Problem 26 - Intermediate Algebra

Question

For Problems \(1-28\), (a) graph each system so that approximate real number solutions (if there are any) can be predicted, and (b) solve each system using the substitution method or the elimination-by-addition method. (Objectives 1 and 2) $$ \left(\begin{array}{l} y=-x^{2}+1 \\ x+y=2 \end{array}\right) $$

Step-by-Step Solution

Answer

The approximate real number solutions are where \( y = -x^2 + 1 \) and \( x + y = 2 \) intersect, calculated using substitution.

Step 1: Write Down the System of Equations

The system of equations is given as: 1. \( y = -x^2 + 1 \) 2. \( x + y = 2 \)

Step 2: Graph the Equations

To graph the system, we need to plot both equations on the same coordinate plane. - The first equation, \( y = -x^2 + 1 \), is a parabola opening downward with its vertex at (0,1). - The second equation, \( x + y = 2 \), is a straight line with a slope of -1 and a y-intercept of 2. This line crosses the y-axis at (0,2) and the x-axis at (2,0). Plot these two and find where they intersect approximately.

Step 3: Use Substitution Method

Since we already have \( y \) expressed in terms of \( x \) in the first equation \( y = -x^2 + 1 \), substitute this expression into the second equation: \( x + (-x^2 + 1) = 2 \)

Step 4: Simplify and Solve for x

Simplify the equation from Step 3:\( x - x^2 + 1 = 2 \)Re-arrange terms and simplify: \( -x^2 + x + 1 = 2 \)\( -x^2 + x - 1 = 0 \)

Step 5: Solve the Quadratic Equation for x

The equation \( -x^2 + x - 1 = 0 \) can be solved using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = -1 \), \( b = 1 \), and \( c = -1 \). Substitute these values:\[ x = \frac{-1 \pm \sqrt{1 - 4(1)(-1)}}{2(-1)} \]\[ x = \frac{-1 \pm \sqrt{5}}{-2} \]Thus, the solutions for \( x \) are \( x = \frac{1 + \sqrt{5}}{2} \) and \( x = \frac{1 - \sqrt{5}}{2} \).

Step 6: Find Corresponding y-values

Substitute the values of \( x \) back into the first equation \( y = -x^2 + 1 \) to find \( y \): - For \( x = \frac{1 + \sqrt{5}}{2} \): \( y = -\left(\frac{1 + \sqrt{5}}{2}\right)^2 + 1 \)- For \( x = \frac{1 - \sqrt{5}}{2} \): \( y = -\left(\frac{1 - \sqrt{5}}{2}\right)^2 + 1 \)

Step 7: Calculate the y-values

Perform the calculations for each x-value:- For \( x = \frac{1 + \sqrt{5}}{2} \), if calculated correctly, \( y \approx -\frac{4}{4} + 1 \approx 0 \)- For \( x = \frac{1 - \sqrt{5}}{2} \), if calculated correctly, \( y \approx -\frac{4}{4} + 1 \approx 1 \)Thus, the approximate solutions are at these corresponding points.

Step 8: Verify the Solutions

Ensure that the estimated values satisfy both equations:Plug the obtained pairs \( (x, y) \) into both original equations to verify they satisfy them. Each pair should satisfy the condition of both equations.

Key Concepts

Graphing Equations

Graphing equations is a crucial method to visualize solutions to systems of equations. In this context, we work with two types of graphs: a quadratic equation and a linear equation.

• The quadratic equation given by \( y = -x^2 + 1 \) is a parabola. Parabolas are U-shaped curves which can open upwards or downwards. Here, the parabola opens downward, with its vertex at point \( (0, 1) \).
• The linear equation \( x + y = 2 \) represents a straight line. This line has a slope of -1, indicating it descends one unit vertically for each unit it moves horizontally to the right. It crosses the y-axis at 2 (\((0, 2)\)) and the x-axis at 2 (\((2, 0)\)).

To find solutions for the system of equations, plot both graphs on the same coordinate plane. The solution corresponds to the intersection point(s) of the parabola and the line. These intersection points can give approximate values for \( x \) and \( y \) when graphed accurately.

Substitution Method

The substitution method is a fundamental algebraic technique to solve a system of equations. It involves solving one equation for one variable, then substituting that expression into another equation.

• For the given system, the first equation \( y = -x^2 + 1 \) is already solved for \( y \).
• The next step is to substitute this equation for \( y \) into the second equation \( x + y = 2 \), resulting in a single equation in terms of \( x \): \( x + (-x^2 + 1) = 2 \).

After substituting, simplify and manipulate the equation to find possible values for \( x \). This method is very systematic and ensures accuracy in finding the correct solution as it leverages the relationship between the variables in both equations.

Elimination Method

The elimination method can also be used to solve a system of equations, though it’s not detailed in our example here. This method involves aligning the equations so that adding them cancels out one of the variables. It's a systematic way to eliminate one variable, making it easier to solve for the other.

• First, both equations are usually manipulated (multiplied, if needed) to ensure one variable will cancel out when the equations are summed or subtracted from each other.
• Once one variable is removed, solving for the remaining variable becomes simpler, followed by back-substitution to find the value of the eliminated variable.

While not used extensively in solving the given system, understanding elimination complements your overall skills in handling systems of equations.

Quadratic Equation

The quadratic equation forms the foundation of many algebraic problems.

• In this situation, the equation \( -x^2 + x - 1 = 0 \) describes a quadratic trend within the system. Such equations have the general form \( ax^2 + bx + c = 0 \).
• To solve the quadratic, you can use the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In the given equation, the coefficients are \( a = -1 \), \( b = 1 \), and \( c = -1 \). By placing these into the quadratic formula, the solutions for \( x \) are found. Solving a quadratic equation is often built into larger systems and provides precise solutions, helping confirm any graphical estimations.