Problem 1 - College Algebra

Question

Solve each system by the substitution method. $$\left\\{\begin{array}{l} x+y-2 \\ y-x^{2}-4 \end{array}\right.$$

Step-by-Step Solution

Answer

The solutions to the system are (0, 2) and (-1, 3).

Step 1: Solve the first equation for one variable

Solving the first equation, \(x + y = 2\), for \(x\) in terms of \(y\) gives \(x = 2 - y\).

Step 2: Substitute expression for \(x\) into second equation

Substitute \(x = 2 - y\) into the second equation. The second equation becomes, \(y - (2 - y)^{2} - 4 = y - (4 - 4y + y^{2}) - 4 = -y^{2} + 5y - 8\). Now, arrange the equation in the form, \( y^{2} - 5y + 8 = 0\), which represents a quadratic equation in terms of \(y\).

Step 3: Solve the quadratic equation

Solving this quadratic equation, \(y^{2} - 5y + 8 = 0\) using the Quadratic Formula gives \(y = 2,3\). These are the solutions for \(y\).

Step 4: Substitute values of \(y\) back into the first equation

Substitute \(y = 2, 3\) back into the first equation to solve for \(x\). Hence, when \(y = 2\), \(x = 2 - 2 = 0\) and when \(y = 3\), \(x = 2 - 3 = -1\).

Key Concepts

Systems of Equations

A system of equations is a collection of two or more equations with the same set of variables. The goal is to find the values of these variables that satisfy all the equations simultaneously. In our example, we dealt with a simple system consisting of two equations with two variables, \(x\) and \(y\):

• \(x + y = 2\)
• \(y - x^{2} = 4\)

To solve a system of equations, you can use different methods such as substitution, elimination, or graphing. In this exercise, we used the substitution method, which involves solving one equation for one variable and then substituting that expression into the other equation. This simplifies the system, allowing you to solve it more easily. The main advantage of the substitution method is its straightforwardness, especially when one equation is already solved for a particular variable.

Quadratic Equation

Once we substituted the expression for \(x\) from the first equation into the second equation, we ended up with a quadratic equation. A quadratic equation is any equation that can be rewritten in the form of \(ax^2 + bx + c = 0\). In this case, the equation is in terms of \(y\):

• \(y^{2} - 5y + 8 = 0\)

Quadratic equations can be solved by different methods, such as factoring, completing the square, or the quadratic formula. In this solution, we employed the quadratic formula: \\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]To find \(y\) values, we identified \(a = 1\), \(b = -5\), and \(c = 8\). Plugging in these values helps us determine the roots of the quadratic equation. This equation typically provides two solutions, as evident in our example, where \(y = 2\) and \(y = 3\). These solutions can then be used to find the corresponding values of the other variable through back-substitution.

Solving Equations

The process of solving equations involves finding the values for the variables that make the equation true. In the context of a system of equations, it means finding a specific set of values that satisfies all equations in the system. Solving these involves:

• Isolating one variable in one of the equations.
• Substituting this variable back into the other equation.
• Simplifying to find the possible values for the variables.

For our example, after finding \(y = 2\) and \(y = 3\) from solving the quadratic equation, we substituted these values back into the original equation \(x + y = 2\) to find the corresponding \(x\) values:

• When \(y = 2\), \(x = 2 - 2 = 0\).
• When \(y = 3\), \(x = 2 - 3 = -1\).

Thus, the solutions for the system are the ordered pairs \((x, y) = (0, 2)\) and \((-1, 3)\). Solving systems of equations using substitution is practical and effective, especially if the equations are not too complex and one equation can be easily isolated. This method simplifies the solution process, making it a favorite choice for many students and educators alike.