Problem 48 - College Algebra

Question

Solve each system by the method of your choice. $$\left\\{\begin{array}{l} 4 x^{2}+x y-30 \\ x^{2}+3 x y--9 \end{array}\right.$$

Step-by-Step Solution

Answer

The possible solutions to the system of equations are \((x, y) = (\sqrt{18}, 3)\) and \((x, y) = (-\sqrt{18}, -3)\)

Step 1: Solve the first equation for one variable

Solve the equation \(4x^2 + xy = 30\) for \(y\). This is achieved by moving \(4x^2\) to the right side to get: \(y = \frac{30 - 4x^2}{x}\)

Step 2: Substitute y in the second equation

Substitute \(y = \frac{30 - 4x^2}{x}\) into the second equation, we get: \(x^2 + 3x(\frac{30 - 4x^2}{x}) = 9\). Simplify it to get: \(5x^2 - 90 = 0\).

Step 3: Solve for x

The simplified equation \(5x^2 - 90 = 0\) then gets rearranged to \(5x^2 = 90\), and dividing by 5 gives \(x^2 = 18\). Solving for x gives \(x = \sqrt{18}\) or \(x = -\sqrt{18}\).

Step 4: Solve for y

Substitute \(x = \sqrt{18}\) and \(x = -\sqrt{18}\) in the equation \(y = \frac{30 - 4x^2}{x}\) to solve for y, giving two possible values of y for the corresponding x solutions.

Key Concepts

Quadratic Equations

Quadratic equations are mathematical expressions that follow the standard form \( ax^2 + bx + c = 0 \). Here, "a", "b", and "c" are constants, with "a" never being zero. Quadratic equations involve variables raised to the second power and are common in various algebraic problems. They form a fundamental component for solving more complex systems of equations.

The unique feature of quadratic equations is their U-shaped graph known as a parabola. Depending on the sign of "a", the parabola can open upwards or downwards. The solutions to a quadratic equation are known as roots, which can be calculated using methods such as factoring, completing the square, or the quadratic formula. In the given exercise, we encounter multiple quadratic terms such as \(4x^2\) and \(x^2\) within the system of equations, which we solve using algebraic manipulation.

Substitution Method

The substitution method is a powerful algebraic tool used to solve systems of equations. It involves solving one of the equations for a single variable and then substituting that expression into the other equation. This method is particularly beneficial when one equation is easily solvable for one variable.

In the provided exercise, we solve the first equation for "y", resulting in the expression \( y = \frac{30 - 4x^2}{x} \). By substituting this expression in the second equation, we effectively eliminate the variable "y", transforming the problem into a single-variable equation in terms of "x". Through substitution, the complexity of working with two equations and two variables is reduced to a simpler equation.

Solving for Variables

Solving for variables is an essential step in algebra where the goal is to find the values of the unknowns that satisfy the equation. In a system of equations, it often involves isolating one variable at a time. This is crucial for finding the specific solutions that make each equation true.

Upon substituting \( y = \frac{30 - 4x^2}{x} \) into the second equation of our system, the equation simplifies to \( 5x^2 - 90 = 0 \). By rearranging and simplifying, we isolate \( x \) and solve it to be \( x = \sqrt{18} \) or \( x = -\sqrt{18} \).

• We substitute each value of "x" back into the expression for "y" to find its possible values.
• This two-step process ensures both variables are accounted for within the system, leading to a complete solution for the problem.