Problem 8 - 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) $$ \left(\begin{array}{l} x+y=4 \\ x^{2}+y^{2}=10 \end{array}\right) $$

Step-by-Step Solution

Answer

The solutions to the system are \((3, 1)\) and \((1, 3)\).

Step 1: Understand the System of Equations

We have two equations here:1. Linear Equation: \( x+y=4 \)2. Non-linear Equation (Circle): \( x^2+y^2=10 \) Our goal is to find the values of \( x \) and \( y \) that satisfy both equations.

Step 2: Graph the Equations

To graph the first equation, \( x+y=4\), rewrite it in the slope-intercept form \( y=-x+4 \). This is a line with y-intercept 4 and slope -1.The second equation, \( x^2+y^2=10 \), is a circle centered at the origin with radius \( \sqrt{10} \).Graph both on the same coordinate plane and look for intersection points, which are the approximate real number solutions.

Step 3: Solve Using Substitution

Solve the linear equation \( x+y=4 \) for \( y \):\[ y = 4 - x \]Substitute \( y = 4 - x \) into the circle's equation:\[ x^2 + (4-x)^2 = 10 \]Expand and simplify:\[ x^2 + 16 - 8x + x^2 = 10 \]\[ 2x^2 - 8x + 16 = 10 \]Simplify further:\[ 2x^2 - 8x + 6 = 0 \]

Step 4: Solve the Quadratic Equation

Divide the entire equation by 2 to simplify:\[ x^2 - 4x + 3 = 0 \]Factor the quadratic:\[ (x-3)(x-1) = 0 \]Set each factor to zero:1. \( x - 3 = 0 \rightarrow x = 3 \)2. \( x - 1 = 0 \rightarrow x = 1 \)

Step 5: Determine y-values

Substitute back to find \( y \):1. If \( x = 3 \), then \( y = 4 - 3 = 1 \).2. If \( x = 1 \), then \( y = 4 - 1 = 3 \).Thus, we have two solutions: \((3, 1)\) and \((1, 3)\).

Step 6: Verify the Solutions

Substitute both solutions into the circle equation to verify.For \((3, 1)\):\[ 3^2 + 1^2 = 9 + 1 = 10 \]For \((1, 3)\):\[ 1^2 + 3^2 = 1 + 9 = 10 \]Both satisfy the circle equation, confirming the solutions are correct.

Key Concepts

Graphical Solutions

When solving systems of equations, a graphical solution provides a visual way to find where the equations intersect. For our system consisting of a linear equation and a nonlinear equation, plotting them reveals the shared points in the coordinate plane.

Consider the linear equation, \( x + y = 4 \), which can be represented as a straight line. By rewriting this as \( y = -x + 4 \), we see the line crosses the y-axis at 4 and slopes downward at 45 degrees.

The equation \( x^2 + y^2 = 10 \) describes a circle centered at the origin with a radius of \( \sqrt{10} \).

By graphing both, you see where the line meets the circle. These intersection points approximate the real number solutions for our equations. It makes it visually obvious and intuitive to notice potential answers.

Substitution Method

The substitution method involves solving one of the equations for one variable and inserting it into the other equation. This allows us to reduce a system into a single equation with one unknown.

Begin with the linear equation: \( x + y = 4 \). Solve for \( y \), giving \( y = 4 - x \). This expression is then substituted into the circle's equation, \( x^2 + y^2 = 10 \), causing it to become \( x^2 + (4 - x)^2 = 10 \).

Expanding and simplifying this yields the quadratic \( 2x^2 - 8x + 6 = 0 \).

The substitution method efficiently transforms a pair of equations into one, often simplifying subsequent steps. It's particularly useful when one equation is linear.

Elimination Method

The elimination method strategically cancels one variable by adding or subtracting equations. Though not used directly in this exercise, understanding its purpose is valuable.

Envision manipulating both equations until one variable has opposite coefficients in both, allowing it to cancel when the equations are combined.

Through elimination, anything that might have been cumbersome about manual substitutions is circumvented, especially if equations are solely linear and cancellation is straightforward.

This method is a boon when dealing with multiple linear equations and provides a clear path to the solutions.

Linear Equations

Linear equations create straight lines when graphed and have no exponent greater than 1 on any variable. They are foundational to solving systems geometrically and analytically.

In our exercise, \( x + y = 4 \) represents such an equation. Written in slope-intercept form, \( y = -x + 4 \), it shows a line slanting downwards through the point (0, 4).

Linear equations are predictable and consistent, making them simple to work with, be it by substitution, elimination, or graphing. Every point on their line satisfies the equation.

Nonlinear Equations

Nonlinear equations, like \( x^2 + y^2 = 10 \), include variables raised to powers greater than one. This often results in curvy graphs, like circles or parabolas.

The equation for our circle is centered at the origin with a radius \( \sqrt{10} \). Unlike lines, nonlinear graphs aren't straight and thus often have variable outcomes or intersection points.

Solving systems involving such equations requires careful procedural application, often leading you to graphical solutions or algebraic manipulation as used in substitution. Nonlinear equations introduce diversity and complexity to problem-solving.