Solving Algebraic Equations And Inequalities A Step By Step Guide

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This comprehensive article provides step-by-step solutions for various algebraic equations and inequalities, focusing on solving for the variable 'x'. Understanding how to manipulate and solve these types of problems is crucial for success in mathematics. We will explore different techniques, including cross-multiplication, clearing denominators, and solving inequalities. By working through these examples, you will gain a solid foundation in algebraic problem-solving.

1. Solving a Linear Equation with Fractions

Solving linear equations often involves dealing with fractions, which can seem intimidating at first. However, by applying the right techniques, we can simplify these equations and find the value of the unknown variable. Our first equation is rac{2w+5}{5} = rac{8}{15}. To solve for w, we will use cross-multiplication, a method that allows us to eliminate the fractions and create a simpler equation. Cross-multiplication involves multiplying the numerator of the first fraction by the denominator of the second fraction and vice versa. This method is based on the fundamental principle that if two fractions are equal, their cross products are equal. In this case, we'll multiply (2w + 5) by 15 and 5 by 8. This step effectively removes the fractions, making the equation easier to solve. Once we've cross-multiplied, we'll have a linear equation without fractions, which we can then solve using standard algebraic techniques such as distribution, combining like terms, and isolating the variable. The key to mastering these types of equations is understanding the properties of equality and how to manipulate equations while maintaining their balance. Remember, whatever operation you perform on one side of the equation, you must also perform on the other side to ensure the equation remains true. This principle is the foundation of all algebraic manipulations and is essential for solving equations accurately.

Step-by-step Solution:

  1. Cross-multiply: (2w+5)×15=8×5{(2w + 5) \times 15 = 8 \times 5}
  2. Expand: 30w+75=40{30w + 75 = 40}
  3. Subtract 75 from both sides: 30w=40−75{30w = 40 - 75} 30w=−35{30w = -35}
  4. Divide both sides by 30: w=−3530{w = \frac{-35}{30}}
  5. Simplify the fraction: w=−76{w = -\frac{7}{6}}

Therefore, the solution to the equation is w = -\frac{7}{6}.

2. Solving a Rational Equation with a Quadratic Term

Rational equations involve fractions where the variable appears in the denominator, making them slightly more complex than simple linear equations. The equation we'll tackle here is rac{x+1}{4} = rac{3}{16x}. To solve this, we'll again use cross-multiplication to eliminate the fractions. This step is crucial as it transforms the rational equation into a more manageable form. After cross-multiplying, we'll obtain an equation that involves a quadratic term, which means we'll likely need to rearrange the equation into a standard quadratic form (ax^2 + bx + c = 0) to solve it effectively. Once the equation is in this standard form, we can employ several methods to find the solutions, including factoring, using the quadratic formula, or completing the square. Factoring is often the quickest method if the quadratic expression can be easily factored. However, the quadratic formula is a universal tool that works for any quadratic equation, regardless of whether it can be factored. The key is to identify the coefficients a, b, and c from the standard form and substitute them into the formula. Remember that quadratic equations can have up to two distinct solutions, reflecting the two possible values of x that satisfy the equation. Solving rational equations requires careful attention to detail, especially when dealing with potential extraneous solutions. These are solutions that arise during the solving process but do not satisfy the original equation. It's always essential to check your solutions by substituting them back into the original equation to ensure they are valid.

Step-by-step Solution:

  1. Cross-multiply: (x+1)×16x=3×4{(x + 1) \times 16x = 3 \times 4}
  2. Expand: 16x2+16x=12{16x^2 + 16x = 12}
  3. Rearrange to the standard quadratic form: 16x2+16x−12=0{16x^2 + 16x - 12 = 0}
  4. Divide the entire equation by 4 to simplify: 4x2+4x−3=0{4x^2 + 4x - 3 = 0}
  5. Factor the quadratic equation: (2x+3)(2x−1)=0{(2x + 3)(2x - 1) = 0}
  6. Set each factor equal to zero and solve for x: 2x+3=0⇒x=−32{2x + 3 = 0 \Rightarrow x = -\frac{3}{2}} 2x−1=0⇒x=12{2x - 1 = 0 \Rightarrow x = \frac{1}{2}}

Therefore, the solutions to the equation are x = -\frac{3}{2} and x = \frac{1}{2}.

