in the quadratic equation. This process simplifies the system into a single equation that can be solved for \(y\).

Quadratic Equation

A quadratic equation is a polynomial equation of degree two. It typically takes the form \(ax^2 + bx + c = 0\). Solving a quadratic equation can involve several methods such as factoring, using the quadratic formula, or completing the square.

In this scenario, after substituting \(x\) into the circle equation and simplifying, we reached a standard quadratic form:\[ 25y^2 + 200y + 400 = 0 \]Dividing this equation by 25 simplified it further to \(y^2 + 8y + 16 = 0\), which could then be factored or solved using the quadratic formula to find the value of \(y\). In this particular case, we noticed the equation could be factored easily as \((y + 4)^2 = 0\).

Linear Equation

Linear equations form the backbone of solving systems using the substitution method. A linear equation is represented in the standard form \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants.

• Linear equations graph as straight lines in Cartesian coordinates.
• They represent constant rates of change.
• To solve, isolate one variable to use in substitution.

In the given problem, we began solving by transforming the linear equation \(3x + 4y = -25\) into a format that made it easy to substitute into the quadratic equation. By rearranging terms, we solved for \(x\), resulting in \(x = \frac{-25 - 4y}{3}\), which efficiently set the stage for substitution into the quadratic equation.

Solving Equations

Solving equations involves finding the values of variables that satisfy all given conditions in a mathematical sentence. In a system of equations, these conditions are represented by multiple equations and their solutions represent points of intersection.

• Equations can be linear, quadratic, or of higher degree.
• The solution method often depends on the type of equation.
• They can represent relationships between variables and real-world situations.

In our exercise, solving the system required handling both a quadratic equation and a linear equation. By using the substitution method, we reduced the complexity by working with one equation at a time. Once \(y\) was determined, we returned to the expression of \(x\) to find its corresponding value. This gives us the solution point - the intersection of the two equations on a graph, which is \((-3, -4)\).