Solve The Equation For The Indicated Variable

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Greels

Apr 18, 2025 · 6 min read

Solve The Equation For The Indicated Variable
Solve The Equation For The Indicated Variable

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    Solve the Equation for the Indicated Variable: A Comprehensive Guide

    Solving equations for a specific variable is a fundamental skill in algebra and a crucial step in various mathematical and scientific applications. This comprehensive guide will walk you through different techniques and strategies to master this skill, covering a range of equation types and complexities. We'll explore the underlying principles, offer practical examples, and equip you with the confidence to tackle even the most challenging problems.

    Understanding the Fundamentals

    Before diving into specific techniques, let's solidify our understanding of the basic principles. Solving an equation for a specific variable means isolating that variable on one side of the equation, expressing it in terms of the other variables and constants. This involves manipulating the equation using algebraic operations, always ensuring that we maintain the equality. The key principle to remember is that whatever operation you perform on one side of the equation, you must perform the same operation on the other side to maintain balance.

    Core Algebraic Operations

    The core algebraic operations used in solving equations include:

    • Addition and Subtraction: Adding or subtracting the same value from both sides of the equation.
    • Multiplication and Division: Multiplying or dividing both sides of the equation by the same non-zero value.
    • Distributive Property: Expanding expressions by multiplying a term with each term inside parentheses. (a(b + c) = ab + ac)
    • Exponents and Roots: Using exponents to simplify expressions or applying roots to eliminate exponents.

    Solving Linear Equations

    Linear equations are equations where the highest power of the variable is 1. These are generally the simplest to solve.

    Example 1: Simple Linear Equation

    Solve for x: 2x + 5 = 11

    1. Subtract 5 from both sides: 2x + 5 - 5 = 11 - 5 => 2x = 6
    2. Divide both sides by 2: 2x / 2 = 6 / 2 => x = 3

    Example 2: Linear Equation with Multiple Variables

    Solve for y: 3x + 2y = 10

    1. Subtract 3x from both sides: 3x + 2y - 3x = 10 - 3x => 2y = 10 - 3x
    2. Divide both sides by 2: 2y / 2 = (10 - 3x) / 2 => y = 5 - (3/2)x

    Solving Quadratic Equations

    Quadratic equations are equations where the highest power of the variable is 2. These require more sophisticated techniques.

    Example 3: Solving a Quadratic Equation for x

    Solve for x: x² + 5x + 6 = 0

    This equation can be factored: (x + 2)(x + 3) = 0

    Therefore, the solutions are x = -2 and x = -3.

    Alternatively, the quadratic formula can be used:

    x = (-b ± √(b² - 4ac)) / 2a where the equation is in the form ax² + bx + c = 0

    For our example, a = 1, b = 5, c = 6. Substituting these values into the quadratic formula will yield the same solutions, x = -2 and x = -3.

    Example 4: Solving a Quadratic Equation for a Different Variable

    Solve for y: x = 2y² + 4y - 6

    This problem requires rearranging the equation to the standard quadratic form before applying the quadratic formula or factoring. It's crucial to note that in this instance, the solution for y will be in terms of x. Rearranging the equation, we get:

    2y² + 4y - (6 + x) = 0

    Now, we can apply the quadratic formula, where a = 2, b = 4, and c = -(6 + x). The solution will be a function of x:

    y = (-4 ± √(16 - 4 * 2 * -(6 + x))) / 4

    Simplifying, we obtain:

    y = (-4 ± √(16 + 48 + 8x)) / 4 = (-4 ± √(64 + 8x)) / 4 = (-1 ± √(16 + 2x)) / 1

    Therefore, y = -1 ± √(16 + 2x).

    Solving Equations with Fractions

    Equations with fractions require careful manipulation to eliminate the denominators.

    Example 5: Equation with Fractions

    Solve for x: (x/2) + (x/3) = 5

    1. Find a common denominator: The common denominator of 2 and 3 is 6.
    2. Multiply both sides by the common denominator: 6 * ((x/2) + (x/3)) = 6 * 5
    3. Simplify: 3x + 2x = 30
    4. Combine like terms: 5x = 30
    5. Solve for x: x = 6

    Solving Equations with Radicals

    Equations containing radicals (square roots, cube roots, etc.) require careful handling.

    Example 6: Equation with a Square Root

    Solve for x: √(x + 2) = 3

    1. Square both sides: (√(x + 2))² = 3² => x + 2 = 9
    2. Solve for x: x = 7 Important Note: Always check your solution(s) to ensure they don't introduce extraneous solutions (solutions that don't satisfy the original equation). In this case, substituting x = 7 back into the original equation confirms that it is a valid solution.

    Example 7: Equation with Multiple Radicals

    Solve for x: √(x + 1) + √(x - 1) = 2

    This requires more strategic manipulation. Let's isolate one radical:

    1. Subtract √(x - 1) from both sides: √(x + 1) = 2 - √(x - 1)
    2. Square both sides: (√(x + 1))² = (2 - √(x - 1))² => x + 1 = 4 - 4√(x - 1) + x - 1
    3. Simplify and isolate the remaining radical: 4√(x - 1) = 2
    4. Divide by 4: √(x - 1) = 1/2
    5. Square both sides: x - 1 = 1/4
    6. Solve for x: x = 5/4 Again, check the solution in the original equation.

    Solving Systems of Equations

    Solving for a variable often involves working with systems of equations. These systems can be solved using various methods, including substitution and elimination.

    Example 8: System of Linear Equations

    Solve for x and y:

    x + y = 5 x - y = 1

    Method 1: Elimination

    Add the two equations together: 2x = 6 => x = 3 Substitute x = 3 into either equation to solve for y: 3 + y = 5 => y = 2

    Method 2: Substitution

    Solve the first equation for x: x = 5 - y Substitute this expression for x into the second equation: (5 - y) - y = 1 Solve for y: 5 - 2y = 1 => 2y = 4 => y = 2 Substitute y = 2 back into either equation to solve for x: x + 2 = 5 => x = 3

    Solving Exponential and Logarithmic Equations

    These equations involve exponents and logarithms, requiring specific techniques.

    Example 9: Exponential Equation

    Solve for x: 2ˣ = 8

    This can be solved by recognizing that 8 = 2³. Therefore, 2ˣ = 2³, which implies x = 3. More complex exponential equations may require the use of logarithms.

    Example 10: Logarithmic Equation

    Solve for x: log₂(x) = 3

    By definition of logarithms, this means 2³ = x. Therefore, x = 8.

    Conclusion

    Solving equations for a specific variable is a crucial skill in mathematics and related fields. This guide has provided a comprehensive overview of different techniques for solving various types of equations, from simple linear equations to more complex exponential and logarithmic equations. Remember to practice regularly, starting with simpler problems and gradually progressing to more challenging ones. Mastering these techniques will significantly enhance your problem-solving abilities and open doors to more advanced mathematical concepts. The key is to understand the underlying principles, apply the appropriate techniques consistently, and always verify your solutions.

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