Problem 14 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solve the system by the method of substitution. Check your solution graphically. $$\left\\{\begin{aligned} -2 x+y &=-5 \\ x^{2}+y^{2} &=25 \end{aligned}\right.$$

Step-by-Step Solution

Answer

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

Step 1: Solve one of the equations for one variable

In order to proceed with the substitution method, one of the equations must be solved for one variable. That equation can then be substituted into the other equation. The first equation \( -2x + y = -5 \) can be rewritten as \( y = 2x -5 \)

Step 2: Substitute y into the second equation

Replace y in the second equation \( x^{2}+y^{2} = 25 \) by the expression obtained in Step 1. The equation becomes \( x^{2}+(2x-5)^{2}=25 \)

Step 3: Solve the obtained equation

Next, solve the obtained equation. The equation becomes a quadratic equation: \(x^{2} + 4x^{2} - 20x + 25 = 25\), which simplifies to \(5x^{2} - 20x = 0\). Solve this equation to find x.

Step 4: Solve for x

The equation \(5x^{2} - 20x = 0\) can be factored to \(5x(x-4) = 0\). Setting each factor equal to zero gives the solutions, x = 0 and x = 4.

Step 5: Substitute each value of x into the first equation to solve for y

When substituting x=0 into the equation \(y = 2x -5\), we get \(y = -5\). When substituting \(x = 4\), it gives \(y = 3\). Thus the solutions of the system of equations are (0, -5) and (4, 3).

Step 6: Check your solution graphically

To check the solution graphically, plot the two equations on the same set of axes. The points where the two graphs intersect should correspond to the solutions obtained. In other words, the graphs of the equations should intersect at the points (0, -5) and (4, 3).

Key Concepts

Substitution Method

The substitution method is an effective technique for solving systems of equations. This method involves solving one of the equations for one variable and then substituting this expression into the other equation. Let's break this down with our example.

• We started with the equations: -2x + y = -5 and x² + y² = 25.
• The first equation can be rearranged to solve for y, giving us y = 2x - 5.
• Once we have y expressed in terms of x, it can be substituted back into the second equation in place of y.

This substitution helps us eliminate one variable, allowing us to solve for the remaining variable. In this example, it reduced the problem to a single equation in terms of x, which we could then solve using algebraic methods.

Quadratic Equations

With the substitution method, we often end up solving quadratic equations. A quadratic equation is generally in the form ax² + bx + c = 0. In our exercise, after substituting y = 2x - 5 into x² + y² = 25, the equation transformed into a quadratic: 5x² - 20x = 0.

• This equation can be factored as 5x(x - 4) = 0.
• Factorization is a useful algebraic technique where we express the quadratic equation as a product of simpler expressions.
• By setting each factor to zero, we find the roots of the equation, which gives x = 0 and x = 4.

These solutions give us the potential x-values for the intersection points of the initial system of equations.

Graphical Solution

The graphical solution offers a visual interpretation of the equations. By plotting both equations on a graph, you can clearly see where they intersect. This method is particularly useful to verify the answers obtained algebraically.

• The first equation −2x + y = −5 is a straight line.
• The second equation x² + y² = 25 is a circle centered at the origin with a radius of 5.

By plotting these, we expect the line to intersect the circle at the points that correspond to our solutions: (0, -5) and (4, 3). Observing these intersection points on the graph confirms that our calculated solutions are indeed correct.

Intersection Points

Intersection points in a system of equations are where the solutions are found. These are the coordinates that satisfy both equations simultaneously. In our example:

• The calculated intersection points were (0, -5) and (4, 3).
• These points are determined by solving the equations and substituting the values back to verify algebraic consistency in both equations.

For each solution, substituting x and y back into the original equations ensures they both hold true, confirming the solution's accuracy. Graphically, these points correspond to where the line intersects the circle on the plot. Hence, intersection points are crucial to verifying solutions both algebraically and graphically.