91Ó°ÊÓ

Graphing equations is a visual way to understand their solutions. By plotting the equations on a graph, the intersections of the graphs represent the solutions to the system. Let’s take the example from our exercise:

For the equation \( y = \sqrt{x} \), you start by selecting values for \( x \) and computing the corresponding \( y \) values. For instance, if \( x = 0 \), then \( y = 0 \). Similarly, if \( x = 1 \), then \( y = 1 \), and so on. Plot these points, like \( (0,0), (1,1), (4,2), (9,3) \), and draw a smooth curve through them. This gives you the graph of \( y = \sqrt{x} \), which looks like the upper half of a sideways parabola opening to the right.

Next, graph the second equation \( x^{2} + y^{2} = 20 \). This equation represents a circle with its center at the origin (0,0) and a radius of \( \sqrt{20} \), roughly 4.47. To graph it, identify the intercepts where the circle cuts the axes: \( (\sqrt{20}, 0), (-\sqrt{20}, 0), (0, \sqrt{20}), (0, -\sqrt{20}) \). Connect these points to form the circle.

• By graphing both equations, you visually see their intersections, indicating potential solutions.
• For a clearer understanding, label the graphs and highlight the intersection points.

The substitution method is used to solve systems of equations by replacing one variable with its equivalent expression involving the other variable. Here's how it works with our example:

Start by expressing one variable in terms of the other from one of the equations. We already have \( y = \sqrt{x} \). This is our first step.

Next, substitute this expression into the second equation: \( x^{2} + y^{2} = 20 \). Replace \( y \) with \( \sqrt{x} \). We get:

• \( x^{2} + (\sqrt{x})^{2} = 20 \)
• Simplify to \( x^{2} + x = 20 \)
• Rearrange to form a standard quadratic equation: \( x^{2} + x - 20 = 0 \).

Now solve this quadratic equation, which leads to finding values for \( x \). The substitution method is effective because it reduces the system to a single equation with one variable, making it easier to solve. Once you find the value of \( x \), substitute it back to find the corresponding \( y \) value.

Quadratic equations are equations that can be written in the form \( ax^{2} + bx + c = 0 \). They often have two solutions, which can be real or complex. Let’s take a closer look using our example:

From our substitution method, we derived the quadratic equation \( x^{2} + x - 20 = 0 \). To solve this, we use the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \)
• Here, \( a = 1 \), \( b = 1 \), and \( c = -20 \).
• Substitute these values into the formula to get:
  • \( x = \frac{-1 \pm \sqrt{1 + 80}}{2} \)
  • which simplifies to \( x = \frac{-1 \pm 9}{2} \)
  • This results in two possible values: \( x = 4 \) and \( x = -5 \).

Since our context requires non-negative \( x \) values (square roots of negative numbers are not real in this setting), we discard \( x = -5 \) and keep x = 4.

Next, substitute \( x = 4 \) back into the original equation \( y = \sqrt{x} \) to find \( y = 2 \). Thus, the solution to our system is \( (4, 2) \).

• Quadratic equations are central in many math problems, and mastering them is key to solving various systems encountered in algebra.