Example Question - subtraction
Here are examples of questions we've helped users solve.
Calculate the operations
<p>Primero, encontramos un denominador común para las fracciones. El común denominador de \(3\) y \(4\) es \(12\).</p> <p>Convertimos las fracciones: \(\frac{2}{3} = \frac{8}{12}\), \(\frac{7}{3} = \frac{28}{12}\), y \(\frac{1}{4} = \frac{3}{12}\).</p> <p>Ahora, realizamos la operación: \(\frac{8}{12} + \frac{28}{12} - \frac{3}{12} = \frac{8 + 28 - 3}{12} = \frac{33}{12}\).</p> <p>Finalmente, simplificamos: \(\frac{33}{12} = \frac{11}{4}\).</p>
Combined Operations in Rational Numbers
<p>Para resolver la operación b) que es:</p> <p>2 + \frac{1}{15} - \frac{2}{9}</p> <p>Primero, encuentra un común denominador para las fracciones. El común denominador de 15 y 9 es 45.</p> <p>Convierte las fracciones:</p> <p>2 = \frac{90}{45}, \frac{1}{15} = \frac{3}{45}, \frac{2}{9} = \frac{10}{45}</p> <p>Ahora, realiza la operación:</p> <p>\frac{90}{45} + \frac{3}{45} - \frac{10}{45} = \frac{90 + 3 - 10}{45} = \frac{83}{45}</p> <p>Por lo tanto, la respuesta es:</p> <p>\frac{83}{45}</p>
Calculating Fractions
<p>a) \frac{2}{3} + \frac{7}{3} - \frac{1}{4} = \frac{2 + 7 \cdot 1 - \frac{3}{4}}{3} = \frac{9 - 0.75}{3} = \frac{8.25}{3} = \frac{33}{12} = \frac{11}{4}</p> <p>b) 2 \frac{1}{3} + \frac{1}{2} - \frac{1}{5} - \frac{1}{15} = \frac{7}{3} + \frac{1}{2} - \frac{1}{5} - \frac{1}{15} = \frac{70 + 15 - 6 - 2}{30} = \frac{77}{30}</p> <p>c) \frac{13}{36} - \frac{5}{4} - \frac{2}{9} = \frac{13}{36} - \frac{45}{36} - \frac{8}{36} = \frac{13 - 45 - 8}{36} = \frac{-40}{36} = -\frac{10}{9}</p> <p>d) 3 + \left( -\frac{5}{4} \right) + 2 \frac{3}{4} = 3 - \frac{5}{4} + \frac{11}{4} = 3 + \frac{6}{4} = 3 + \frac{3}{2} = \frac{6 + 3}{2} = \frac{9}{2}</p> <p>e) \frac{5}{6} - \frac{2}{9} = \frac{15}{18} - \frac{4}{18} = \frac{11}{18}</p> <p>f) 7 \frac{1}{4} - \left( 4 - \frac{1}{2} \right) = \frac{29}{4} - \left( \frac{8}{2} - \frac{1}{2} \right) = \frac{29}{4} - \frac{7}{2} = \frac{29 - 14}{4} = \frac{15}{4}</p>
Combined Addition and Subtraction with Grouping Symbols
<p>Primero se calcula cada término interno:</p> <p>\(\frac{5}{20} = \frac{1}{4}\)</p> <p>\( -\frac{1}{10} \)</p> <p>\(+ \frac{2}{5}\)</p> <p>\(\left[\frac{3}{4} + \left(2 - \frac{5}{3}\right) + 4\right]\)</p> <p>Para el término \((2 - \frac{5}{3})\):</p> <p>\(2 = \frac{6}{3}\)</p> <p>Entonces, \(2 - \frac{5}{3} = \frac{6}{3} - \frac{5}{3} = \frac{1}{3}\)</p> <p>Ahora se resuelve \(\frac{3}{4} + \frac{1}{3} + 4\):</p> <p>El mínimo común múltiplo de 4 y 3 es 12:</p> <p>\(\frac{3}{4} = \frac{9}{12}\)</p> <p>\(\frac{1}{3} = \frac{4}{12}\)</p> <p>Luego:</p> <p>\(\frac{9}{12} + \frac{4}{12} + 4 = \frac{13}{12} + 4\)</p> <p>Transformamos 4 en fracción:</p> <p> \(4 = \frac{48}{12}\)</p> <p>Finalmente:</p> <p>\(\frac{13}{12} + \frac{48}{12} = \frac{61}{12}\)</p> <p>Ahora juntamos todos los términos:</p> <p>\(\frac{1}{4} - \frac{1}{10} + \frac{61}{12}\)</p> <p>Encontramos un MCM para 4, 10 y 12, que es 60:</p> <p>\(\frac{1}{4} = \frac{15}{60}\)</p> <p>\(-\frac{1}{10} = -\frac{6}{60}\)</p> <p>\(\frac{61}{12} = \frac{305}{60}\)</p> <p>Ahora sumamos:</p> <p>\(\frac{15}{60} - \frac{6}{60} + \frac{305}{60} = \frac{314}{60}\)</p> <p>Finalmente simplificamos:</p> <p>\(\frac{314}{60} = \frac{157}{30}\)</p> <p>Respuesta final: \(\frac{157}{30}\)</p>
