Example Question - simultaneous equations
Here are examples of questions we've helped users solve.
Solving a System of Linear Equations
\[ \begin{align*} \text{Дано уравнение:} \\ &\frac{x}{2} + \frac{x - 3y}{4} = \frac{3(y - 2x)}{6} \\ \text{Приведем к общему знаменателю:} \\ &\frac{3x}{6} + \frac{3(x - 3y)}{12} = \frac{3(y - 2x)}{6} \\ \text{Упростим уравнение, домножив все члены на 12:} \\ &6x + 3(x - 3y) = 2(y - 2x) \\ &6x + 3x - 9y = 2y - 4x \\ \text{Перенесем все члены уравнения с переменной y в одну сторону, а с x – в другую:} \\ &6x + 3x + 4x = 9y + 2y \\ &13x = 11y \\ \text{Выразим переменную y через x:} \\ &y = \frac{13}{11}x \end{align*} \] \[ \begin{align*} \text{Теперь подставим выражение для y во второе уравнение:} \\ &2x - 3y = \frac{3(x - y)}{2} \\ &2x - 3\left(\frac{13}{11}x\right) = \frac{3}{2}(x - \frac{13}{11}x) \\ \text{Упростим уравнение, приведем его к общему знаменателю:} \\ &\frac{22x - 3 \cdot 13x}{11} = \frac{3}{2} \cdot \frac{11x - 13x}{11} \\ &\frac{22x - 39x}{11} = \frac{3(11x - 13x)}{22} \\ &\frac{-17x}{11} = \frac{3(-2x)}{22} \\ \text{Умножим обе стороны на 22:} \\ &-34x = 3(-2x) \\ &-34x = -6x \\ &-34x + 6x = 0 \\ &-28x = 0 \\ \text{Теперь найдем x:} \\ &x = 0 \end{align*} \] \[ \begin{align*} \text{Так как x равно нулю, подставим x в выражение для y:} \\ &y = \frac{13}{11} \cdot 0 \\ &y = 0 \end{align*} \] \[ \begin{align*} \text{Ответ:} \\ &x = 0 \\ &y = 0 \end{align*} \]
Solving a System of Linear Equations
<p>La imagen proporciona un sistema de ecuaciones lineales que parece incompleto. Solo una ecuación es visible, por lo que no se puede resolver un sistema de ecuaciones sin la segunda ecuación completa. La ecuación visible es:</p> <p>2x + 3y = 7</p> <p>Para cualquier intento de resolución, necesitaríamos la segunda ecuación que parece empezar con "y = ..." pero no está completa para proceder con la solución. Normalmente, se necesitarían dos ecuaciones completas para resolver un sistema de dos variables (x e y). Por lo tanto, no podemos proporcionar una solución sin la segunda ecuación.</p>
Solve the System of Linear Equations with Four Variables
<p>Let's denote the equations as follows:</p> <p>\[ \begin{align*} I: & \quad 2x + 3y - z + 4w = 10 \\ II: & \quad x - y + 2z - w = -3 \\ III: & \quad 3x + y + z + 2w = 12 \\ IV: & \quad 4x - 2y + 3z + w = 7 \\ \end{align*} \]</p> <p>Use matrix notation \( A\vec{x} = \vec{b} \), where matrix \( A \) is the coefficients of the variables, \( \vec{x} \) is the column vector of the variables, and \( \vec{b} \) is the column vector of constants:</p> <p>\[ A = \begin{bmatrix} 2 & 3 & -1 & 4 \\ 1 & -1 & 2 & -1 \\ 3 & 1 & 1 & 2 \\ 4 & -2 & 3 & 1 \\ \end{bmatrix}, \quad \vec{x} = \begin{bmatrix} x \\ y \\ z \\ w \\ \end{bmatrix}, \quad \vec{b} = \begin{bmatrix} 10 \\ -3 \\ 12 \\ 7 \\ \end{bmatrix} \]</p> <p>Calculate the inverse of \( A \) and multiply it by \( \vec{b} \) to find \( \vec{x} \):</p> <p>\[ \vec{x} = A^{-1}\vec{b} \]</p> <p>This requires finding the inverse of \( A \), which is a 4x4 matrix and can be done using a calculator or software capable of matrix operations. After finding \( A^{-1} \), perform the matrix multiplication to obtain the values of \( x \), \( y \), \( z \), and \( w \).</p> <p>Since solving a 4x4 matrix inverse by hand is complicated, I cannot provide the precise numerical solution here. Typically, you would use computer software such as MATLAB, Python (with numpy library), or a graphing calculator to compute the inverse matrix and product to obtain the solution for \( x \), \( y \), \( z \), and \( w \).</p>