3. Solving a Rational Equation with Multiple Fractions

Rational equations often involve multiple fractions, and the equation rac{12}{5y} - rac{2}{5} = rac{2}{y} is a prime example. To solve for y, we'll first need to eliminate the fractions, which can be achieved by finding the least common denominator (LCD) of all the fractions in the equation. In this case, the denominators are 5y, 5, and y. The LCD is the smallest multiple that all the denominators divide into evenly, which in this instance is 5y. Once we've identified the LCD, we'll multiply every term in the equation by it. This step is crucial because it clears the fractions, transforming the equation into a more manageable form without denominators. After multiplying through by the LCD, we'll simplify the equation by performing any necessary algebraic operations, such as combining like terms and isolating the variable y. Remember, the goal is to get y by itself on one side of the equation. As with all rational equations, it's essential to check our solutions to ensure they are valid. We'll substitute the values we find for y back into the original equation to see if they satisfy it. This step is particularly important because sometimes we can obtain solutions that don't actually work in the original equation; these are known as extraneous solutions. By verifying our solutions, we can avoid making errors and ensure we have the correct answers.

Step-by-step Solution:

  1. Identify the least common denominator (LCD): The LCD is 5y.
  2. Multiply both sides of the equation by the LCD: 5y(125y−25)=5y(2y){5y \left( \frac{12}{5y} - \frac{2}{5} \right) = 5y \left( \frac{2}{y} \right)}
  3. Distribute and simplify: 12−2y=10{12 - 2y = 10}
  4. Add 2y to both sides and subtract 10 from both sides: 12−10=2y{12 - 10 = 2y} 2=2y{2 = 2y}
  5. Divide both sides by 2: y=1{y = 1}

Therefore, the solution to the equation is y = 1.

4. Solving a Rational Inequality

Rational inequalities, like the one we're about to solve, rac{2x-7}{x+1} ≤{\leq} -7, involve comparing a rational expression to a value. To solve these, we first need to manipulate the inequality so that one side is zero. This is a crucial step because it allows us to identify the critical points of the inequality. Critical points are the values of x that make the expression equal to zero or undefined. They are the points where the expression might change its sign, which is essential for solving the inequality. Once we have the inequality in the form where one side is zero, we'll find the critical points by setting both the numerator and the denominator of the rational expression equal to zero and solving for x. These critical points divide the number line into intervals, and the sign of the expression within each interval remains constant. To determine the solution to the inequality, we'll test a value from each interval in the original inequality. If the test value satisfies the inequality, then the entire interval is part of the solution. If the test value does not satisfy the inequality, then the interval is not part of the solution. Remember to pay close attention to the inequality sign. If the inequality includes an equals sign (≤{\leq} or ≥{\geq}), then the critical points that make the expression equal to zero are included in the solution. However, critical points that make the denominator zero are never included in the solution, as they make the expression undefined. Expressing the solution in interval notation is a concise way to represent all the values of x that satisfy the inequality.

Step-by-step Solution:

  1. Add 7 to both sides to get zero on one side: 2x−7x+1+7≤0{\frac{2x - 7}{x + 1} + 7 \leq 0}
  2. Find a common denominator and combine terms: 2x−7+7(x+1)x+1≤0{\frac{2x - 7 + 7(x + 1)}{x + 1} \leq 0} 2x−7+7x+7x+1≤0{\frac{2x - 7 + 7x + 7}{x + 1} \leq 0} 9xx+1≤0{\frac{9x}{x + 1} \leq 0}
  3. Find the critical points: Set the numerator and the denominator equal to zero: 9x=0⇒x=0{9x = 0 \Rightarrow x = 0} x+1=0⇒x=−1{x + 1 = 0 \Rightarrow x = -1}
  4. Create a sign chart using the critical points: The critical points are -1 and 0. We test the intervals (-∞, -1), (-1, 0), and (0, ∞).
    • For (-∞, -1), test x = -2: 9(−2)−2+1=−18−1=18>0{\frac{9(-2)}{-2 + 1} = \frac{-18}{-1} = 18 > 0} (not part of the solution)
    • For (-1, 0), test x = -0.5: 9(−0.5)−0.5+1=−4.50.5=−9<0{\frac{9(-0.5)}{-0.5 + 1} = \frac{-4.5}{0.5} = -9 < 0} (part of the solution)
    • For (0, ∞), test x = 1: 9(1)1+1=92>0{\frac{9(1)}{1 + 1} = \frac{9}{2} > 0} (not part of the solution)
  5. Determine the solution interval: The inequality is satisfied in the interval (-1, 0]. We include 0 because the inequality is less than or equal to, but we exclude -1 because it makes the denominator zero.

Therefore, the solution to the inequality is -1 < x ≤ 0, which can be written in interval notation as (-1, 0].

In this article, we've explored various types of algebraic equations and inequalities and provided detailed solutions for each. From solving linear equations with fractions to tackling rational inequalities, the key to success lies in understanding the fundamental principles of algebra and applying them systematically. Remember to always check your solutions, especially when dealing with rational equations and inequalities, to avoid extraneous solutions. By mastering these techniques, you'll be well-equipped to solve a wide range of algebraic problems.