Basic Arithmetic Operations
<p>1. Calculate \( 35.6 - 27.92 \): </p> <p> \( 35.6 - 27.92 = 7.68 \) </p> <p>2. Calculate \( 35.6 + 5.67 \): </p> <p> \( 35.6 + 5.67 = 41.27 \) </p> <p>3. Calculate \( 56.78 \times 7.5 \): </p> <p> \( 56.78 \times 7.5 = 426.85 \) </p> <p>4. Calculate \( 91.8 \div 3.6 \): </p> <p> \( 91.8 \div 3.6 = 25.5 \) </p>
Solving for a number based on given arithmetic operations
Seja \( x \) o número desconhecido. Primeiro, a pessoa subtrai 4 do número desconhecido: <p>\( x - 4 \)</p> Em seguida, ela divide a diferença por 3 e o quociente é 2: <p>\( \frac{x - 4}{3} = 2 \)</p> Para encontrar o valor de \( x \), multiplicamos ambos os lados da equação por 3: <p>\( x - 4 = 2 \times 3 \)</p> <p>\( x - 4 = 6 \)</p> Agora, adicionamos 4 a ambos os lados para isolar \( x \): <p>\( x = 6 + 4 \)</p> <p>\( x = 10 \)</p> Portanto, o número desconhecido é 10.
Solving a Simple Algebraic Equation
<p>Seja \( x \) o número desconhecido. Temos a seguinte equação baseada na descrição:</p> <p>\( (x - 4) / 3 = 2 \)</p> <p>Multiplicamos ambos os lados da equação por 3 para isolar \( x - 4 \):</p> <p>\( x - 4 = 3 \times 2 \)</p> <p>\( x - 4 = 6 \)</p> <p>Adicionamos 4 a ambos os lados da equação para encontrar \( x \):</p> <p>\( x = 6 + 4 \)</p> <p>\( x = 10 \)</p> <p>Portanto, o número é 10.</p>
Finding the Number Based on Subtraction and Division
<p>Seja \( x \) o número desconhecido.</p> <p>Paola subtrai 4 do número: \( x - 4 \).</p> <p>Ela divide a diferença por 3: \( \frac{x - 4}{3} \).</p> <p>O quociente é 2, então temos a equação: \( \frac{x - 4}{3} = 2 \).</p> <p>Multiplicamos ambos os lados por 3 para isolar \( x \): \( x - 4 = 2 \cdot 3 \).</p> <p>Agora simplificamos e resolvemos para \( x \): \( x - 4 = 6 \).</p> <p>Adicionamos 4 a ambos os lados da equação: \( x = 6 + 4 \).</p> <p>Portanto, o número é \( x = 10 \).</p>
Problem Involving Subtraction and Multiplication to Find a Number
<p>Vamos chamar o número desconhecido de \( x \).</p> <p>Segundo a questão, Lily subtrai 8 do número. Então, temos \( x - 8 \).</p> <p>Depois, ela multiplica a diferença por 3, ou seja \( 3(x - 8) \).</p> <p>É dado que o produto é 14, então temos a equação: \( 3(x - 8) = 14 \).</p> <p>Resolvendo a equação:</p> <p>\( 3(x - 8) = 14 \)</p> <p>\( 3x - 24 = 14 \)</p> <p>\( 3x = 14 + 24 \)</p> <p>\( 3x = 38 \)</p> <p>\( x = \frac{38}{3} \)</p> <p>\( x = 12\frac{2}{3} \)</p>
Finding the Original Number Before Subtraction and Multiplication
Seja \( x \) o número original. De acordo com o enunciado, se você subtrai 8 do número e depois multiplica por 2, o produto é 14. Montamos a seguinte equação para representar a situação: <p>\( (x - 8) \cdot 2 = 14 \)</p> Dividimos ambos os lados da equação por 2: <p>\( x - 8 = \frac{14}{2} \)</p> <p>\( x - 8 = 7 \)</p> Agora, adicionamos 8 a ambos os lados da equação para isolar \( x \): <p>\( x = 7 + 8 \)</p> <p>\( x = 15 \)</p> Portanto, o número original é 15.
