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Magazine Chapter 9: Problem 6
Solve each equation algebraically and graphically. Compare the solutions found using each method. a) \(\frac{3 x}{x-2}+5 x=\frac{x+4}{x-2}\) b) \(2 x+3=\frac{3 x^{2}+14 x+8}{x+4}\) c) \(\frac{6 x}{x-3}+3 x=\frac{2 x^{2}}{x-3}-5\) d) \(\frac{2 x-1}{x^{2}-x}+4=\frac{x}{x-1}\)
Short Answer
Expert verified
The solutions are: \(a) x = 2, -0.4\), \(b) x = 1, -4\), \(c) x=3, -5\), \(d) x = 1 \pm \frac{2\sqrt{3}}{3} \). Verify by graphing.
Step by step solution
01
Simplify Equation a
Given: \( \frac{3 x}{x-2}+5 x=\frac{x+4}{x-2} \). Combine into a single fraction on both sides: \( \frac{3x + 5x(x-2)}{x-2} = \frac{x + 4}{x-2} \). Simplify the left-hand side: \( \frac{3x + 5x^2 - 10x}{x-2} = \frac{x + 4}{x-2} \). Combine like terms: \( \frac{5x^2 - 7x}{x-2} = \frac{x + 4}{x-2} \). Now multiply both sides by \( x-2 \) and solve: 5x^2 - 7x = x + 4. Rearrange: 5x^2 - 8x - 4 = 0.
02
Solve Quadratic Equation
Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve. Here, \( a = 5 \), \( b = -8 \), and \( c = -4 \): \( x = \frac{8 \pm \sqrt{(-8)^2 - 4(5)(-4)}}{2(5)} \). Simplify: \( x = \frac{8 \pm \sqrt{64 + 80}}{10} \). Further simplify: \( x = \frac{8 \pm \sqrt{144}}{10} \). So, \( x = \frac{8 \pm 12}{10} \). Hence, the solutions are \( x = 2 \) or \( x = -0.4 \).
03
Graphical Solution for Part a
Graph both sides of the equation separately: Graph \( \frac{3x}{x-2} + 5x \) and \( \frac{x + 4}{x-2} \). Find the x-values where the two graphs intersect. These x-values are the solutions. Verify that the graphical solutions match the algebraic solutions \( x = 2 \) and \( x = -0.4 \).
04
Simplify Equation b
Given: \( 2 x + 3 = \frac{3 x^{2} + 14 x + 8}{x+4} \). Multiply both sides by \( x+4 \) to eliminate the denominator: \( (2x+3)(x+4) = 3x^2 + 14x + 8 \). Expand and simplify: \( 2x^2 + 3x + 8x + 12 = 3x^2 + 14x + 8 \). Combine like terms: \( 2x^2 + 11x + 12 = 3x^2 + 14x + 8 \). Subtract the right-hand side from both sides: \( -x^2 - 3x + 4 = 0 \) or \( x^2 + 3x - 4 = 0 \).
05
Solve Quadratic Equation
Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve. Here, \( a = 1 \), \( b = 3 \), and \( c = -4 \): \( x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-4)}}{2(1)} \). Simplify: \( x = \frac{-3 \pm \sqrt{9 + 16}}{2} \). Further simplify: \( x = \frac{-3 \pm \sqrt{25}}{2} \). So, \( x = \frac{-3 \pm 5}{2} \). Hence, the solutions are \( x = 1 \) and \( x = -4 \).
06
Graphical Solution for Part b
Graph both sides of the equation separately: Graph \( 2x + 3 \) and \( \frac{3x^{2} + 14x + 8}{x+4} \). Find the x-values where the two graphs intersect. These x-values are the solutions. Verify that the graphical solutions match the algebraic solutions \( x = 1 \) and \( x = -4 \).
