Problem 31 - College Algebra and Calculus: An Applied Approach

Question

Solve the system by the method of substitution. $$\left\\{\begin{array}{l}3 x-7 y+6=0 \\ x^{2}-y^{2}=4\end{array}\right.$$

Step-by-Step Solution

Answer

The solution pair for the system of equations are (-2, 0) and (1, 2.1)

Step 1: Solve the linear equation for one variable

The linear equation \(3x - 7y + 6 = 0\) can be solved for \(x\) which gives \(x = (7y - 6) / 3\).

Step 2: Substitute the solved value into quadratic equation

Substituting \(x\) in \(x^2 - y^2 = 4\) with \(x = (7y - 6) / 3\), we will get \((7y - 6)^2/9 - y^2 = 4\), which simplifies to \(49y^2 - 84y + 36 - 9y^2 = 36\), and then to \(40y^2 - 84y = 0\).

Step 3: Solve the quadratic equation for y

We can factor out \(4y\) from the equation \(40y^2 - 84y = 0\), which gives \(4y(10y - 21) = 0\). The solutions for \(y\) are therefore \(y = 0\) and \(y = 21/10 = 2.1\).

Step 4: Substitute the value of y back into the linear equation to find x

Substituting \(y = 0\) into \(x = (7y - 6) / 3\), we find \(x = -2\). Substituting \(y = 2.1\) into the same equation, we find \(x = 1\).

Key Concepts

substitution method

The substitution method is a powerful algebraic technique used to find solutions for systems of equations. It involves solving one of the equations for one of its variables, and then substituting this expression into the other equation(s). This helps to eliminate one variable, making it simpler to solve the system.

For example, in the exercise provided, we are tasked with solving a system of equations using this method. The first step is to isolate one variable in one of the equations. From the linear equation \(3x - 7y + 6 = 0\), we solve for \(x\) and find \(x = \frac{7y - 6}{3}\).

Next, we substitute this expression for \(x\) into the quadratic equation \(x^2 - y^2 = 4\). This substitutes \(x\) and transforms the quadratic equation into one that depends solely on \(y\), allowing us to simplify and solve for \(y\). Once \(y\) is determined, we substitute back to find \(x\).

• Isolate one variable from any equation.
• Substitute this expression into the other equation(s).
• Solve the resulting equation for one variable.
• Substitute back to find the remaining variable(s).

linear equations

A linear equation represents a straight line when plotted on a coordinate plane, and it is usually of the form \(ax + by + c = 0\). This equation consists of variables raised to the first power, and their graph is a flat, straight line.

In the context of our exercise, the linear equation given is \(3x - 7y + 6 = 0\). Solving this equation for \(x\) gives us \(x = \frac{7y - 6}{3}\). Linear equations are straightforward to manipulate algebraically, making them a good starting point in the substitution method to express one variable in terms of another.

Linear equations give a constant rate of change, which means that there is a direct, proportional relationship between the variables. In a system of equations featuring both linear and quadratic equations, starting with the linear part often simplifies the problem.

• Variable powers are no more than one.
• Graphs as a straight line on a coordinate plane.
• Simpler to work with algebraically.

quadratic equations

Quadratic equations take the form \(ax^2 + bx + c = 0\) and their solutions form a parabola when graphed. Unlike linear equations, these involve the variable being squared, adding a layer of complexity.

In our exercise, the quadratic equation given is \(x^2 - y^2 = 4\). After substituting the expression for \(x\) from the linear equation, we simplify it to a quadratic equation solely in terms of \(y\). The task then becomes to solve for \(y\), which means finding the values that satisfy the quadratic equation.

To solve the quadratic equation \(40y^2 - 84y = 0\), factoring is an effective technique. We factored out common terms to simplify and obtain the solutions \(y = 0\) and \(y = 2.1\). Quadratic equations require careful handling, particularly since they can have two solutions.

• Involve squared terms, forming parabolas on graphs.
• Can be solved by factoring, completing the square, or using the quadratic formula.
• May have two real solutions, one solution, or no real solutions.