Solving a German Math Problem on Place Values and Subtraction
<p>Let's denote the unknown number as X where X has units in the form of thousands (T), hundreds (H), tens (Z), and ones (E). According to the question, X can be written in the form of THZE. We will solve each part of the question separately.</p> <p>(a) For the first part, we are given that:</p> <p>X has 2 tens of thousands, hence T = 2</p> <p>Half as many tens as tens of thousands, hence Z = \frac{T}{2} = \frac{2}{2} = 1</p> <p>Twice as many ones as thousands, hence E = 2 \cdot T = 2 \cdot 2 = 4</p> <p>Seven hundreds, hence H = 7</p> <p>Therefore, X = 21714</p> <p>(b) If we subtract 3 from the tens place, 3 from the hundreds place, and 4 from the thousands place in X, the result is 2086. Now let's adjust X accordingly:</p> <p>New thousands (T) = 2 - 4 = -2 (which is not possible because a negative number of thousands does not make sense, so there's a logical contradiction here. The number we obtained from part (a) cannot satisfy the condition in part (b). We need to revise our assumptions or there might be a misinterpretation of the question. Since we cannot have T < 0, the problem might be structured incorrectly or the question might be missing some details. Therefore, we've found an issue at this stage and cannot proceed with the calculation. Given this ambiguity, a correct answer cannot be calculated at this time for part (b).)</p> <p>(c) Without a valid number from parts (a) or (b), we cannot proceed to divide it by 5 for part (c), as the preceding parts contain a logical contradiction or an inaccuracy in problem setup or interpretation.</p> <p>Due to the contradiction encountered in part (b) of the question, we need to check the original problem statement for any errors or missing information to provide a proper solution.</p>
Addition and Subtraction of Algebraic Expressions
<p>Para sumar y restar las siguientes expresiones algebraicas, lo haremos término a término:</p> <p>(-2) + (-4) = -2 - 4 = -6</p> <p>(-1) + 2 = -1 + 2 = 1</p> <p>1 + (-2) = 1 - 2 = -1</p> <p>-1 + (-4) = -1 - 4 = -5</p> <p>1 + 3 = 1 + 3 = 4</p> <p>-1 + (-1) = -1 - 1 = -2</p> <p>(-2) + 2 = -2 + 2 = 0</p> <p>Combinando todos los resultados obtenemos la suma de las expresiones:</p> <p>-6 + 1 - 1 - 5 + 4 - 2 + 0 = -6 + (-1) - 5 + 2 = -6 - 1 - 5 + 2 = -9. Por lo tanto, la respuesta final es \(-9\).</p>
Simple Arithmetic Problems
<p>а) \( 55 - 47 = 8 \)</p> <p>б) \( 75 - 66 = 9 \)</p> <p>в) \( 84 - 78 = 6 \)</p> <p>г) \( 34 - 19 = 15 \)</p> <p>д) \( 27 + 18 = 45 \)</p> <p>е) \( 18 - 18 = 0 \)</p> <p>ж) \( 8 + 12 = 20 \)</p> <p>з) \( 5 + 30 = 35 \)</p>
Basic Arithmetic Problem on Summer Reading
<p>Let the total number of books Ada wants to read be \( T \).</p> <p>Let the number of books about cats be \( C \).</p> <p>Let the number of books about horses be \( H \).</p> <p>According to the problem, \( T = 12 \) and \( C = 5 \).</p> <p>To find \( H \), we subtract the number of books about cats from the total number of books:</p> <p>\( H = T - C \)</p> <p>\( H = 12 - 5 \)</p> <p>\( H = 7 \)</p> <p>Therefore, \( 7 \) books are stories about horses.</p>
Calculating the Number of Mollies in a Pet Fish Collection
<p>\text{Total pet fish} = 17</p> <p>\text{Goldfish} = 7</p> <p>\text{Let m be the number of mollies.}</p> <p>\text{Total pet fish} = \text{Goldfish} + \text{mollies}</p> <p>17 = 7 + m</p> <p>m = 17 - 7</p> <p>m = 10</p> <p>\text{Levi has 10 mollies.}</p>
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