07
Simplify Equation c
Given: \( \frac{6 x}{x-3}+3 x=\frac{2 x^{2}}{x-3}-5 \). Combine into a single fraction on both sides: \( \frac{6x + 3x(x-3)}{x-3} = \frac{2x^2 - 5(x-3)}{x-3} \). Simplify both sides: \( \frac{6x + 3x^2 - 9x}{x-3} = \frac{2x^2 - 5x + 15}{x-3} \). Combine like terms: \( \frac{3x^2 - 3x}{x-3} = \frac{2x^2 - 5x + 15}{x-3} \). Now multiply both sides by \( x-3 \) and solve: 3x^2 - 3x = 2x^2 - 5x + 15. Rearrange: x^2 + 2x - 15 = 0.
08
Solve Quadratic Equation
Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve. Here, \( a = 1 \), \( b = 2 \), and \( c = -15 \): \( x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-15)}}{2(1)} \). Simplify: \( x = \frac{-2 \pm \sqrt{4 + 60}}{2} \). Further simplify: \( x = \frac{-2 \pm \sqrt{64}}{2} \). So, \( x = \frac{-2 \pm 8}{2} \). Hence, the solutions are \( x = 3 \) and \( x = -5 \).
09
Graphical Solution for Part c
Graph both sides of the equation separately: Graph \( \frac{6x}{x-3} + 3x \) and \( \frac{2x^2}{x-3} - 5 \). Find the x-values where the two graphs intersect. These x-values are the solutions. Verify that the graphical solutions match the algebraic solutions \( x = 3 \) and \( x = -5 \).
10
Simplify Equation d
Given: \( \frac{2 x-1}{x^{2}-x}+4=\frac{x}{x-1} \). Find a common denominator for the fractions: Rewrite as \( \frac{2x-1 + 4(x^2-x)}{x^2-x} = \frac{x}{x-1} \). Multiply both sides by \( x^2 - x \): \( (2x-1 + 4(x^2 - x)) = x(x) \). Simplify: \( 2x - 1 + 4x^2 - 4x = x^2 \). Combine like terms: \( 3x^2 - 6x - 1 = 0 \).
11
Solve Quadratic Equation
Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve. Here, \( a = 3 \), \( b = -6 \), and \( c = -1 \): \( x = \frac{6 \pm \sqrt{(-6)^2 - 4(3)(-1)}}{2(3)} \). Simplify: \( x = \frac{6 \pm \sqrt{36 + 12}}{6} \). Further simplify: \( x = \frac{6 \pm \sqrt{48}}{6} \). Simplify further: \( x = \frac{6 \pm 4\sqrt{3}}{6} \) or \( x = 1 \pm \frac{2\sqrt{3}}{3} \). Hence, the solutions are \( x = 1 + \frac{2\sqrt{3}}{3} \) and \( x = 1 - \frac{2\sqrt{3}}{3} \).
12
Graphical Solution for Part d
Graph both sides of the equation separately: Graph \( \frac{2x-1}{x^2-x} + 4 \) and \( \frac{x}{x-1} \). Find the x-values where the two graphs intersect. These x-values are the solutions. Verify that the graphical solutions match the algebraic solutions \( x = 1 + \frac{2\sqrt{3}}{3} \) and \( x = 1 - \frac{2\sqrt{3}}{3} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equations
Solving rational equations often leads to quadratic equations. A quadratic equation is any equation that can be rearranged to the form \(ax^2 + bx + c = 0\). In this form, \(a\), \(b\), and \(c\) are constants where \(a e 0\). To find the values of \(x\), you can use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
If the equation simplifies to a quadratic form, applying this formula helps you find the exact solutions by substituting the specific values of \(a\), \(b\), and \(c\). The solutions to the quadratic equation are the points where the original rational equation is satisfied.
Understanding how to reformat an equation into this form and apply the quadratic formula is key to solving many algebraic problems. In the context provided, several of the rational equations reduce to quadratic forms which are then solved using this formula.
If the equation simplifies to a quadratic form, applying this formula helps you find the exact solutions by substituting the specific values of \(a\), \(b\), and \(c\). The solutions to the quadratic equation are the points where the original rational equation is satisfied.
Understanding how to reformat an equation into this form and apply the quadratic formula is key to solving many algebraic problems. In the context provided, several of the rational equations reduce to quadratic forms which are then solved using this formula.