Solving a System of Linear Equations
<p>Given the system of equations:</p> <p>\[\begin{align*} 5x + 4y &= 58 \quad (1)\\ 3x + 7y &= 67 \quad (2) \end{align*}\]</p> <p>To solve it, we can use the method of substitution or elimination. Here, we will use the elimination method for this solution:</p> <p>Multiply equation (1) by 3 and equation (2) by 5 to make the coefficients of \( x \) identical:</p> <p>\[\begin{align*} 3*(5x + 4y) &= 3*58 \\ 5*(3x + 7y) &= 5*67 \end{align*}\]</p> <p>\[\begin{align*} 15x + 12y &= 174 \quad (3)\\ 15x + 35y &= 335 \quad (4) \end{align*}\]</p> <p>Subtract equation (3) from equation (4) to eliminate \( x \):</p> <p>\[\begin{align*} (15x + 35y) - (15x + 12y) &= 335 - 174\\ 23y &= 161 \end{align*}\]</p> <p>Solving for \( y \):</p> <p>\[y = \frac{161}{23} = 7\]</p> <p>Now, substitute \( y = 7 \) into equation (1):</p> <p>\[5x + 4(7) = 58\]</p> <p>\[5x + 28 = 58\]</p> <p>\[5x = 58 - 28\]</p> <p>\[5x = 30\]</p> <p>Solving for \( x \):</p> <p>\[x = \frac{30}{5} = 6\]</p> <p>Thus, the solution to the system of equations is \( x = 6 \) and \( y = 7 \).</p>
Solving a System of Linear Equations Using Substitution or Elimination
<p>The system of equations is:</p> \[ \begin{align*} 5x + 4y &= 58 \quad \text{(1)} \\ 3x + 7y &= 67 \quad \text{(2)} \end{align*} \] <p>To solve the system, we can use either substitution or elimination. Here, we will use the elimination method. The goal is to eliminate one variable by making the coefficients of either \( x \) or \( y \) the same in both equations.</p> <p>Multiply equation (1) by 3 and equation (2) by 5:</p> \[ \begin{align*} 3(5x + 4y) &= 3(58) \quad \Rightarrow \quad 15x + 12y &= 174 \quad \text{(3)} \\ 5(3x + 7y) &= 5(67) \quad \Rightarrow \quad 15x + 35y &= 335 \quad \text{(4)} \end{align*} \] <p>Subtract equation (3) from equation (4):</p> \[ \begin{align*} (15x + 35y) - (15x + 12y) &= 335 - 174 \\ 15x + 35y - 15x - 12y &= 161 \\ 23y &= 161 \end{align*} \] <p>Divide by 23 to find \( y \):</p> \[ \begin{align*} y &= \frac{161}{23} \\ y &= 7 \end{align*} \] <p>Substitute \( y = 7 \) into equation (1):</p> \[ \begin{align*} 5x + 4(7) &= 58 \\ 5x + 28 &= 58 \\ 5x &= 58 - 28 \\ 5x &= 30 \end{align*} \] <p>Divide by 5 to find \( x \):</p> \[ \begin{align*} x &= \frac{30}{5} \\ x &= 6 \end{align*} \] <p>The solution set is \( (x, y) = (6, 7) \).</p>
Solving a System of Linear Equations Using Elimination Method
