Given That $x \ \textless \ Em\ \textgreater \ Y = 3x + \frac{x}{y}$, Solve For $x$ In The Following Cases:(a) $x \ \textless \ /em\ \textgreater \ 3 = 20$(b) \$x \ \textless \ Em\ \textgreater \ 2 =

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Introduction

In mathematics, solving equations is a fundamental concept that involves finding the value of a variable that satisfies a given equation. Equations can be linear or non-linear, and they can be solved using various methods such as algebraic manipulation, graphical methods, or numerical methods. In this article, we will focus on solving linear and non-linear equations, specifically the equation $x \ \textless \ em\ \textgreater \ y = 3x + \frac{x}{y}$, in the following cases: (a) $x \ \textless \ /em\ \textgreater \ 3 = 20$ and (b) $x \ \textless \ em\ \textgreater \ 2 = 15$.

Case (a): Solving for x when x < 3 = 20

To solve for x in the equation $x \ \textless \ /em\ \textgreater \ 3 = 20$, we need to isolate the variable x. The first step is to subtract 3 from both sides of the equation to get:

x=20−3x = 20 - 3

Simplifying the right-hand side of the equation, we get:

x=17x = 17

Therefore, the value of x that satisfies the equation $x \ \textless \ /em\ \textgreater \ 3 = 20$ is x = 17.

Case (b): Solving for x when x < 2 = 15

To solve for x in the equation $x \ \textless \ em\ \textgreater \ 2 = 15$, we need to isolate the variable x. The first step is to subtract 2 from both sides of the equation to get:

x=15−2x = 15 - 2

Simplifying the right-hand side of the equation, we get:

x=13x = 13

Therefore, the value of x that satisfies the equation $x \ \textless \ em\ \textgreater \ 2 = 15$ is x = 13.

Solving the Original Equation

Now that we have solved for x in the two cases, we can substitute the values of x into the original equation $x \ \textless \ em\ \textgreater \ y = 3x + \frac{x}{y}$ to verify the solutions.

For case (a), substituting x = 17 into the original equation, we get:

17 \textless em \textgreater y=3(17)+17y17 \ \textless \ em\ \textgreater \ y = 3(17) + \frac{17}{y}

Simplifying the equation, we get:

17 \textless em \textgreater y=51+17y17 \ \textless \ em\ \textgreater \ y = 51 + \frac{17}{y}

To solve for y, we can multiply both sides of the equation by y to get:

17y=51y+1717y = 51y + 17

Subtracting 17y from both sides of the equation, we get:

0=34y+170 = 34y + 17

Subtracting 17 from both sides of the equation, we get:

−17=34y-17 = 34y

Dividing both sides of the equation by 34, we get:

y=−1734y = -\frac{17}{34}

Therefore, the value of y that satisfies the equation $x \ \textless \ /em\ \textgreater \ 3 = 20$ is y = -17/34.

For case (b), substituting x = 13 into the original equation, we get:

13 \textless em \textgreater y=3(13)+13y13 \ \textless \ em\ \textgreater \ y = 3(13) + \frac{13}{y}

Simplifying the equation, we get:

13 \textless em \textgreater y=39+13y13 \ \textless \ em\ \textgreater \ y = 39 + \frac{13}{y}

To solve for y, we can multiply both sides of the equation by y to get:

13y=39y+1313y = 39y + 13

Subtracting 13y from both sides of the equation, we get:

0=26y+130 = 26y + 13

Subtracting 13 from both sides of the equation, we get:

−13=26y-13 = 26y

Dividing both sides of the equation by 26, we get:

y=−1326y = -\frac{13}{26}

Therefore, the value of y that satisfies the equation $x \ \textless \ em\ \textgreater \ 2 = 15$ is y = -13/26.

Conclusion

In this article, we have solved for x in the equation $x \ \textless \ em\ \textgreater \ y = 3x + \frac{x}{y}$ in the following cases: (a) $x \ \textless \ /em\ \textgreater \ 3 = 20$ and (b) $x \ \textless \ em\ \textgreater \ 2 = 15$. We have also verified the solutions by substituting the values of x into the original equation. The values of x that satisfy the equations are x = 17 and x = 13, respectively. The corresponding values of y are y = -17/34 and y = -13/26, respectively.

References

  • [1] "Algebra and Trigonometry" by Michael Sullivan
  • [2] "Calculus" by Michael Spivak
  • [3] "Linear Algebra and Its Applications" by Gilbert Strang

Future Work

Q: What is the difference between a linear equation and a non-linear equation?

A: A linear equation is an equation in which the highest power of the variable(s) is 1. For example, the equation 2x + 3 = 5 is a linear equation. A non-linear equation, on the other hand, is an equation in which the highest power of the variable(s) is greater than 1. For example, the equation x^2 + 2x + 1 = 0 is a non-linear equation.

Q: How do I solve a linear equation?

A: To solve a linear equation, you can use the following steps:

  1. Simplify the equation by combining like terms.
  2. Isolate the variable by adding or subtracting the same value to both sides of the equation.
  3. Divide both sides of the equation by the coefficient of the variable.

Q: How do I solve a non-linear equation?

A: To solve a non-linear equation, you can use the following steps:

  1. Simplify the equation by combining like terms.
  2. Use algebraic manipulation to isolate the variable.
  3. Use numerical methods or graphical methods to approximate the solution.

Q: What is the difference between a quadratic equation and a cubic equation?

A: A quadratic equation is a non-linear equation in which the highest power of the variable is 2. For example, the equation x^2 + 2x + 1 = 0 is a quadratic equation. A cubic equation, on the other hand, is a non-linear equation in which the highest power of the variable is 3. For example, the equation x^3 + 2x^2 + x + 1 = 0 is a cubic equation.

Q: How do I solve a quadratic equation?

A: To solve a quadratic equation, you can use the following steps:

  1. Simplify the equation by combining like terms.
  2. Use the quadratic formula to find the solutions: x = (-b ± √(b^2 - 4ac)) / 2a.
  3. Check the solutions to make sure they are valid.

Q: How do I solve a cubic equation?

A: To solve a cubic equation, you can use the following steps:

  1. Simplify the equation by combining like terms.
  2. Use algebraic manipulation to isolate the variable.
  3. Use numerical methods or graphical methods to approximate the solution.

Q: What is the difference between a system of linear equations and a system of non-linear equations?

A: A system of linear equations is a set of linear equations that are solved simultaneously. For example, the system of equations:

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

is a system of linear equations. A system of non-linear equations, on the other hand, is a set of non-linear equations that are solved simultaneously. For example, the system of equations:

x^2 + 2y^2 = 4 x + 2y = 3

is a system of non-linear equations.

Q: How do I solve a system of linear equations?

A: To solve a system of linear equations, you can use the following steps:

  1. Simplify the equations by combining like terms.
  2. Use algebraic manipulation to isolate the variables.
  3. Use substitution or elimination methods to solve the system.

Q: How do I solve a system of non-linear equations?

A: To solve a system of non-linear equations, you can use the following steps:

  1. Simplify the equations by combining like terms.
  2. Use algebraic manipulation to isolate the variables.
  3. Use numerical methods or graphical methods to approximate the solution.

Conclusion

In this article, we have answered some common questions about solving linear and non-linear equations. We have discussed the difference between linear and non-linear equations, and how to solve them using various methods. We have also discussed the difference between a system of linear equations and a system of non-linear equations, and how to solve them using various methods.