Resuelva las siguientes preguntas. valor de la variable de la siguiente ecuación. [tex]$\frac{13}{15} z+\frac{1}{2}+2 z-\frac{6}{5}=\frac{5}{12}-\frac{1}{2} z+\frac{1}{15}-\frac{4}{3} z$[/tex]
Answer (1)
Combine like terms on both sides of the equation.
Isolate the terms with z on one side and the constant terms on the other side.
Solve for z by dividing both sides of the equation by the coefficient of z .
The solution for z is 282 71 .
Explanation
Understanding the Problem We are given the equation 15 13 z + 2 1 + 2 z − 5 6 = 12 5 − 2 1 z + 15 1 − 3 4 z Our goal is to find the value of z that satisfies this equation. To do this, we will first combine like terms on both sides of the equation. This means grouping the terms with z together and the constant terms together.
Grouping Like Terms First, let's group the terms with z on the left side and the constant terms on the right side. We have: 15 13 z + 2 z + 2 1 z + 3 4 z = 12 5 + 15 1 + 5 6 − 2 1 Now, we need to find a common denominator for the fractions involving z and for the constant fractions.
Finding Common Denominators For the terms with z , the common denominator is 30. So we rewrite the fractions: 30 26 z + 30 60 z + 30 15 z + 30 40 z = 12 5 + 15 1 + 5 6 − 2 1 Combining these terms, we get: 30 26 + 60 + 15 + 40 z = 30 141 z For the constant terms, the common denominator is 60. So we rewrite the fractions: 60 25 + 60 4 + 60 72 − 60 30 = 60 25 + 4 + 72 − 30 = 60 71 So our equation now looks like this: 30 141 z = 60 71
Solving for z Now, we solve for z by multiplying both sides of the equation by 141 30 :
z = 60 71 × 141 30 = 2 × 30 71 × 141 30 = 2 × 141 71 = 282 71 Therefore, the value of z is 282 71 .
Final Answer The value of the variable z in the given equation is 282 71 , which is approximately equal to 0.2518.
Examples
In electrical engineering, when analyzing circuits, you often encounter equations with fractions and variables representing currents or voltages. Solving these equations allows you to determine the values of these currents or voltages, which is crucial for understanding the behavior of the circuit and designing it effectively. For example, if you have a circuit with resistors and voltage sources, you might need to solve an equation similar to the one above to find the current flowing through a particular resistor. This helps in ensuring that the components operate within their specified limits and the circuit functions as intended.
In chemistry, when dealing with chemical reactions and equilibrium, you often need to solve equations involving concentrations of reactants and products. These equations can also involve fractions and variables, and solving them allows you to determine the equilibrium concentrations of the substances involved in the reaction. This is important for understanding the extent to which a reaction will proceed and for optimizing reaction conditions to obtain the desired products.
