Example Question - system of equations
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Solve the System of Equations with Square Root and Quadratic Equations
<p>Определим, что \( \sqrt{x - x^2} = 4 - x y \). Возводим обе стороны в квадрат:</p> <p>\( x - x^2 = (4 - xy)^2 \)</p> <p>\( x - x^2 = 16 - 8xy + x^2y^2 \)</p> <p>Получим выражение для \( y^2 \) из второго уравнения системы:</p> <p>\( y^2 - 4xy + 4 = 0 \)</p> <p>\( y^2 = 4xy - 4 \)</p> <p>Подставим это выражение для \( y^2 \) в предыдущее уравнение:</p> <p>\( x - x^2 = 16 - 8xy + x(4xy - 4) \)</p> <p>\( x - x^2 = 16 - 8xy + 4x^2y - 4x \)</p> <p>Теперь упростим и решим получившееся уравнение относительно \( x \):</p> <p>\( x - x^2 = 16 - 8xy + 4x^2y - 4x \)</p> <p>\( 0 = 16 - 8xy + 4x^2y - 4x + x^2 - x \)</p> <p>\( 0 = 16 - 4x(2y - 1) + x^2(4y - 1) \)</p> <p>Далее, учитывая, что система имеет несколько решений, можно проанализировать и использовать подходы для решения квадратных уравнений, чтобы найти допустимые значения \( x \) и \( y \). В конечном итоге, необходимо решить систему уравнений с учетом возможных ограничений, введенных квадратным корнем и квадратными уравнениями.</p>
Multiple Mathematics Problems Involving Algebra, Geometry, and Complex Numbers
<p>First question: Find the 4th term from the end in the expansion of \( \left( \frac{3}{x^2} - x^3 \right)^7 \).</p> <p>The 4th term from the end is the same as the 4th term from the beginning, which corresponds to \( T_4 \) in the expansion:</p> <p>\( T_k = \binom{n}{k-1} \cdot (a)^{n-(k-1)} \cdot (b)^{k-1} \) where \( n = 7 \), \( a = \frac{3}{x^2} \), and \( b = -x^3 \).</p> <p>\( T_4 = \binom{7}{3} \cdot \left(\frac{3}{x^2}\right)^{7-3} \cdot (-x^3)^3 \)</p> <p>\( T_4 = 35 \cdot \left(\frac{3}{x^2}\right)^4 \cdot (-x^3)^3 \)</p> <p>\( T_4 = 35 \cdot \frac{81}{x^8} \cdot (-x^9) \)</p> <p>\( T_4 = 35 \cdot 81 \cdot \frac{-x^9}{x^8} \)</p> <p>\( T_4 = -2835 \cdot \frac{1}{x} \)</p> <p>Second question: Find the equation of lines passing through (1,2) and making angle 30° with y-axis.</p> <p>The slope of the line making a 30° angle with the y-axis is the tangent of (90°-30°), which is \( \tan(60°) \).</p> <p>Slope (m) of the desired line: \( m = \tan(60°) = \sqrt{3} \)</p> <p>Use the point-slope form \( y - y_1 = m(x - x_1) \) to find the equation of the line passing through (1,2).</p> <p>\( y - 2 = \sqrt{3}(x - 1) \)</p> <p>Equation of the line: \( y = \sqrt{3}x + (2 - \sqrt{3}) \)</p> <p>Third question: Find the domain and range of the function \( f(x) = \frac{1}{\sqrt{9 - x^2}} \).</p> <p>The domain is where the function is defined and the denominator is not zero.</p> <p>\( 9 - x^2 > 0 \)</p> <p>\( -3 < x < 3 \), so the domain is \( (-3, 3) \).</p> <p>Since the function is the reciprocal of a square root, its range is all positive real numbers, \( (0, \infty) \).</p> <p>Fourth question: Solve the system of equations \( Re(z) = 0, |z| = 2 \).</p> <p>A complex number \( z = x + yi \) satisfies \( Re(z) = 0 \) when \( x = 0 \).</p> <p>Using \( |z| = 2 \), we get \( |0 + yi| = 2 \), which means \( \sqrt{0^2 + y^2} = 2 \).</p> <p>\( y = \pm 2 \), so \( z = 0 \pm 2i \).</p> <p>Fifth question: Write the complex number \( 1 + 7i \) in polar form.</p> <p>Let \( z = 1 + 7i \).</p> <p>Magnitude \( r = |z| = \sqrt{1^2 + 7^2} = \sqrt{50} = 5\sqrt{2} \).</p> <p>Angle \( \theta = \tan^{-1}\left(\frac{7}{1}\right) \), which is in the first quadrant.</p> <p>Polar form: \( z = r(\cos \theta + i \sin \theta) \)</p> <p>\( z = 5\sqrt{2}\left(\cos \left(\tan^{-1} 7\right) + i \sin \left(\tan^{-1} 7\right)\right) \)</p>
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 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>
Matrix Equations and Investment Problem Solving
