Problem 24 - Intermediate Algebra

Question

Solve each system of equations by substitution for real values of \(x\) and \(y.\) See Examples 2 and 3. $$ \left\\{\begin{array}{l} x^{2}-x-y=2 \\ 4 x-3 y=0 \end{array}\right. $$

Step-by-Step Solution

Answer

Solutions are \((x, y) = (3, 4)\) and \((x, y) = (-\frac{2}{3}, -\frac{8}{9})\).

Step 1: Solve for y in the Second Equation

Start with the second equation, which is linear and easier to manipulate: \(4x - 3y = 0\). Solve for \(y\) by isolating it on one side. \[ 4x = 3y \] \[ y = \frac{4x}{3} \].

Step 2: Substitute y into the First Equation

Take the expression we found for \(y\) in Step 1 and substitute it into the first equation: \(x^2 - x - y = 2\). This becomes \(x^2 - x - \frac{4x}{3} = 2\).

Step 3: Simplify and Solve the Quadratic Equation

Combine the terms on the left-hand side, keeping the equation in terms of \(x\). \[x^2 - x - \frac{4x}{3} = 2 \] Multiply every term by 3 to clear the fraction: \[ 3x^2 - 3x - 4x = 6 \]. This simplifies to \[3x^2 - 7x - 6 = 0\], a quadratic equation.

Step 4: Factor or Use the Quadratic Formula

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -7\), and \(c = -6\): \[ x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \times 3 \times (-6)}}{2 \times 3} \]. This results in \( x = \frac{7 \pm \sqrt{121}}{6} \). \(\sqrt{121} = 11\).

Step 5: Calculate the Values of x

Solve for the two possible \(x\) values: \[ x = \frac{7 + 11}{6} = 3 \] and \[ x = \frac{7 - 11}{6} = -\frac{2}{3} \]. So, \(x = 3\) or \(x = -\frac{2}{3}\).

Step 6: Find Corresponding y Values

Use the expression for \(y\) from Step 1, \(y = \frac{4x}{3}\), to find \(y\) when \(x = 3\): \[ y = \frac{4(3)}{3} = 4 \]. For \(x = -\frac{2}{3}\): \[ y = \frac{4(-\frac{2}{3})}{3} = -\frac{8}{9} \].

Step 7: State the Solution to the System

The solutions to the system are \((x,y) = (3, 4)\) and \((x,y) = \left(-\frac{2}{3}, -\frac{8}{9}\right)\).

Key Concepts

Substitution Method

The substitution method is a straightforward way to solve systems of equations. It involves solving one of the equations for one variable, then substituting that expression into the other equation. This effectively reduces a system of equations to a single equation, which is often easier to solve.

• **Identify an Equation**: Choose the equation that is easiest to manipulate. In many cases, this will be a linear equation due to its simpler operations.
• **Solve for One Variable**: Rearrange the equation to express one variable in terms of others. For instance, if given the equation \(4x - 3y = 0\), solving for \(y\) gives \(y = \frac{4x}{3}\).
• **Substitute the Expression**: Replace the variable in the other equation with the expression obtained. This results in an equation with only one variable, simplifying the process of solving the system.

This method is particularly useful for systems involving a linear and a non-linear equation, like in our exercise.

Quadratic Equations

Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable. Solving these equations can reveal the values of \(x\) that satisfy the equation.

• **Factoring**: This involves rewriting the quadratic in the form \((x - p)(x - q) = 0\), where \(p\) and \(q\) are the roots of the quadratic. Once factored, the solutions can be easily found by setting each factor equal to zero.
• **Quadratic Formula**: This universal method can solve any quadratic equation: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). The discriminant, \(b^2 - 4ac\), determines the nature and number of roots (real and distinct, real and the same, or complex).

In our example, we encountered the quadratic equation \(3x^2 - 7x - 6 = 0\). Using the quadratic formula, it yields two values for \(x\), illustrating the multiple roots characteristic of these equations.

Solving Systems of Equations

Solving systems of equations involves finding all solutions that satisfy every equation in the system simultaneously. This can involve linear equations, like \(ax + by = c\), or include non-linear equations, as seen in our exercise.

• **Graphical Method**: Plotting each equation on a coordinate plane reveals the intersection points, which are the solutions to the system.
• **Substitution**: As discussed earlier, this method involves substituting one variable in terms of another, simplifying the system to a single equation.
• **Elimination Method**: This method involves adding or subtracting equations to eliminate a variable, reducing the system to a simpler form.

In the original exercise, combining the substitution method with solving a quadratic equation allows us to find the set of \((x, y)\) pairs that solve the system. The process combines algebraic manipulation with strategic solving techniques for different types of equations.