Example Question - equation solving
Here are examples of questions we've helped users solve.
Solving a Basic Algebra Problem
Vamos resolver a questão dada na imagem. A questão diz que "Henry adiciona 5 a um número. Ele divide a soma por 2 e o quociente é 6". Seja \( x \) o número desconhecido. Então, a expressão para o problema é: \[ \frac{x + 5}{2} = 6 \] Para encontrar o valor de \( x \), siga os passos: <p>\( x + 5 = 6 \cdot 2 \)</p> <p>\( x + 5 = 12 \)</p> <p>\( x = 12 - 5 \)</p> <p>\( x = 7 \)</p> Portanto, o número desconhecido é 7.
Simple Algebraic Problem Involving Multiplication and Addition
<p>Vamos considerar o número desconhecido como \( x \).</p> <p>De acordo com o problema, Sam multiplica este número por 2 e então adiciona 3 ao produto, e o resultado dessa operação é 7. Logo, podemos expressar isso como uma equação: </p> <p>\( 2x + 3 = 7 \)</p> <p>Para resolver a equação, primeiro subtraímos 3 de ambos os lados:</p> <p>\( 2x + 3 - 3 = 7 - 3 \)</p> <p>\( 2x = 4 \)</p> <p>Agora dividimos ambos os lados por 2 para obter \( x \):</p> <p>\( \frac{2x}{2} = \frac{4}{2} \)</p> <p>\( x = 2 \)</p> <p>Portanto, o número que Sam escolheu é 2.</p>
Solving a Rational Equation
<p>Given equation:</p> <p>\(\frac{x}{2} = \frac{5x - 24}{x - 4}\)</p> <p>To solve for \( x \), cross-multiply to eliminate the denominators:</p> <p>\(x(x - 4) = 2(5x - 24)\)</p> <p>Expand both sides:</p> <p>\(x^2 - 4x = 10x - 48\)</p> <p>Move all terms to one side to set the equation to zero:</p> <p>\(x^2 - 4x - 10x + 48 = 0\)</p> <p>Simplify by combining like terms:</p> <p>\(x^2 - 14x + 48 = 0\)</p> <p>Factor the quadratic equation:</p> <p>\((x - 6)(x - 8) = 0\)</p> <p>Set each factor equal to zero:</p> <p>\(x - 6 = 0 \quad \text{or} \quad x - 8 = 0\)</p> <p>Solve for \( x \):</p> <p>\(x = 6 \quad \text{or} \quad x = 8\)</p>
Solving an Algebraic Equation Involving Basic Arithmetic Operations
<p>The image displays an equation with a missing number which needs to be found:</p> <p>\( (5 \times 8) + (\boxed{?} \times 8) = 40 + \boxed{?} \)</p> <p>Since \( 5 \times 8 = 40 \), the equation simplifies to:</p> <p>\( 40 + (\boxed{?} \times 8) = 40 + \boxed{?} \)</p> <p>To find the missing number, let's denote it as \( x \):</p> <p>\( 40 + (x \times 8) = 40 + x \)</p> <p>Subtract \( 40 \) from both sides:</p> <p>\( (x \times 8) = x \)</p> <p>Divide both sides by \( x \) (assuming \( x \neq 0 \)):</p> <p>\( 8 = 1 \)</p> <p>This leads to a contradiction, as \( 8 \) does not equal \( 1 \). The problem seems to be constructed erroneously, as the equation provided does not hold true for any real number \( x \). We would need additional information or a correction to the equation to proceed.</p>
Trigonometric Equation Solving