<p>The image quality does not allow for a complete transcription of the problems, but I can guide you through the general steps for solving a problem similar to the first one, which involves matrix equations:</p> <p>To solve a matrix equation of the form \( AX = B \), where \( A \) and \( B \) are known matrices and \( X \) is the unknown matrix:</p> <p>1. Find the inverse of matrix \( A \), denoted as \( A^{-1} \), assuming that it exists.</p> <p>2. Multiply both sides of the equation by \( A^{-1} \) on the left to obtain \( A^{-1}AX = A^{-1}B \).</p> <p>3. Simplify the left side to get \( IX = A^{-1}B \), where \( I \) is the identity matrix.</p> <p>4. Since multiplying by the identity matrix does not change the matrix, we have \( X = A^{-1}B \).</p> <p>5. Compute \( A^{-1}B \) to find the matrix \( X \).</p> <p>For the problem related to investment, which seems to be a word problem with a financial context:</p> <p>1. Set up the problem by defining the variables and writing down the given information.</p> <p>2. Write down the equations that represent the investment situation described in the problem.</p> <p>3. Solve the system of equations using appropriate techniques such as substitution, elimination, or matrix methods if applicable.</p> <p>4. Interpret the solution in the context of the problem to find the required information about the investment.</p> <p>Without the full text of the problems, I cannot provide an exact solution. If you can provide a clearer image or transcribe the text, I'll be able to assist you better.</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>
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>
Cramer's Rule Application to Solve a System of Equations
Primero determinemos las matrices para aplicar la regla de Cramer. La matriz de los coeficientes es: \[ A = \begin{pmatrix} 4 & 2\\ 2 & 2 \end{pmatrix} \] El determinante de \( A \) es: \[ \det(A) = (4)(2) - (2)(2) = 8 - 4 = 4 \] Ahora, para la variable \( x \), reemplazamos la primera columna de \( A \) con el vector de términos constantes y calculamos su determinante: \[ A_x = \begin{pmatrix} 13 & 2\\ 5 & 2 \end{pmatrix} \] \[ \det(A_x) = (13)(2) - (5)(2) = 26 - 10 = 16 \] Y para la variable \( y \), reemplazamos la segunda columna de \( A \) con el vector de términos constantes y calculamos su determinante: \[ A_y = \begin{pmatrix} 4 & 13\\ 2 & 5 \end{pmatrix} \] \[ \det(A_y) = (4)(5) - (13)(2) = 20 - 26 = -6 \] Finalmente, utilizando la regla de Cramer, encontramos los valores de \( x \) y \( y \): \[ x = \frac{\det(A_x)}{\det(A)} = \frac{16}{4} = 4 \] \[ y = \frac{\det(A_y)}{\det(A)} = \frac{-6}{4} = -\frac{3}{2} \] Por lo tanto, la solución del sistema es \( x = 4 \) y \( y = -\frac{3}{2} \).
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>
Determining the Constant Value for a System of Equations Without Solutions
The system of equations is given by: \[ \begin{align*} 3x + \frac{1}{2}y &= \frac{3}{2} \\ 2x + \frac{3}{4}y &= 1 \\ \frac{1}{2}x + \frac{1}{4}y &= \frac{p}{2} + \frac{1}{2} \end{align*} \] To determine a value of \( p \) that leads to no solution, the equations must be inconsistent. This happens when the ratios of the coefficients of \( x \) and \( y \) are equal for two equations, but the ratio of the constants is different. First, let's make the coefficients of \( y \) similar in the first two equations: \[ \begin{align*} 3x + \frac{1}{2}y &= \frac{3}{2} \ |\times2 \\ 2x + \frac{3}{4}y &= 1 \ |\times4 \\ \end{align*} \] This results in: \[ \begin{align*} 6x + y &= 3 \\ 8x + 3y &= 4 \end{align*} \] Now, let's make the coefficient of \( x \) in the third equation similar to the first equation: \[ \begin{align*} \frac{1}{2}x + \frac{1}{4}y &= \frac{p}{2} + \frac{1}{2} \ |\times6 \\ \end{align*} \] This results in: \[ 3x + \frac{3}{2}y = 3p + 3 \] Comparing the first and last equations, the coefficient ratio of \( x \) to \( y \) is \( 6:1 \) and \( 3:\frac{3}{2} \). Simplifying the second ratio gives us \( 2:1 \), which is the same as \( 6:3 \). For the system to have no solution, the constant term (after the simplification) should not have the same ratio. Comparing the constants from the first and the last equations, \( 3 \) and \( 3p + 3 \) must not have the same ratio as the coefficients of \( x \) and \( y \), which is \( 2:1 \). Therefore, \( 2(3) \) should not equal \( 3p + 3 \). Equating and solving gives us: \[ 6 \neq 3p + 3 \] Subtracting \( 3 \) from both sides gives: \[ 3 \neq 3p \] Dividing by \( 3 \) we get: \[ 1 \neq p \] Thus, the value of \( p \) that results in no solution is any value except \( p = 1 \).
