Problem 18 - Precalculus: Functions and Graphs

Question

Use the method of substitution to solve the system. $$\left\\{\begin{array}{l} x^{2}+y^{2}=1 \\ y+2 x=-3 \end{array}\right.$$

Step-by-Step Solution

Answer

There is no real solution to the system as the quadratic equation has a negative discriminant.

Step 1: Solve the Linear Equation for y

Start with the linear equation from the system: \( y + 2x = -3 \). Solve for \( y \) by subtracting \( 2x \) from both sides, resulting in \( y = -3 - 2x \).

Step 2: Substitute Expression for y into First Equation

Substitute \( y = -3 - 2x \) into the first equation \( x^2 + y^2 = 1 \). This gives us: \( x^2 + (-3 - 2x)^2 = 1 \).

Step 3: Expand and Simplify the Equation

Expand \( (-3 - 2x)^2 \) to get \( 9 + 12x + 4x^2 \). Insert this into the equation: \( x^2 + 9 + 12x + 4x^2 = 1 \), resulting in \( 5x^2 + 12x + 9 = 1 \).

Step 4: Rearrange into a Quadratic Equation

Subtract 1 from both sides to arrange the equation as \( 5x^2 + 12x + 8 = 0 \).

Step 5: Solve the Quadratic Equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 5 \), \( b = 12 \), and \( c = 8 \). Compute the discriminant: \( b^2 - 4ac = 144 - 160 = -16 \), indicating no real solution.

Key Concepts

Substitution Method

The substitution method is a technique used to solve systems of equations where one equation is substituted into another. This allows you to solve for one variable and then use that solution to find the other variables. In our case, we start with a system that includes both a linear equation and a non-linear equation. The system is:

• Linear: \( y + 2x = -3 \)
• Non-linear: \( x^2 + y^2 = 1 \)

Using substitution entails isolating one variable in one equation and substituting its expression into the other equation. Here, we solved the linear equation for \( y \) to get \( y = -3 - 2x \). This expression was then substituted back into the non-linear equation, \( x^2 + y^2 = 1 \), allowing us to work with only one variable (x) in a quadratic form.

Quadratic Formula

The quadratic formula helps us find the roots of quadratic equations. These are equations of the form \( ax^2 + bx + c = 0 \). The quadratic formula is expressed as:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \( a \), \( b \), and \( c \) are coefficients from the quadratic equation. For our example, once we substituted the expression for \( y \) into the non-linear equation, we rearranged it into the form \( 5x^2 + 12x + 8 = 0 \). The coefficients identified were:

• \( a = 5 \)
• \( b = 12 \)
• \( c = 8 \)

Plugging these into the quadratic formula helps determine the potential real solutions for the variable \( x \).

No Real Solutions

When solving quadratic equations through the quadratic formula, the discriminant \( b^2 - 4ac \) can give important insights. The discriminant can tell us how many solutions exist and what kind of solutions they are:

• If \( b^2 - 4ac > 0 \), there are two distinct real solutions.
• If \( b^2 - 4ac = 0 \), there is exactly one real solution.
• If \( b^2 - 4ac < 0 \), there are no real solutions.

In our problem, we found that the discriminant \( 144 - 160 = -16 \). This negative discriminant indicates there are no real solutions for the equation. Thus, the system of equations does not intersect in the real number domain.

Linear and Non-Linear Systems

A system of equations typically has more than one equation that can be either linear or non-linear. Understanding the differences helps in choosing the right method for solving them:

• Linear equations have variables of the first degree, often represented as lines when graphed.
• Non-linear equations include variables raised to powers higher than one, or involve multiple terms, often represented as curves when graphed.

In this exercise, our linear equation is \( y + 2x = -3 \), and the non-linear one is \( x^2 + y^2 = 1 \), which is in fact the equation of a circle. Solving such a system may involve seeing how a line intersects with a curve. The concepts of substitution and quadratic equations help in finding these points of intersection, or confirming their absence as is seen here.