Graphical Solutions
Graphical solutions involve plotting the equations on a coordinate plane. Each side of the rational equation is graphed separately, and the points where the graphs intersect represent the solutions.
When you graph the two sides of the rational equation, you are looking for the x-values where the y-values of the two graphs are equal. These x-values are the solutions to the equation. This technique provides a visual representation and verification of the solutions obtained algebraically.
For example, graphing \(\frac{3x}{x-2} + 5x\) and \(\frac{x + 4}{x-2}\) will show intersections at \(x = 2\) and \(x = -0.4\), confirming the algebraic solutions. Using a graphing calculator or software can make this process faster and more accurate. It helps in verifying solutions and understanding the behavior of the functions visually.
When you graph the two sides of the rational equation, you are looking for the x-values where the y-values of the two graphs are equal. These x-values are the solutions to the equation. This technique provides a visual representation and verification of the solutions obtained algebraically.
For example, graphing \(\frac{3x}{x-2} + 5x\) and \(\frac{x + 4}{x-2}\) will show intersections at \(x = 2\) and \(x = -0.4\), confirming the algebraic solutions. Using a graphing calculator or software can make this process faster and more accurate. It helps in verifying solutions and understanding the behavior of the functions visually.
Algebraic Simplification
Algebraic simplification involves transforming equations into a simpler form to make solving them easier. This can involve combining like terms, eliminating fractions by multiplying by common denominators, and factoring expressions.
Consider the expression \(\frac{6x}{x-3} + 3x = \frac{2x^2}{x-3} - 5\). Simplify by combining the fractions: \(\frac{6x + 3x(x-3)}{x-3} = \frac{2x^2 - 5(x-3)}{x-3}\). Simplify both sides to \(\frac{3x^2 - 3x}{x-3} = \frac{2x^2 - 5x + 15}{x-3}\). This process removes the fractions and allows you to solve the equation algebraically.
Simplifying makes the underlying structure of the equation clearer, enabling the application of techniques like the quadratic formula to find the solutions. It's essential to perform each step carefully to avoid mistakes and ensure the solutions are correct.
Consider the expression \(\frac{6x}{x-3} + 3x = \frac{2x^2}{x-3} - 5\). Simplify by combining the fractions: \(\frac{6x + 3x(x-3)}{x-3} = \frac{2x^2 - 5(x-3)}{x-3}\). Simplify both sides to \(\frac{3x^2 - 3x}{x-3} = \frac{2x^2 - 5x + 15}{x-3}\). This process removes the fractions and allows you to solve the equation algebraically.
Simplifying makes the underlying structure of the equation clearer, enabling the application of techniques like the quadratic formula to find the solutions. It's essential to perform each step carefully to avoid mistakes and ensure the solutions are correct.
Common Denominators
Finding a common denominator is crucial when working with rational equations. This process involves identifying a shared base in the denominators and using it to combine fractions.
For example, in the equation \(\frac{2x-1}{x^2-x}+4=\frac{x}{x-1}\), find a common denominator for the fractions. Rewrite as: \(\frac{2x-1 + 4(x^2-x)}{x^2-x} = \frac{x}{x-1}\). Multiply both sides by \(x^2 - x\) to eliminate denominators. This method simplifies the equation, making it easier to solve.
Common denominators help in combining fractions and eliminating complex fractions from the equations. This step is necessary for clearing denominators and transitioning to simpler polynomial or quadratic forms before applying further algebraic techniques.
For example, in the equation \(\frac{2x-1}{x^2-x}+4=\frac{x}{x-1}\), find a common denominator for the fractions. Rewrite as: \(\frac{2x-1 + 4(x^2-x)}{x^2-x} = \frac{x}{x-1}\). Multiply both sides by \(x^2 - x\) to eliminate denominators. This method simplifies the equation, making it easier to solve.
Common denominators help in combining fractions and eliminating complex fractions from the equations. This step is necessary for clearing denominators and transitioning to simpler polynomial or quadratic forms before applying further algebraic techniques.