\[ \begin{align*} &\text{给定方程式为} \quad 2\tan(\theta) = 3\cos(\theta) \\ &\text{解这个方程式，我们首先将其转换成同一个三角函数，} \\ &\text{使用基本恒等式} \quad \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}: \\ &2\frac{\sin(\theta)}{\cos(\theta)} = 3\cos(\theta) \\ &2\sin(\theta) = 3\cos^2(\theta) \\ &\text{因为} \quad \cos^2(\theta) = 1 - \sin^2(\theta), \text{代入上面方程得到} \\ &2\sin(\theta) = 3(1 - \sin^2(\theta)) \\ &3\sin^2(\theta) + 2\sin(\theta) - 3 = 0 \\ &\text{将}\sin(\theta)\text{代换为} x \\ &3x^2 + 2x - 3 = 0 \\ &\text{使用求根公式得到:} \\ &x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ &x = \frac{-2 \pm \sqrt{(2)^2 - 4(3)(-3)}}{2(3)} \\ &x = \frac{-2 \pm \sqrt{4 + 36}}{6} \\ &x = \frac{-2 \pm \sqrt{40}}{6} \\ &x = \frac{-2 \pm 2\sqrt{10}}{6} \\ &\text{因为} -1 \leq \sin(\theta) \leq 1, \text{所以我们舍弃} x > 1 \text{和} x < -1 \text{的根。} \\ &x = -1 \text{（不合理）或 } x = \frac{\sqrt{10}-1}{3} \\ &\text{所以}\sin(\theta) = \frac{\sqrt{10}-1}{3} \\ &\text{利用计算器得到} \theta \text{的两个可能的值(确保在 0° 至 360°之间)} \\ &\theta \approx 50.2^\circ \text{或} 360^\circ - 50.2^\circ \approx 309.8^\circ \\ &\text{最终答案为：} \theta \approx 50.2^\circ \text{或} 309.8^\circ \end{align*} \]
Verification of a Proposed Solution for a Linear Equation
<p>To verify if \( x = -2 \) is a solution to the linear equation \( 3x + 5 = -1 \), substitute \( x \) with -2:</p> <p>\( 3(-2) + 5 = -1 \)</p> <p>\( -6 + 5 = -1 \)</p> <p>\( -1 = -1 \)</p> <p>Since the left-hand side equals the right-hand side after the substitution, \( x = -2 \) is indeed a solution to the given linear equation.</p>
Determining the Value of a for Curves Intersection
Langkah 1: Susun persamaan f(x) = h(x) dan g(x) = h(x). Langkah 2: Selesaikan untuk x dari kedua persamaan tersebut. Persamaan f(x) = h(x): -2x^2 + 1 = -5x^2 + 1 3x^2 = 0 x = 0 Persamaan g(x) = h(x): -2x^2 + 1 = -ax^2 + 1 2x^2 - ax^2 = 0 x^2(2 - a) = 0 Sebab x tidak boleh sentiasa 0 (asta yang akan jadi jika kita bawa 2 - a = 0), maka 2 - a = 0 a = 2 Langkah 3: Tentukan julat nilai a. Kerana tiada pembatasan lain yang diberikan, kita boleh simpulkan bahawa nilai a mesti 2 untuk memastikan semua kurva bersilang pada titik yang sama.
Finding Supplementary Angles in Leaf Veins
The question states that the veins in a leaf form a pair of supplementary angles. Supplementary angles are two angles whose measures add up to 180 degrees. Given the measures of the two angles as \( m\angle1 = (7x + 13)^\circ \) and \( m\angle2 = (25x + 7)^\circ \), we can set up an equation that reflects their supplementary nature: \[ (7x + 13) + (25x + 7) = 180 \] Simplify the equation: \[ 7x + 25x + 13 + 7 = 180 \] \[ 32x + 20 = 180 \] Now, solve for \( x \): \[ 32x = 180 - 20 \] \[ 32x = 160 \] \[ x = \frac{160}{32} \] \[ x = 5 \] Now that we have the value of \( x \), we can find the measure of each angle: \[ m\angle1 = 7x + 13 \] \[ m\angle1 = 7(5) + 13 \] \[ m\angle1 = 35 + 13 \] \[ m\angle1 = 48^\circ \] And for the other angle: \[ m\angle2 = 25x + 7 \] \[ m\angle2 = 25(5) + 7 \] \[ m\angle2 = 125 + 7 \] \[ m\angle2 = 132^\circ \] So the measures of the two angles are 48 degrees and 132 degrees, respectively.
