Example Question - two variables
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Solving a System of Equations with Two Variables
미안하지만, 이미지에 텍스트가 일부 가려져 있어 모든 정보를 정확히 알 수 없습니다. 하지만 제가 볼 수 있는 부분에서 주어진 두 방정식은 아래와 같습니다. 1) \( ax - by = 8 \) 2) \( bx + ay = 7 \) 이 두 방정식으로 두 변수 x와 y에 대한 해를 구해야 합니다. 이를 해결하기 위해서는 일반적으로 연립방정식의 해법을 사용합니다. 방정식을 풀기 위해서는, 먼저 한 변수에 대하여 다른 변수를 표현하거나, 두 방정식을 더하거나 빼서 한 변수를 제거하는 방법을 사용할 수 있습니다. 예를 들어, 첫 번째 방정식을 a로 나누고 두 번째 방정식을 b로 나눈 뒤 두 식을 더하거나 빼서 한 변수를 소거할 수 있습니다. 더 자세한 정보나 이미지가 제대로 보일 수 있도록 하여 다시 질문해주시면 더 정확한 풀이를 제공할 수 있을 것입니다.
Solving Linear Equations with Two Variables
Para resolver esta ecuación lineal para "x" e "y", necesitamos una ecuación adicional ya que tenemos dos variables. Con una sola ecuación y dos incógnitas, hay infinitas soluciones posibles. La ecuación podría ser representada por una línea en un plano cartesiano; sin embargo, para encontrar valores específicos para "x" e "y", necesitaríamos otra ecuación que intersecte con la primera, formando un sistema de ecuaciones. Si solo necesitas expresar "y" en términos de "x" usando la ecuación proporcionada, puedes aislar "y" de la siguiente manera: \[ 6x + 2y = 4 \] Primero, restamos \(6x\) de ambos lados de la ecuación: \[ 2y = 4 - 6x \] Luego dividimos ambos lados de la ecuación por 2 para despejar "y": \[ y = \frac{4 - 6x}{2} \] Simplificando, obtenemos: \[ y = 2 - 3x \] Ahora "y" está expresado en términos de "x". Pero recuerda, sin otra ecuación, no podemos encontrar valores específicos de estas variables.
Solving System of Equations with Two Variables
It seems you have provided an image of two system of equations marked as 175. (a) and 176. (a) from a textbook. To solve the system in 176. (a), which is: \[ \begin{cases} 3(x - 1) = 4y + 1, \\ 5(y - 1) = x + 1, \end{cases} \] Let's solve it step by step. 1. Expand the first equation: \[ 3x - 3 = 4y + 1 \] 2. Now, arrange the terms to isolate x in terms of y: \[ 3x = 4y + 4 \] \[ x = \frac{4y + 4}{3} \] Now let's take this expression for x and substitute it into the second equation. 3. Expand the second equation: \[ 5y - 5 = x + 1 \] Now substitute \( x \) with \( \frac{4y + 4}{3} \): \[ 5y - 5 = \frac{4y + 4}{3} + 1 \] 4. Multiply through by 3 to get rid of the fraction: \[ 3(5y - 5) = 4y + 4 + 3 \] \[ 15y - 15 = 4y + 7 \] 5. Subtract 4y from both sides and add 15 to both sides: \[ 15y - 4y = 7 + 15 \] \[ 11y = 22 \] 6. Divide by 11 to solve for y: \[ y = \frac{22}{11} \] \[ y = 2 \] Now that we have y, we can substitute it back into the expression we derived for x: \[ x = \frac{4(2) + 4}{3} \] \[ x = \frac{8 + 4}{3} \] \[ x = \frac{12}{3} \] \[ x = 4 \] Therefore, the solution to the system is \( x = 4 \) and \( y = 2 \).
Solving an Exponential Equation with Two Variables
The image shows an exponential equation: e^(t + 7x) + 11 = 20. To solve for the variables t and x in this equation, follow these steps: 1. Isolate the exponential term on one side of the equation: e^(t + 7x) = 20 - 11 e^(t + 7x) = 9 2. Since e^(t + 7x) = 9, take the natural logarithm of both sides to get rid of the exponential base 'e': ln(e^(t + 7x)) = ln(9) 3. By the property of logarithms, ln(e^y) = y for any y, we have: t + 7x = ln(9) From this point, the equation cannot be solved for a unique solution because we have a single equation with two variables. To find specific values for t and x, we would need a second independent equation. If there is additional information or another equation provided, you can use it to solve for the unknowns. Otherwise, you have an infinite set of solutions depending on the value of one of the variables.
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