Solve The Following Equations For X: (2x+1)(x+2)=0, X - √(12-4x) = 0, X² + 8x + 15 = 0. Solve The Following Equations For Y: 5√(y-3) = 2, (y² + 6y - 35)/(y² + 6y) = 2. Solve The Following System Of Equations For X And Y: 4x - Y = 3 And [Missing Equation].

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This article provides a detailed walkthrough of solving various algebraic equations, covering different types such as quadratic equations, equations involving square roots, and simultaneous equations. Each solution is presented step-by-step to ensure clarity and understanding. Whether you're a student looking to improve your algebra skills or someone needing a refresher, this guide will help you grasp the fundamental techniques for solving equations.

1.1 Solving for x

1.1.1 Solve (2x+1)(x+2)=0

Quadratic equations are a fundamental topic in algebra, and solving them often involves finding the roots of the equation. When faced with an equation in factored form, like this one, we can use the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. This principle is the cornerstone for solving equations presented in factored form.

To solve the equation (2x+1)(x+2)=0, we set each factor equal to zero:

  1. Set the first factor equal to zero: 2x+1=0

    • Subtract 1 from both sides: 2x = -1
    • Divide by 2: x = -1/2

  2. Set the second factor equal to zero: x+2=0

    • Subtract 2 from both sides: x = -2

Therefore, the solutions for the equation (2x+1)(x+2)=0 are x = -1/2 and x = -2. These values of x make the original equation true, which can be verified by substituting them back into the equation. This method is highly efficient for solving quadratic equations that are already factored, providing a straightforward path to the solution.

The practical applications of solving quadratic equations are vast, ranging from physics and engineering to economics and computer science. For instance, in physics, quadratic equations can be used to model projectile motion, calculating the trajectory of an object thrown into the air. In engineering, they are crucial for designing structures and systems, ensuring stability and efficiency. Economists use quadratic equations to model cost and revenue functions, helping businesses make informed decisions. In computer science, they appear in algorithms and optimization problems. Mastering the techniques for solving quadratic equations, especially those in factored form, is a valuable skill that extends far beyond the classroom.

1.1.2 Solve x - √(12-4x) = 0

Equations involving square roots require a different approach to ensure we eliminate the radical while maintaining the equation's balance. The primary strategy is to isolate the square root term on one side of the equation and then square both sides. This process eliminates the square root, allowing us to solve for the variable. However, it's crucial to check the solutions obtained, as squaring both sides can introduce extraneous solutions—values that satisfy the transformed equation but not the original.

To solve the equation x - √(12-4x) = 0, we follow these steps:

  1. Isolate the square root term:

    • Add √(12-4x) to both sides: x = √(12-4x)

  2. Square both sides of the equation:

    • x² = (√(12-4x))²
    • x² = 12 - 4x

  3. Rearrange the equation into a quadratic equation:

    • x² + 4x - 12 = 0

  4. Factor the quadratic equation:

    • (x+6)(x-2) = 0

  5. Set each factor equal to zero:

    • x + 6 = 0 -> x = -6
    • x - 2 = 0 -> x = 2

Now, we need to check both solutions in the original equation:

  • For x = -6:
    • -6 - √(12 - 4(-6)) = -6 - √(12 + 24) = -6 - √36 = -6 - 6 = -12 ≠ 0
    • So, x = -6 is an extraneous solution.
  • For x = 2:
    • 2 - √(12 - 4(2)) = 2 - √(12 - 8) = 2 - √4 = 2 - 2 = 0
    • So, x = 2 is a valid solution.

Therefore, the only valid solution for the equation x - √(12-4x) = 0 is x = 2. This verification step is vital in equations involving radicals to ensure the accuracy of the solutions.

The application of equations with square roots is common in various scientific and engineering contexts. For example, in physics, these equations are used to calculate velocities and distances in scenarios involving motion under constant acceleration. In engineering, they can appear in structural analysis and design, particularly when dealing with stress and strain calculations. The process of isolating the radical term and squaring both sides is a fundamental technique that can be adapted to solve a wide range of problems, making it an essential tool in mathematical problem-solving.

1.1.3 Solve x² + 8x + 15 = 0

The quadratic equation x² + 8x + 15 = 0 is a classic example of a polynomial equation of degree two. To solve it, we aim to find the values of x that make the equation true. One common method for solving quadratic equations is factoring. Factoring involves expressing the quadratic expression as a product of two binomials. If the quadratic equation cannot be easily factored, other methods such as completing the square or using the quadratic formula can be employed.

To solve x² + 8x + 15 = 0 by factoring, we look for two numbers that multiply to 15 and add to 8. These numbers are 3 and 5:

  1. Factor the quadratic equation:

    • (x + 3)(x + 5) = 0

  2. Set each factor equal to zero:

    • x + 3 = 0 -> x = -3
    • x + 5 = 0 -> x = -5

Therefore, the solutions for the equation x² + 8x + 15 = 0 are x = -3 and x = -5. These values of x are the roots of the quadratic equation, and substituting them back into the original equation will confirm that they make the equation true.

Quadratic equations have widespread applications in various fields. In mathematics, they are fundamental in calculus and coordinate geometry. In physics, they describe projectile motion, energy, and oscillation phenomena. For example, the height of a projectile thrown upwards can be modeled using a quadratic equation, allowing us to calculate its maximum height and time of flight. In engineering, quadratic equations are used in the design of bridges, buildings, and electronic circuits. They also play a critical role in economics, where they can model cost, revenue, and profit functions. The ability to solve quadratic equations is thus a crucial skill in many academic and professional disciplines.

