Example Question - unknown variables
Here are examples of questions we've helped users solve.
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 Equations for Unknown Variables
Claro, puedo ayudarte con eso. La pregunta muestra varias ecuaciones donde se debe resolver para encontrar el valor de la letra \( a \) en cada caso. Vamos a resolver cada una de ellas: a) \( 25 : a = 5 \) Para resolver para \( a \), se necesita despejar \( a \) multiplicando ambos lados de la ecuación por \( a \) y luego dividiendo ambos lados por 5: \[ a = \frac{25}{5} \] \[ a = 5 \] b) \( 18 : a = 1 \) En este caso, como cualquier número dividido por sí mismo da como resultado 1, \( a \) debe ser igual al número en el otro lado de la ecuación: \[ a = 18 \] c) \( 14 : (a) = 2 \) Despejamos \( a \) de forma similar: \[ a = \frac{14}{2} \] \[ a = 7 \] d) \( -7 + a = -11 \) Para despejar \( a \), sumamos 7 a ambos lados de la ecuación: \[ a = -11 + 7 \] \[ a = -4 \] Resumiendo, las respuestas son: a) \( a = 5 \) b) \( a = 18 \) c) \( a = 7 \) d) \( a = -4 \)
Solving Quadrilateral Angles Equation
The image you provided shows a quadrilateral ABCD with its interior angles labeled as follows: - Angle A = x degrees - Angle B = y degrees - Angle C = (2z) degrees - Angle D = (z - 15) degrees The sum of the interior angles of any quadrilateral is 360 degrees. Therefore, to solve for x, y, and z, we can set up the following equation: x + y + (2z) + (z - 15) = 360 Combine like terms: x + y + 3z - 15 = 360 This gives us one equation with three unknowns, which means we cannot uniquely determine the values of x, y, and z without additional information. To solve for these variables, we would need two more independent equations relating them. However, if the question is to express x, y, and z in terms of each other, you could solve one variable in terms of the other two. If that's the case or if there are additional constraints or information, please provide them so that we can assist you further.
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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