Problem 11 - College Algebra

Question

Solve each system by the substitution method. $$\begin{aligned} &y^{2}=x^{2}-9\\\ &2 y=x-3 \end{aligned}$$

Step-by-Step Solution

Answer

The solutions to the system of equations are (3, 0) and (-5, -4).

Step 1: Solve the second equation for x

The second equation is 2y = x - 3. We add 3 to both sides, yielding x = 2y + 3.

Step 2: Substitute x in the first equation

Substitute x = 2y + 3 in the first equation, which is \(y^{2} = x^{2} - 9\). This yields \(y^{2} = (2y + 3)^{2} - 9\).

Step 3: Simplify the resultant equation

Expanding the term on the right gives \(y^{2} = 4y^{2} + 12y + 9 - 9\). This simplifies to \(y^{2} = 4y^{2} + 12y\).

Step 4: Rearrange and solve for y

If we rearrange the equation to 0 = 3y^{2} + 12y to use the quadratic formula, we obtain \(y = 0\) and \(y = -4\).

Step 5: Solve for x using these y values

Using \(x = 2y + 3\), when \(y = 0\) we get \(x = 3\), and when \(y = -4\) we get \(x = -5\).

Key Concepts

System of Equations

Dealing with multiple equations involving the same variables? That's a system of equations for you. Particularly, in algebra, when you have two or more equations that are linked by common variables, you enter the realm of solving systems of equations. The goal is to find a solution that satisfies all the equations simultaneously.

There are several methods for tackling these systems, such as graphing, substitution, elimination, and matrix operations. Each method has its own merits, and the choice depends on the specifics of the system you're working with. In our exercise, the substitution method becomes the star of the show: we cleverly use one equation to solve for one variable and then replace its value in another equation. This is a strategic move to reduce the system's complexity and inch our way towards the solution.

Quadratic Formula

Ah, the quadratic formula, where would algebra be without you? It's the go-to tool for solving quadratic equations, which look like \(ax^{2} + bx + c = 0\) where \(a\), \(b\), and \(c\) are constants, and \(a\) is not zero. The formula itself \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \) looks a bit daunting, but it's a reliable formula that guarantees a solution to any quadratic equation — assuming the solutions are real numbers.

The beauty of this formula lies in its universality. You plug your coefficients in, do the math, and voilà, you've got your solutions! The plus-minus sign (\(\pm\)) indicates that you'll normally get two solutions, which is exactly what happened in our step-by-step solution.

Solving Quadratic Equations

Solving quadratic equations can sometimes feel like a magic act, uncovering the x's hidden beneath the squared terms and coefficient veils. When you're given an equation in the form \(ax^{2} + bx + c = 0\), there are several methods to reveal these mysterious x-values. You can use factoring (when possible), completing the square, graphing the equation to find where it touches the x-axis, or deploying the ever-reliable quadratic formula.

Each method has its peculiar charms. Factoring is neat and quick, completing the square shows the structure of the equation, and graphing gives you a visual approach. For our textbook problem, after substituting and rearranging, we ended up with a quadratic equation that called for the quadratic formula. This is often the case when the other methods are not feasible or would take too much time.

Algebraic Substitution

In the algebraic arena, substitution is like a strategic game of chess. It's about making smart moves: replacing variables with other expressions to simplify the complexity and crack the problem open. We can think of it as a hand-off in a relay race — one equation gives the baton to another.

When we talk about substitution in the context of solving systems of equations, it's about expressing one variable in terms of another and then 'substituting' this expression into another equation. This is precisely what we did in the step-by-step solution. We expressed \(x\) in terms of \(y\) and then put that expression in place of \(x\) in the other equation. The result? A single equation with one variable, much easier to solve than the original system. Remember to always check the solutions in the original equations to confirm their validity.