1.2 Solving for y

1.2.1 Solve 5√(y-3) = 2 (leave your answer to 2 decimal places)

Equations involving square roots, such as 5√(y-3) = 2, require careful manipulation to isolate the variable. The initial step involves isolating the square root term, followed by squaring both sides of the equation to eliminate the radical. This method allows us to transform the equation into a simpler form that can be solved for y. Remember, it's essential to check the solution in the original equation to avoid extraneous roots, which can arise from the squaring process.

To solve 5√(y-3) = 2, we proceed as follows:

  1. Isolate the square root term:

    • Divide both sides by 5: √(y-3) = 2/5

  2. Square both sides of the equation:

    • (√(y-3))² = (2/5)²
    • y - 3 = 4/25

  3. Solve for y:

    • Add 3 to both sides: y = 4/25 + 3
    • Convert 3 to a fraction with a denominator of 25: y = 4/25 + 75/25
    • y = 79/25

  4. Convert the fraction to a decimal (rounded to 2 decimal places):

    • y ≈ 3.16

Now, let's check the solution in the original equation:

  • 5√(3.16 - 3) = 5√0.16 ≈ 5 * 0.4 = 2

The solution checks out. Therefore, the solution for the equation 5√(y-3) = 2, rounded to two decimal places, is y ≈ 3.16.

Equations involving square roots are prevalent in various scientific and engineering applications. For example, they can be used to model the speed of sound in different media or to calculate the period of a pendulum. In physics, these equations often appear in problems related to energy and motion. In engineering, they might be used in structural analysis or fluid dynamics. The ability to accurately solve these equations is critical for making precise calculations and predictions in these fields. The step-by-step approach of isolating the square root, squaring both sides, and verifying the solution is a fundamental technique that extends to more complex problems involving radicals.

1.2.2 Solve (y² + 6y - 35)/(y² + 6y) = 2

Rational equations, such as (y² + 6y - 35)/(y² + 6y) = 2, require a systematic approach to eliminate the fractions and solve for the variable. The initial step involves multiplying both sides of the equation by the denominator to clear the fraction. This transforms the equation into a more manageable form, often a polynomial equation. It's crucial to identify any restrictions on the variable due to the denominator, as these values would make the original expression undefined. Additionally, checking the solutions in the original equation is essential to ensure they are valid and not extraneous.

To solve the equation (y² + 6y - 35)/(y² + 6y) = 2, we follow these steps:

  1. Multiply both sides by the denominator (y² + 6y):

    • (y² + 6y - 35) = 2(y² + 6y)

  2. Expand and simplify the equation:

    • y² + 6y - 35 = 2y² + 12y

  3. Rearrange the equation into a quadratic equation:

    • 0 = y² + 6y + 35

  4. Solve the quadratic equation. Since it does not factor easily, we can use the quadratic formula: y = [-b ± √(b² - 4ac)] / (2a)

    • Where a = 1, b = 6, and c = 35
    • y = [-6 ± √(6² - 4(1)(35))] / (2(1))
    • y = [-6 ± √(36 - 140)] / 2
    • y = [-6 ± √(-104)] / 2

Since the discriminant (b² - 4ac) is negative, the solutions are complex numbers. Thus, there are no real solutions for this equation.

The absence of real solutions indicates that there are no real values of y that satisfy the original equation. This outcome highlights the importance of the discriminant in determining the nature of the roots of a quadratic equation. When the discriminant is negative, the quadratic equation has complex roots, meaning the solutions involve imaginary numbers. Rational equations and quadratic equations are widely used in various fields, including engineering, physics, and economics, to model and solve a variety of real-world problems. Understanding how to handle such equations, including recognizing when solutions are complex, is a critical skill in mathematical problem-solving.

1.3 Solving for x and y Simultaneously

Solving simultaneous equations is a fundamental skill in algebra, often encountered in various mathematical and scientific contexts. The goal is to find the values of the variables that satisfy all equations in the system. One common method is the substitution method, where one equation is solved for one variable in terms of the other, and this expression is substituted into the other equation. This reduces the system to a single equation in one variable, which can then be solved. The solution is then substituted back into one of the original equations to find the value of the other variable.

Given the system of equations:

  1. 4x - y = 3
  2. There is a missing second equation in the original problem. To provide a comprehensive solution, I will add a second equation: x + y = 2

Now we have the system:

  1. 4x - y = 3
  2. x + y = 2

To solve this system using the substitution method, we can follow these steps:

  1. Solve the second equation for y:

    • y = 2 - x

  2. Substitute this expression for y into the first equation:

    • 4x - (2 - x) = 3

  3. Simplify and solve for x:

    • 4x - 2 + x = 3
    • 5x = 5
    • x = 1

  4. Substitute the value of x back into the equation y = 2 - x:

    • y = 2 - 1
    • y = 1

Therefore, the solution for the system of equations is x = 1 and y = 1. This pair of values satisfies both equations simultaneously.

Simultaneous equations arise in numerous applications across various fields. In physics, they are used to solve problems involving multiple forces or velocities acting on an object. In economics, they are used to model supply and demand curves, finding the equilibrium price and quantity. In engineering, they are essential for circuit analysis and structural design. For instance, in electrical engineering, Kirchhoff's laws lead to systems of linear equations that need to be solved to determine currents and voltages in a circuit. The substitution method, along with other techniques such as elimination, provides a powerful tool for tackling such problems, making it a core concept in applied mathematics and sciences.