Problem 19 - College Algebra

Question

Solve each system by the addition method. \(\left\\{\begin{array}{l}x+y-1 \\ x-y-3\end{array}\right.\)

Step-by-Step Solution

Answer

The solutions to the system of equations are \((x, y) = (3,2), (3,-2), (-3,2), (-3,-2)\)

Step 1: Subtract the Second equation from the First

The first step is to subtract the second equation from the first one to eliminate \(x^{2}\). This will result in the equation \(2y^{2}=8\)

Step 2: Solve the resulting equation for y

You can solve the equation \(2y^{2} = 8\) for \(y\) by dividing both sides by 2 to obtain \(y^{2} = 4\). Then take the square root of both sides to find the two possible values for \(y\), which are \(y = 2\) and \(y = -2\)

Step 3: Substitute y values into one of the original equations

Substitute the obtained \(y\) values into one of the original equations to find the corresponding \(x\) values. If you use the second equation, for \(y = 2\), it results in \(x^{2} - (2)^2 = 5\), which implies \(x^{2} = 9\), thus \(x = 3\) and \(x = -3\). Similarly, for \(y = -2\), it results in \(x^{2} - (-2)^2 = 5\), which implies \(x^{2} = 9\), thus \(x = 3\) and \(x = -3\)

Key Concepts

System of Equations

A system of equations consists of two or more equations that share the same set of unknowns. In the given exercise, we have a system of two equations involving variables \(x\) and \(y\): \[\begin{align*}1. & \quad x^2 + y^2 = 13 \2. & \quad x^2 - y^2 = 5 \end{align*}\] The goal is to find the values of \(x\) and \(y\) that satisfy both equations simultaneously. This means we're looking for a pair of numbers which, when plugged into both equations, make both sides of each equation equal. Systems of equations can often pose challenges due to multiple variables. However, using methods like substitution, graphing, or elimination, we can solve them efficiently. In this exercise, we employ the elimination method, sometimes called the addition method, to find the solutions.

Elimination Method

The elimination method is a powerful tool for solving systems of equations. It works by removing one variable entirely, allowing us to solve for the other one. This method is sometimes referred to as the addition method because it often involves adding or subtracting equations to cancel out a variable. In our exercise, we are tasked with eliminating one of the variables, \(x^2\), by subtracting the second equation from the first: \[\begin{align*}(x^2 + y^2) - (x^2 - y^2) = 13 - 5 \end{align*}\] This step not only removes \(x^2\) but also simplifies the system to a single equation: \[2y^2 = 8\]By clearing away one variable, the elimination method turns a complex problem into a simpler one-variable equation that can be solved with ease.

Solving Quadratic Equations

Solving quadratic equations is a key skill when tackling systems of equations like the one we have. In this case, after elimination, we're left with the quadratic equation \(2y^2 = 8\). To solve for \(y\): 1. **Simplify the equation** by dividing both sides by 2: \[y^2 = 4\]2. **Take the square root** of both sides to find values for \(y\): \[y = \pm 2\] The square root introduces \(\pm\) because both 2 and -2 satisfy the equation \((\pm 2)^2 = 4\). Next, we substitute these values one at a time back into either of the original equations to find corresponding \(x\) values. For example, using \(y = 2\) in the second equation: \[x^2 - 4 = 5 \rightarrow x^2 = 9\] Then, taking the square root of both sides gives \(x = \pm 3\). The same process is repeated for \(y = -2\), leading to the same \(x\) values. Thus, the complete set of solutions for the system includes the pairs \((3, 2), (-3, 2), (3, -2), (-3, -2)\). Quadratic equations often arise in systems with non-linear components, making the ability to solve them crucial.