<p>To solve the system of equations:</p> <p>\[\begin{align*} 5x + 4y &= 58 \\ 3x + 7y &= 67 \end{align*}\]</p> <p>We can use the elimination method by multiplying the first equation by 7 and the second equation by -4 to eliminate \(y\):</p> <p>\[\begin{align*} 7(5x + 4y) &= 7 \cdot 58 \\ -4(3x + 7y) &= -4 \cdot 67 \end{align*}\]</p> <p>\[\begin{align*} 35x + 28y &= 406 \\ -12x - 28y &= -268 \end{align*}\]</p> <p>Add the two equations to eliminate \(y\):</p> <p>\[\begin{align*} (35x + 28y) + (-12x - 28y) &= 406 + (-268) \\ 23x &= 138 \end{align*}\]</p> <p>Divide by 23 to solve for \(x\):</p> <p>\[x = \frac{138}{23} = 6\]</p> <p>Now substitute \(x = 6\) into one of the original equations to find \(y\). We'll use the first equation:</p> <p>\[5(6) + 4y = 58\]</p> <p>\[30 + 4y = 58\]</p> <p>\[4y = 58 - 30\]</p> <p>\[4y = 28\]</p> <p>\[y = \frac{28}{4} = 7\]</p> <p>Therefore, the solution set is \((x, y) = (6, 7)\).</p>
System of Linear Equations Problem
<p>The given system of equations is:</p> <p>\[\begin{cases} 3(2x - 3y) - 4(x + y) = 7x - 3y - 5 \\ 7(x - 2y) - 5(2x - y) = -9x + 9y + 7 \end{cases}\]</p> <p>First, expand and simplify both equations:</p> <p>Equation 1: \(6x - 9y - 4x - 4y = 7x - 3y - 5\)</p> <p>Rearrange and combine like terms:</p> <p>\(2x - 13y = 7x - 3y - 5 \Rightarrow -5x + 10y = -5\)</p> <p>Divide by -5 to simplify the equation:</p> <p>\(x - 2y = 1\) ... (i)</p> <p>Equation 2: \(7x - 14y - 10x + 5y = -9x + 9y + 7\)</p> <p>Rearrange and combine like terms:</p> <p>\(-3x - 9y = -9x + 9y + 7 \Rightarrow 6x - 18y = 7\)</p> <p>Divide by 6 to simplify the equation:</p> <p>\(x - 3y = \dfrac{7}{6}\) ... (ii)</p> <p>Now we can solve the system using equations (i) and (ii):</p> <p>From equation (i):</p> <p>\(x = 2y + 1\)</p> <p>Substitute \(x\) from equation (i) into equation (ii):</p> <p>\(2y + 1 - 3y = \dfrac{7}{6}\)</p> <p>\(-y + 1 = \dfrac{7}{6}\)</p> <p>\(-y = \dfrac{7}{6} - 1\)</p> <p>\(-y = \dfrac{1}{6}\)</p> <p>\(y = -\dfrac{1}{6}\)</p> <p>Now substitute \(y\) into equation (i):</p> <p>\(x = 2(-\dfrac{1}{6}) + 1\)</p> <p>\(x = -\dfrac{1}{3} + 1\)</p> <p>\(x = \dfrac{2}{3}\)</p> <p>The solution to the system of equations is \(x = \dfrac{2}{3}\), \(y = -\dfrac{1}{6}\).</p>
Solving a System of Linear Equations Using Cramer's Rule
Para resolver el sistema de ecuaciones lineales utilizando la regla de Cramer, primero escribimos las ecuaciones en la forma \( Ax = B \), donde \( A \) es la matriz de coeficientes, \( x \) es la matriz de variables, y \( B \) es la matriz de términos constantes. Las ecuaciones dadas son: \[ \begin{align*} 4x + 2y &= 13 \\ 2x + 2y &= 5 \end{align*} \] La matriz de coeficientes \( A \) es: \[ A = \begin{bmatrix} 4 & 2 \\ 2 & 2 \end{bmatrix} \] La matriz de términos constantes \( B \) es: \[ B = \begin{bmatrix} 13 \\ 5 \end{bmatrix} \] Para hallar \( x \), reemplazamos la primera columna de \( A \) por \( B \) y calculamos el determinante: \[ A_x = \begin{bmatrix} 13 & 2 \\ 5 & 2 \end{bmatrix} \] \[ \text{det}(A_x) = (13)(2) - (5)(2) = 26 - 10 = 16 \] Para hallar \( y \), reemplazamos la segunda columna de \( A \) por \( B \) y calculamos el determinante: \[ A_y = \begin{bmatrix} 4 & 13 \\ 2 & 5 \end{bmatrix} \] \[ \text{det}(A_y) = (4)(5) - (2)(13) = 20 - 26 = -6 \] El determinante de \( A \) es: \[ \text{det}(A) = (4)(2) - (2)(2) = 8 - 4 = 4 \] Entonces, las soluciones son: \[ x = \frac{\text{det}(A_x)}{\text{det}(A)} = \frac{16}{4} = 4 \] \[ y = \frac{\text{det}(A_y)}{\text{det}(A)} = \frac{-6}{4} = -1.5 \] Por lo tanto, las soluciones del sistema son \( x = 4 \) y \( y = -1.5 \).
