Example Question - linear equations
Here are examples of questions we've helped users solve.
Linear Equation in Two Variables
<p>To solve the equation \(3x - 5y - 21 = 0\), we can isolate \(y\).</p> <p>Add \(5y\) and \(21\) to both sides:</p> <p>\(3x = 5y + 21\)</p> <p>Now, subtract \(21\) from both sides:</p> <p>\(3x - 21 = 5y\)</p> <p>Divide each term by \(5\) to solve for \(y\):</p> <p>\(y = \frac{3x - 21}{5}\)</p>
Vector Addition Problem
<p>Let the two vectors be </p> <p> \[ \begin{bmatrix} 2 \\ 3 \\ 0 \end{bmatrix} + \begin{bmatrix} -1 & y \\ x & -2 \end{bmatrix} = \begin{bmatrix} 1 \\ -4 \\ 9 \\ -2 \end{bmatrix} \] </p> <p>Set up the equation for each component:</p> <p>1. \(2 - 1 = 1\)</p> <p>2. \(3 + y = -4\)</p> <p>3. \(0 + x = -2\)</p> <p>From the first equation: \(1\) is valid.</p> <p>From the second equation: \(y = -4 - 3 = -7\).</p> <p>From the third equation: \(x = -2\).</p> <p>Thus, \(x = -2\) and \(y = -7\).</p>
Solving Linear Equations Involving Area
Da zwei Gleichungen gegeben sind und das Ziel ist, \( y \) zu lösen, kann das folgendermaßen angegangen werden: <p>\( A = 12 \cdot y \) (1)</p> <p>\( A = 52 \cdot y \) (2)</p> Diese beiden Gleichungen scheinen einen gemeinsamen Flächenwert \( A \) zu haben, aber unterschiedliche Multiplikatoren für \( y \). Wenn das der Fall ist, wären die Multiplikatoren gleich, d.h. \( 12 = 52 \), was offensichtlich nicht wahr ist. Daher muss es einen Fehler in der Fragestellung geben, oder es fehlt Kontext bzw. zusätzliche Information, um die Gleichungen angemessen zu lösen. Ohne zusätzliche Information kann nicht bestimmt werden, was \( y \) ist, oder ob \( A \) in beiden Gleichungen gleich ist. Wenn \( A \) nicht gleich sein soll, kann \( y \) für jede Gleichung unterschiedlich sein, aber ohne spezifische Werte für \( A \) kann \( y \) nicht gelöst werden.
Algebraic Equation Simplification
<p>La ecuación proporcionada en la imagen es: \(5x - (-8) + (-9) = 9x - (-7 + 1)\)</p> <p>Primero simplificamos la ecuación combinando términos semejantes y eliminando los paréntesis:</p> <p>\(5x + 8 - 9 = 9x - (-7 + 1)\)</p> <p>\(5x - 1 = 9x - (6)\)</p> <p>Ahora procedemos a aislar la variable \(x\):</p> <p>\(5x - 9x = -6 + 1\)</p> <p>\(-4x = -5\)</p> <p>\(x = \frac{-5}{-4}\)</p> <p>\(x = \frac{5}{4}\)</p> <p>Por lo tanto, la solución de la ecuación es \(x = \frac{5}{4}\).</p>
Comparison of Food Package Prices
Let the price per pack of macaroni be \(x\) RM and the price per pack of pizza be \(y\) RM. For the Ruby package: <p>\(7x + 3y = \text{Price of Ruby Package}\)</p> For the Emerald package: <p>\(4x + 6y = \text{Price of Emerald Package}\)</p> Solve these two linear equations simultaneously to find \(x\) and \(y\), the prices per pack of macaroni and pizza respectively.