Equation Solving: Determining Constant Value
To solve the question, we will add the two equations together: The first equation is: -2x + 4y = 15 The second equation is: -12x - 4y = -8 Adding the left hand sides of the equations: (-2x + 4y) + (-12x - 4y) (-2x - 12x) + (4y - 4y) -14x + 0 And adding the right hand sides of the equations: 15 + (-8) 7 So after adding the two equations together we have: -14x + 0 = 7 The question is asking for the value that belongs in the green box, which corresponds to the constant term on the right side of the equation after adding the equations together. Therefore, the value that belongs in the green box is 7. The correct answer: B. 7
Solving a Two-Job Earnings Problem
I can help you solve the problem. To answer the question, we will follow these steps: ### Part a: We need to write an equation with the given information. Maggie has one job that pays $9 per hour and another that pays $11 per hour. We'll use the variable \( x \) to denote the number of hours she works at the $9 per hour job and \( y \) for the number of hours she works at the $11 per hour job. Given that Maggie earns a total of $297 in one week, the equation based on this information will be: \[ 9x + 11y = 297 \] ### Part b: To find the x-intercept, set \( y = 0 \) and solve for \( x \). To find the y-intercept, set \( x = 0 \) and solve for \( y \). #### X-intercept \( (x, 0) \): \[ 9x + 11(0) = 297 \] \[ 9x = 297 \] \[ x = \frac{297}{9} \] \[ x = 33 \] So the x-intercept is \( (33, 0) \). #### Y-intercept \( (0, y) \): \[ 9(0) + 11y = 297 \] \[ 11y = 297 \] \[ y = \frac{297}{11} \] \[ y = 27 \] So the y-intercept is \( (0, 27) \). Therefore, the x-intercept is 33 and the y-intercept is 27.
Solving for v in terms of u and finding a specific value
To express \( v \) in terms of \( u \), we will start with the given equation \( u = \sqrt{2v - 3} \) and solve for \( v \). Here are the steps: (a) Express \( v \) in terms of \( u \): 1. Square both sides of the equation to eliminate the square root: \( u^2 = (\sqrt{2v - 3})^2 \) 2. This simplifies to: \( u^2 = 2v - 3 \) 3. Now add 3 to both sides to isolate the \( v \) term: \( u^2 + 3 = 2v \) 4. Finally, divide both sides by 2 to solve for \( v \): \( v = \frac{u^2 + 3}{2} \) (b) Find the value of \( v \) if \( u = 5 \): To find the value of \( v \) when \( u = 5 \), substitute \( u \) with 5 in the equation from part (a): \( v = \frac{5^2 + 3}{2} \) \( v = \frac{25 + 3}{2} \) \( v = \frac{28}{2} \) \( v = 14 \) So, the value of \( v \) when \( u = 5 \) is \( 14 \).
Understanding Equation Solving With Like Terms
To solve the equation \( 403s + 1002s = 1410s \), we can combine like terms on the left side by adding the coefficients of \( s \) together. Here's how you do it step by step: 1. Combine like terms (the terms with \( s \)): \[ 403s + 1002s = (403 + 1002)s = 1405s \] 2. Now that we have \( 1405s \) on the left side and \( 1410s \) on the right side, we set them equal to each other: \[ 1405s = 1410s \] 3. In order to find the solution for \( s \), you generally would isolate \( s \) by dividing both sides of the equation by the coefficient of \( s \). However, since the coefficients of \( s \) are equal and there is no constant term, this would result in an identity where any value of \( s \) satisfies the equation. But since the coefficients are actually different (1405 vs. 1410), there seems to be a mistake in the original problem or a transcription error. If the equation were correctly stated as \( 1405s = 1410s \), the only solution would be \( s = 0 \) because that would be the only way the two sides could be equal. Any non-zero value of \( s \) would result in the two sides not being equal. Since the equation you provided is already balanced with \( 403s + 1002s = 1410s \), which simplifies to \( 1405s = 1410s \), this equation has no solution for \( s \) unless there is a mistake in the terms provided. If the equation is correct, then the solution is that there is no value for \( s \) that would satisfy this equation.
CamTutor
In regards to math, we are professionals.
Get In Touch
Email: camtutor.ai@gmail.com