Solving a System of Linear Equations
Para resolver el sistema de ecuaciones lineales, podemos usar el método de eliminación o sustitución. En este caso, utilizaremos el método de eliminación. <p>\( \begin{cases} 3x + y = 1 \\ -7x - 2y = -1 \end{cases} \)</p> <p>Multiplicamos la primera ecuación por 2 para poder eliminar la variable \(y\):</p> <p>\( 2(3x + y) = 2(1) \)</p> <p>\( 6x + 2y = 2 \)</p> <p>Ahora sumamos esta nueva ecuación a la segunda ecuación del sistema:</p> <p>\( (6x + 2y) + (-7x - 2y) = 2 + (-1) \)</p> <p>\( 6x - 7x + 2y - 2y = 2 - 1 \)</p> <p>\( -x = 1 \)</p> <p>Ahora resolvemos para \(x\):</p> <p>\( x = -1 \)</p> <p>Sustituimos el valor de \(x\) en la primera ecuación original para encontrar \(y\):</p> <p>\( 3(-1) + y = 1 \)</p> <p>\( -3 + y = 1 \)</p> <p>\( y = 1 + 3 \)</p> <p>\( y = 4 \)</p> <p>Por lo tanto, el sistema tiene la solución:</p> <p>\( x = -1 \)</p> <p>\( y = 4 \)</p>
Solving a System of Equations
이미지에 주어진 연립방정식을 풀기 위해서 다음과 같은 단계를 따르면 됩니다. 주어진 연립방정식은 다음과 같습니다. 1. ax - by = 8 2. bx + ay = 7 이 두 방정식으로부터 x와 y의 값을 구해야 합니다. 먼저, 첫 번째 방정식을 a로 나누면 x - (b/a)y = 8/a가 되고, 두 번째 방정식을 b로 나누면 x + (a/b)y = 7/b가 됩니다. 두 방정식을 더하면 다음과 같습니다. x - (b/a)y + x + (a/b)y = 8/a + 7/b 괄호 안의 y항을 상쇄시키기 위해 첫 번째 식에 b를, 두 번째 식에 a를 곱해서 y항의 계수를 같게 만듭니다. bx - by^2/a + ax + ay^2/b = 8b/a + 7a/b 이제 두 y항을 상쇄시키기 위해 더합니다. (b + a)x = 8b/a + 7a/b 단위 분수 법으로 우변을 계산하기 위해 통분을 합니다. (b + a)x = (8b^2 + 7a^2)/ab 이제 x에 대해 풉니다. x = (8b^2 + 7a^2)/ab(b + a) x = (8b + 7a)/(ab) 이제 x의 값을 구하려면 실제 계산을 하면 됩니다. y의 값을 구하기 위해서는 유사한 방법을 사용하면 됩니다. 첫 번째 방정식에서 x항을 피하고 두 번째 방정식에서 y항을 피하는 식으로 재배열하고, 각각 a, b로 나눈 후 더해서 y의 값을 구할 수 있습니다.
Inconsistent System of Equations
제시된 문제를 풀기 위해 여기에 제공된 두 연립방정식을 이용할 것입니다. \( \frac{ax - by}{bx + ay} = 8 \) 및 \( \frac{bx - ay}{ax - by} = \frac{7}{8} \) 해를 구하기 위해 먼저 두 번째 방정식의 분모와 분자를 바꾸기 위해 역수를 취합시다: \( \frac{ax - by}{bx + ay} = \frac{8}{7} \) (두 번째 방정식의 역수) 이제 우리는 두 방정식이 같다는 것을 알고 있으니: \( 8 = \frac{8}{7} \) 이는 모순이므로, 주어진 시스템에 해가 없음을 의미합니다. 그러므로 x와 y에 대한 해는 존재하지 않습니다.
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.