Solving a System of Linear Equations Using Augmented Matrix Method
<p>The given image shows a system of linear equations represented in an augmented matrix form, which we aim to solve. To find the solution, we must reduce this matrix to its reduced row echelon form (RREF).</p> <p>Here's the step-by-step solution:</p> <p>Starting matrix:</p> \[ \begin{bmatrix} 3 & -4 & | & 1 & 0 & 0 \\ 2 & 4 & | & 0 & 1 & 0 \\ -4 & 2 & | & 0 & 0 & 1 \\ \end{bmatrix} \] <p>Step 1: Make the first element of the first row a 1 (if it's not already) by dividing the entire first row by 3.</p> \[ \begin{bmatrix} 1 & -\frac{4}{3} & | & \frac{1}{3} & 0 & 0 \\ 2 & 4 & | & 0 & 1 & 0 \\ -4 & 2 & | & 0 & 0 & 1 \\ \end{bmatrix} \] <p>Step 2: Eliminate the first element of the second and third row to make them 0 by adding (-2) times the first row to the second row and adding 4 times the first row to the third row.</p> \[ \begin{bmatrix} 1 & -\frac{4}{3} & | & \frac{1}{3} & 0 & 0 \\ 0 & \frac{14}{3} & | & -\frac{2}{3} & 1 & 0 \\ 0 & \frac{2}{3} & | & \frac{4}{3} & 0 & 1 \\ \end{bmatrix} \] <p>Step 3: Make the second element of the second row a 1 by dividing the entire second row by $\frac{14}{3}$.</p> \[ \begin{bmatrix} 1 & -\frac{4}{3} & | & \frac{1}{3} & 0 & 0 \\ 0 & 1 & | & -\frac{1}{7} & \frac{3}{14} & 0 \\ 0 & \frac{2}{3} & | & \frac{4}{3} & 0 & 1 \\ \end{bmatrix} \] <p>Step 4: Eliminate the second element of the first and third row to make them 0 by adding $\frac{4}{3}$ times the second row to the first row and adding $-\frac{2}{3}$ times the second row to the third row.</p> \[ \begin{bmatrix} 1 & 0 & | & \frac{5}{21} & \frac{4}{21} & 0 \\ 0 & 1 & | & -\frac{1}{7} & \frac{3}{14} & 0 \\ 0 & 0 & | & \frac{8}{7} & -\frac{1}{7} & 1 \\ \end{bmatrix} \] <p>This matrix is in RREF. The solutions to the system of equations can be read directly from the matrix:</p> \[ x = \frac{5}{21}, y = -\frac{1}{7}, z = \frac{8}{7} \]
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 Linear Equations by Combining Like Terms
<p>\text{Учитывая уравнение: } 2x - 3y = \frac{2}{5}(x - 2y) + \frac{3}{2}(x - y)</p> <p>\text{Первый шаг - избавиться от знаменателей, умножив каждый член на 10 (наименьшее общее кратное знаменателей 5 и 2): }</p> <p>10 \cdot 2x - 10 \cdot 3y = 10 \cdot \frac{2}{5}(x - 2y) + 10 \cdot \frac{3}{2}(x - y)</p> <p>20x - 30y = 4(x - 2y) + 15(x - y)</p> <p>\text{Раскроем скобки: }</p> <p>20x - 30y = 4x - 8y + 15x - 15y</p> <p>\text{Сложим подобные слагаемые справа: }</p> <p>20x - 30y = (4x + 15x) - (8y + 15y)</p> <p>20x - 30y = 19x - 23y</p> <p>\text{Теперь перенесем все члены с переменными на одну сторону, а числа - на другую: }</p> <p>20x - 19x = 23y - 30y</p> <p>x = -7y</p> <p>\text{Финальное уравнение в простейшей форме: } x = -7y.</p>
Solving Algebraic Equations
<p>Для решения уравнений вида \(\frac{x}{a} = b\) и \(\frac{m}{n} = c\), где \(x, m, n, a, b, c\) - числа, нужно выполнить следующие шаги:</p> <p>а) \(\frac{x}{9} = 13\)</p> <p>Шаг 1: Умножьте обе части уравнения на 9.</p> <p>\(x = 13 \times 9\)</p> <p>\(x = 117\)</p> <p>б) \(\frac{132}{k} = 11\)</p> <p>Шаг 1: Умножьте обе части уравнения на \(k\).</p> <p>132 = 11k</p> <p>Шаг 2: Разделите обе части уравнения на 11.</p> <p>\(k = \frac{132}{11}\)</p> <p>\(k = 12\)</p> <p>в) \(\frac{m}{12} = 28\)</p> <p>Шаг 1: Умножьте обе части уравнения на 12.</p> <p>\(m = 28 \times 12\)</p> <p>\(m = 336\)</p> <p>г) \(\frac{528}{n} = 66\)</p> <p>Шаг 1: Умножьте обе части уравнения на \(n\).</p> <p>528 = 66n</p> <p>Шаг 2: Разделите обе части уравнения на 66.</p> <p>\(n = \frac{528}{66}\)</p> <p>\(n = 8\)</p>
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>
Graphing Linear Equations
