Example Question - solve for y

Here are examples of questions we've helped users solve.

Solving a Linear Equation with Fractions

<p>La ecuación original es \( y = 7 - 2 \cdot (\frac{2}{3}) \).</p> <p>Primero, simplificamos la operación dentro del paréntesis:</p> <p>\( y = 7 - 2 \cdot \frac{2}{3} \)</p> <p>\( y = 7 - \frac{4}{3} \)</p> <p>Luego, convertimos el número entero 7 a una fracción con denominador 3 para poder hacer la resta:</p> <p>\( y = \frac{21}{3} - \frac{4}{3} \)</p> <p>Finalmente, realizamos la resta de fracciones con el mismo denominador:</p> <p>\( y = \frac{21 - 4}{3} \)</p> <p>\( y = \frac{17}{3} \)</p>

Logarithmic Equation Problem Solving

\[ \begin{align*} \log_7(7^{2y} - 18) & = y + 1 \\ 7^{2y} - 18 & = 7^{y+1} \\ 7^{2y} - 7^{y}7 - 18 & = 0 \\ \text{令 } z = 7^{y}, \text{得到} \\ z^{2} - 7z - 18 & = 0 \\ (z - 9)(z + 2) & = 0 \\ z = 9 \text{ 或 } z = -2, & \text{但是} z = 7^{y} > 0, \text{所以不考虑} z = -2. \\ 7^{y} & = 9 \\ y & = \log_7 9 \end{align*} \] 所以正确答案是 \( C. \log_7 9 \).

Mathematical Fraction Calculation

<p>\(\frac{4}{y} = 7^{-3}\)</p> <p>To solve for \(y\), first note that \(7^{-3} = \frac{1}{7^3} = \frac{1}{343}\). Now, set the fraction equal to \(y\).</p> <p>\(\frac{4}{y} = \frac{1}{343}\)</p> <p>Cross multiply to solve for \(y\):</p> <p>\(4 \cdot 343 = y \cdot 1\)</p> <p>\(y = 4 \cdot 343\)</p> <p>\(y = 1372\)</p>

Solving Linear Equation for y

La ecuación que aparece en la imagen es \(3x - 3y = -3\). Para resolverla en términos de una de las variables, podríamos resolver para \(y\) o para \(x\). Voy a resolver para \(y\): Primero, sumamos \(3y\) a ambos lados de la ecuación para aislar los términos que contienen a \(x\) en un lado: \[3x - 3y + 3y = -3 + 3y\] Esto simplifica a: \[3x = 3y - 3\] Ahora, para obtener \(y\) por sí solo, dividimos ambos lados de la ecuación por \(3\): \[\frac{3x}{3} = \frac{3y - 3}{3}\] Lo que nos da: \[x = y - 1\] Finalmente, para escribir la ecuación en términos de \(y\), simplemente despejamos \(y\): \[y = x + 1\] Entonces, la solución de la ecuación en términos de \(y\) es \(y = x + 1\).

Solving System of Equations for y

To solve the given system of equations for \( y \), follow these steps: The system of equations given are: \[ y = \frac{4}{7}x \] \[ \frac{2}{3}x = y + \frac{5}{7} \] First, we can use the first equation to substitute \( x \) in terms of \( y \) in the second equation. To find \( x \) in terms of \( y \) from the first equation, we rearrange the equation as: \[ y = \frac{4}{7}x \] \[ x = \frac{7}{4}y \] Now we substitute \( x \) into the second equation: \[ \frac{2}{3}\left(\frac{7}{4}y\right) = y + \frac{5}{7} \] To solve for \( y \), we now multiply the terms: \[ \frac{14}{12}y = y + \frac{5}{7} \] Simplify the left side by reducing the fraction: \[ \frac{7}{6}y = y + \frac{5}{7} \] To solve for y, get all the terms involving y on one side of the equation: \[ \frac{7}{6}y - y = \frac{5}{7} \] Since \( y \) is the same as \( \frac{6}{6}y \), we can rewrite the equation as: \[ \frac{7}{6}y - \frac{6}{6}y = \frac{5}{7} \] Combine the y terms: \[ \frac{1}{6}y = \frac{5}{7} \] Finally, solve for \( y \) by multiplying both sides by the reciprocal of \( \frac{1}{6} \), which is 6: \[ y = \frac{5}{7} \times 6 \] \[ y = \frac{5 \times 6}{7} \] \[ y = \frac{30}{7} \] So the value of \( y \) is \( \frac{30}{7} \). The correct answer is 30/7.

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