Graphing Linear Equations to Find Intersection Points
To graph each pair of lines from the provided sets A, B, and C, follow these steps for each set: **Set A:** Equations: 1. \(y = 2x + 1\) 2. \(y = -x + 7\) For each equation, find two points by choosing values for \(x\) and calculating the corresponding \(y\) value. Then plot these points on a graph and draw a line through them. Equation 1: Let \(x = 0\), then \(y = 2(0) + 1 = 1\). So, one point is (0,1). Let \(x = 1\), then \(y = 2(1) + 1 = 3\). Another point is (1,3). These two points will help you graph the first line. Equation 2: Let \(x = 0\), then \(y = -(0) + 7 = 7\). One point is (0,7). Let \(x = 1\), then \(y = -(1) + 7 = 6\). Another point is (1,6). These two points will help you graph the second line. **Set B:** Equations: 1. \(2y + 8 = x\) 2. \(4y - 2x = -16\) First, rearrange these equations into slope-intercept form (\(y = mx + b\)). Equation 1: \(x = 2y + 8\) can be written as \(y = \frac{1}{2}x - 4\) Equation 2: \(4y - 2x = -16\) can be simplified to \(2y - x = -8\), which can be written as \(y = \frac{1}{2}x + 4\) Now, find two points for each of these equations using the method shown in set A, and graph them. **Set C:** Equations: 1. \(6x - 2y = 20\) 2. \(y + 3x - 5 = 0\) First, rearrange these equations into slope-intercept form. Equation 1: \(6x - 2y = 20\) can be simplified to \(3x - y = 10\), which can be written as \(y = 3x - 10\) Equation 2: \(y + 3x - 5 = 0\) can be rearranged to \(y = -3x + 5\) Again, find two points for each equation and graph each line. Please note that I cannot directly graph them here, but this explanation should help you plot them on your own graph paper. Once graphed, you can identify the intersection point of each pair of lines, which is the solution to each set of simultaneous equations.
Solving a System of Equations using Elimination Method
The image shows three different systems of simultaneous equations. Let's solve the first one (a) as requested: \[ \begin{align*} x - 3y - z &= 2 \\ 2x + 5y + 6z &= 3 \\ 3x + 2y + z &= 1 \\ \end{align*} \] To solve this system, we can use the method of substitution or elimination. I will use the elimination method. First, let's multiply the third equation by 6 and subtract the second equation from it to eliminate \( z \). \[ \begin{align*} 6(3x + 2y + z) &= 6(1) \\ 18x + 12y + 6z &= 6 \\ \end{align*} \] Now subtract the second equation: \[ \begin{align*} 18x + 12y + 6z - (2x + 5y + 6z) &= 6 - 3 \\ 16x + 7y &= 3 \quad \text{[Equation (4)]} \\ \end{align*} \] Next, let's eliminate \( z \) by multiplying the first equation by 6 and the third equation by 1 and subtracting them: \[ \begin{align*} 6(x - 3y - z) - (3x + 2y + z) &= 6(2) - 1 \\ 6x - 18y - 6z - 3x - 2y - z &= 12 - 1 \\ 3x - 20y - 7z &= 11 \quad \text{[Equation (5)]} \\ \end{align*} \] Now we have two equations [Equation (4) and Equation (5)] with two unknowns, \( x \) and \( y \). Let's multiply Equation (4) by 20 and Equation (5) by 7 to eliminate \( y \): \[ \begin{align*} 20(16x + 7y) &= 20(3) \\ 7(3x - 20y) &= 7(11) \\ \end{align*} \] \[ \begin{align*} 320x + 140y &= 60 \\ 21x - 140y &= 77 \\ \end{align*} \] Adding these two equations together to eliminate \( y \): \[ \begin{align*} 320x + 21x &= 60 + 77 \\ 341x &= 137 \\ x &= \frac{137}{341} \\ x &= \frac{1}{2.5} \\ x &= 0.4 \\ \end{align*} \] Now, plug the value of \( x \) back into Equation (4): \[ \begin{align*} 16(0.4) + 7y &= 3 \\ 6.4 + 7y &= 3 \\ 7y &= 3 - 6.4 \\ 7y &= -3.4 \\ y &= -\frac{3.4}{7} \\ y &= -\frac{34}{70} \\ y &= -\frac{17}{35} \\ y &= -0.4857... \\ \end{align*} \] Now that we have \( x \) and \( y \), we can solve for \( z \) using the first equation: \[ \begin{align*} 0.4 - 3(-0.4857...) - z &= 2 \\ 0.4 + 1.4571 - z &= 2 \\ z &= 0.4 + 1.4571 - 2 \\ z &= -0.1429... \\ \end{align*} \] Therefore, the solution to the system is \( x \approx 0.4 \), \( y \approx -0.4857 \), and \( z \approx -0.1429 \). Please note that the approximations for \( y \) and \( z \) are rounded.
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