<p>Para graficar la ecuación lineal \( y = 3x - 2 \), primero evaluamos \( y \) para cada valor de \( x \) en la tabla proporcionada.</p> <p>Si \( x = 3 \):</p> <p>\( y = 3(3) - 2 = 9 - 2 = 7 \)</p> <p>Si \( x = 2 \):</p> <p>\( y = 3(2) - 2 = 6 - 2 = 4 \)</p> <p>Si \( x = -2 \):</p> <p>\( y = 3(-2) - 2 = -6 - 2 = -8 \)</p> <p>Y cuando \( x = 0 \), que ya está dado en la tabla:</p> <p>\( y = 3(0) - 2 = 0 - 2 = -2 \)</p> <p>Ahora podemos graficar los puntos (3,7), (2,4), (-2,-8), y (0,-2) en el plano coordenado y dibujar la línea que los une, que será la gráfica de la ecuación lineal \( y = 3x - 2 \).</p>
Plotting Linear Equations on a Graph
<p>Para resolver la ecuación lineal \( y = 3x - 2 \) y trazarla en un plano coordenado, primero necesitamos calcular los valores de \( y \) para los valores dados de \( x \):</p> <p>\( x = 3 \) \\ \( y = 3(3) - 2 \) \\ \( y = 9 - 2 \) \\ \( y = 7 \)</p> <p>\( x = 2 \) \\ \( y = 3(2) - 2 \) \\ \( y = 6 - 2 \) \\ \( y = 4 \)</p> <p>\( x = 1 \) \\ \( y = 3(1) - 2 \) \\ \( y = 3 - 2 \) \\ \( y = 1 \)</p> <p>\( x = 0 \) \\ \( y = 3(0) - 2 \) \\ \( y = 0 - 2 \) \\ \( y = -2 \)</p> <p>Con estos puntos, podemos trazar la línea en el gráfico.</p>
Solving a System of Linear Equations Using the Gauss-Jordan Method
<p>Given the system of equations:</p> <p>\[\begin{cases} 3x_1 - 0.1x_2 - 0.2x_3 = 7.85 \\ 0.1x_1 + 7x_2 - 0.3x_3 = -19.3 \\ 0.3x_1 - 0.2x_2 + 10x_3 = 71.4 \end{cases}\]</p> <p>First, we form the augmented matrix:</p> <p>\[\begin{bmatrix} 3 & -0.1 & -0.2 & | & 7.85 \\ 0.1 & 7 & -0.3 & | & -19.3 \\ 0.3 & -0.2 & 10 & | & 71.4 \end{bmatrix}\]</p> <p>Next, apply Gauss-Jordan elimination steps to reduce the matrix to reduced row echelon form.</p> <p>1. Make the first element of the first row a 1 by dividing the first row by 3:</p> <p>\[\begin{bmatrix} 1 & -0.0333 & -0.0667 & | & 2.6167 \\ 0.1 & 7 & -0.3 & | & -19.3 \\ 0.3 & -0.2 & 10 & | & 71.4 \end{bmatrix}\]</p> <p>2. Make the first element of the second and third rows a 0:</p> <p>\[\begin{bmatrix} 1 & -0.0333 & -0.0667 & | & 2.6167 \\ 0 & 7.0033 & -0.2933 & | & -19.5617 \\ 0 & -0.1 & 9.98 & | & 70.61 \end{bmatrix}\]</p> <p>3. Make the second element of the second row a 1 by dividing the second row by 7.0033:</p> <p>\[\begin{bmatrix} 1 & -0.0333 & -0.0667 & | & 2.6167 \\ 0 & 1 & -0.0418 & | & -2.7936 \\ 0 & -0.1 & 9.98 & | & 70.61 \end{bmatrix}\]</p> <p>4. Make the second element of the first and third rows a 0:</p> <p>\[\begin{bmatrix} 1 & 0 & -0.0677 & | & 2.5234 \\ 0 & 1 & -0.0418 & | & -2.7936 \\ 0 & 0 & 9.982 & | & 70.8886 \end{bmatrix}\]</p> <p>5. Make the third element of the third row a 1 by dividing the third row by 9.982:</p> <p>\[\begin{bmatrix} 1 & 0 & -0.0677 & | & 2.5234 \\ 0 & 1 & -0.0418 & | & -2.7936 \\ 0 & 0 & 1 & | & 7.1047 \end{bmatrix}\]</p> <p>6. Make the third element of the first and second rows a 0:</p> <p>\[\begin{bmatrix} 1 & 0 & 0 & | & 2.9969 \\ 0 & 1 & 0 & | & -2.5035 \\ 0 & 0 & 1 & | & 7.1047 \end{bmatrix}\]</p> <p>We now have the reduced row echelon form, which corresponds to the solutions for \(x_1\), \(x_2\), and \(x_3\):</p> <p>\[x_1 \approx 2.9969\]</p> <p>\[x_2 \approx -2.5035\]</p> <p>\[x_3 \approx 7.1047\]</p> <p>So, the solution to the system of equations is approximately \(x_1 \approx 2.9969\), \(x_2 \approx -2.5035\), \(x_3 \approx 7.1047\).</p>
Solving a System of Linear Equations Using the Gauss-Seidel Method
<p>To use the Gauss-Seidel method, we first express each equation in the form \(x = ... , y = ... , z = ...\).</p> <p>From \(5x + 2y + z = 12\), we get \[x = \frac{12 - 2y - z}{5}\]</p> <p>From \(x + 4y + 2z = 15\), we get \[y = \frac{15 - x - 2z}{4}\]</p> <p>From \(x + 2y + 5z = 20\), we get \[z = \frac{20 - x - 2y}{5}\]</p> <p>We then assume initial values for \(x, y, z\), usually zeros: \(x_0 = 0, y_0 = 0, z_0 = 0\).</p> <p>Iterate the equations using the previous values:</p> <p>\[x_{n+1} = \frac{12 - 2y_n - z_n}{5}\]</p> <p>\[y_{n+1} = \frac{15 - x_{n+1} - 2z_n}{4}\]</p> <p>\[z_{n+1} = \frac{20 - x_{n+1} - 2y_{n+1}}{5}\]</p> <p>The iterations are repeated until the solutions converge to a set level of accuracy.</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>
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