Problem 35 - Algebra and Trigonometry

Question

\(33-40=\) Use the graphical method to find all solutions of the system of equations, rounded to two decimal places. $$ \begin{array}{l}{x^{2}+y^{2}=25} \\ {x+3 y=2}\end{array} $$

Step-by-Step Solution

Answer

Solutions are \((-4.51, 2.17)\) and \((4.91, -0.97)\).

Step 1: Understand the equations

We have two equations. The first one, \(x^2 + y^2 = 25\), is the equation of a circle with a radius of 5 centered at the origin (0,0). The second one, \(x + 3y = 2\), is a linear equation representing a straight line.

Step 2: Graph the circle

Graph the circle defined by \(x^2 + y^2 = 25\). This circle is centered at the origin with a radius of 5. We can plot points to ensure accuracy, like (5,0), (0,5), (-5,0), and (0,-5).

Step 3: Graph the line

Graph the line \(x + 3y = 2\). To do this, find the intercepts: when \(x=0\), \(3y=2\) which gives \(y=\frac{2}{3}\); when \(y=0\), \(x=2\). Plot these intercepts on the graph and draw the line.

Step 4: Find intersection points

Determine where the line \(x + 3y = 2\) intersects the circle \(x^2 + y^2 = 25\). These intersection points are the solutions to the system of equations.

Step 5: Calculate solutions using substitution

Substitute \(x = 2 - 3y\) into \(x^2 + y^2 = 25\) to get:\((2 - 3y)^2 + y^2 = 25\) which simplifies to:\(4 - 12y + 9y^2 + y^2 = 25\) leading to:\(10y^2 - 12y - 21 = 0\).Solve this quadratic equation using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

Step 6: Solve using quadratic formula

For the equation \(10y^2 - 12y - 21 = 0\), coefficients are \(a=10\), \(b=-12\), \(c=-21\). Plug these into the quadratic formula:\[y = \frac{12 \pm \sqrt{(-12)^2 - 4(10)(-21)}}{2(10)}\]Simplify to find the values of \(y\).

Step 7: Compute y-values

Calculate the discriminant: \((-12)^2 - 4(10)(-21) = 144 + 840 = 984\). Thus:\[y = \frac{12 \pm \sqrt{984}}{20}\].\(\sqrt{984} \approx 31.36\), so:\[y_1 = \frac{12 + 31.36}{20} \approx 2.17\]\[y_2 = \frac{12 - 31.36}{20} \approx -0.97\].

Step 8: Find corresponding x-values

For each found \(y\), substitute back into \(x = 2 - 3y\):For \(y_1 = 2.17\), \(x = 2 - 3(2.17) \approx -4.51\).For \(y_2 = -0.97\), \(x = 2 - 3(-0.97) \approx 4.91\).

Key Concepts

System of Equations

In mathematics, a **system of equations** involves two or more equations that are solved simultaneously to find common solutions. In our scenario, we have two equations: a circle, represented by \(x^2 + y^2 = 25\), and a line, defined by \(x + 3y = 2\). Each equation describes a different geometric shape. The essence of solving a system of equations is to find a point or points where the graphs of these equations intersect. This is where the graphical method shines, offering a visual approach to verifying potential solutions.

Intersection Points

**Intersection points** are the solutions to a system of equations where the graphs of these equations meet. When graphing the circle \(x^2 + y^2 = 25\) and the line \(x + 3y = 2\), the intersections are where both conditions are satisfied simultaneously. Finding intersection points graphically may not always be precise, especially if solutions involve irrational numbers, but it provides insight into the possible solutions. To verify the intersections, we fine-tune the answer using algebraic methods such as substitution or elimination.

Quadratic Formula

The **quadratic formula** is a powerful tool in algebra used to solve quadratic equations of the form \(ax^2 + bx + c = 0\). In our solution steps, after substituting from the line equation into the circle equation, we end up with a quadratic equation: \(10y^2 - 12y - 21 = 0\). This is where the quadratic formula comes into play:

• Identify the coefficients: \(a = 10\), \(b = -12\), and \(c = -21\).
• Plug them into the formula: \[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

By computing the discriminant part \(b^2 - 4ac\), we determine the nature of the roots. For this case, the discriminant is positive, indicating two distinct real solutions for \(y\).

Circle and Line Intersection

Intersections between a **circle and a line** can be fascinating because they often result in multiple potential points of contact. Our circle described by \(x^2 + y^2 = 25\) and line given by \(x + 3y = 2\) describe different paths within a coordinate plane. To find where they intersect:

• Substitute one equation into the other to eliminate variables, enabling us to solve for the remaining ones.
• The resulting quadratic equation gives the \(y\)-values, and we substitute back to find corresponding \(x\)-values.
• These points \((-4.51, 2.17)\) and \((4.91, -0.97)\) represent the exact intersection locations between the circle and line.

Understanding these intersections helps in visualizing the shared solutions of the equations in the coordinate system.