Finding the Coefficients of a Cubic Polynomial
Nous avons un polynôme \( P(x) = ax^3 + bx^2 - 18x + c \), où \( a \), \( b \), et \( c \) sont des nombres réels. Nous devons déterminer les valeurs de \( a \), \( b \), et \( c \) en sachant que \( P\left(\frac{1}{2}\right) = 0 \), \( P(0) = 8 \) et \( P(2) = 0 \). En utilisant ces informations, nous pouvons former un système d'équations : 1. \( P\left(\frac{1}{2}\right) = 0 \) donne \( a\left(\frac{1}{2}\right)^3 + b\left(\frac{1}{2}\right)^2 - 18\left(\frac{1}{2}\right) + c = 0 \) En simplifiant, nous obtenons \( \frac{a}{8} + \frac{b}{4} - 9 + c = 0 \) ou \( \frac{a}{8} + \frac{b}{4} + c = 9 \) (équation 1). 2. \( P(0) = 8 \) donne \( c = 8 \) (équation 2). 3. \( P(2) = 0 \) donne \( a(2)^3 + b(2)^2 - 18(2) + c = 0 \) En simplifiant, nous avons \( 8a + 4b - 36 + c = 0 \) Mais nous savons déjà que \( c = 8 \), donc \( 8a + 4b - 36 + 8 = 0 \) ou \( 8a + 4b = 28 \) ou \( 2a + b = 7 \) (équation 3). Maintenant, nous utilisons l'équation 2 pour remplacer \( c \) par \( 8 \) dans l'équation 1 : \( \frac{a}{8} + \frac{b}{4} + 8 = 9 \) donne \( \frac{a}{8} + \frac{b}{4} = 1 \) ou \( a + 2b = 8 \) (équation 4). Nous avons maintenant un système de deux équations avec deux inconnues (a, b) : - \( 2a + b = 7 \) (équation 3) - \( a + 2b = 8 \) (équation 4) Pour résoudre le système, nous pouvons utiliser une méthode d'élimination ou de substitution. Utilisons la méthode de substitution. Multiplions la deuxième équation par 2 et soustrayons la première équation de celle-ci : \( 2(a + 2b) = 2(8) \) \( 2a + 4b = 16 \) Et maintenant soustrayons équation 3 de cette nouvelle équation : \( (2a + 4b) - (2a + b) = 16 - 7 \) \( 3b = 9 \) \( b = 3 \) Maintenant, remplaçons \( b \) dans l'équation 3 pour trouver \( a \) : \( 2a + 3 = 7 \) \( 2a = 4 \) \( a = 2 \) Nous connaissons déjà \( c \) grâce à l'équation 2 : \( c = 8 \) Donc, les valeurs de \( a \), \( b \), et \( c \) sont respectivement 2, 3, et 8.
Determining Number of Solutions for a System of Equations from Intersection Point on a Graph
The image provided shows a graph on a rectangular coordinate system with two lines that intersect. The question asks to determine the number of solutions for the system of equations represented by these lines. If two lines intersect on a graph, it means they have exactly one point in common, which signifies there is exactly one solution to the system of equations. The coordinates of that solution are located at the intersection point. However, due to the low resolution of the image, I'm not able to clearly see the intersection point's coordinates. Normally, to answer this question, I would identify the exact point where the two lines intersect and provide those coordinates, which would represent the solution to the system of equations. Since the question prompts to give the coordinates if there is only one solution, you should look on the graph for where the two lines cross, and provide the x and y values of that point. It's also worth mentioning the other possible answer choices in the question: - "The only solution is ________________." (the blank would normally be filled with the coordinates of the intersection point) - "There is no solution." - "There is an infinite number of solutions." Given that the lines intersect, the correct answer is not "There is no solution" nor "There is an infinite number of solutions," but the first option, specifying the coordinates of the intersection point. Since I cannot provide the coordinates due to the image's quality, you can visually estimate or determine them using the provided graph in the image. If you need help with that, please provide a clearer image or the coordinates of points close to the intersection.
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