Example Question - inequality solutions
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Solving Inequalities with Variable q
Let's solve the inequality given in the image: 5q + 3 < 37 Our first step is to isolate the term with the variable q. To do that, we subtract 3 from both sides of the inequality. 5q + 3 - 3 < 37 - 3 This simplifies to: 5q < 34 Next, we divide both sides by 5 to solve for q: 5q / 5 < 34 / 5 q < 6.8 The value of q must be less than 6.8. Now let's check the provided options to see which one is a solution to the inequality: a) q = 8 Since 8 is greater than 6.8, this is not a solution. b) q = 5 Since 5 is less than 6.8, this is a solution. c) q = 2 Since 2 is less than 6.8, this is also a solution. d) q = -1 Since -1 is less than 6.8, this is also a solution. The question asks for a solution to the inequality. The correct answers are the ones where q is less than 6.8. Therefore, the solutions from the given options are q = -1, q = 2, and q = 5. However, since only one option can be chosen and all three are correct, without additional context or instructions, it is impossible to determine which single option the question intends as the answer. However, if the question requires to choose all possible solutions from the options given, then q = 5, q = 2, and q = -1 would all be acceptable solutions.
Solving Equations and Inequalities in Mathematics
Pour résoudre le problème présenté dans l'image, commençons par le premier point : 1. Résoudre dans R l'équation : \[ \frac{1}{3}x + 3 = 0 \] \[ \frac{2x + 3}{x + 1} = \frac{(x + 1)(x - 3)}{x + 1} \] Résolvons d'abord l’équation simple \( \frac{1}{3}x + 3 = 0 \): Multiplications des deux côtés par 3 pour se débarrasser du facteur \( \frac{1}{3} \): \[ x + 9 = 0 \] Substrayons 9 des deux côtés de l’équation : \[ x = -9 \] Le côté droit de l’équation a une expression fractionnaire que nous pouvons simplifier en annulant les termes communs (x + 1) en supposant que x n'est pas égal à -1 (car cela annulerait le dénominateur, ce qui n'est pas permis). Cela nous donne directement (x - 3). Ensuite, vérifions si les deux côtés de l’équation sont égaux pour x = -9 : À gauche nous avons \( \frac{2(-9) + 3}{-9 + 1} = \frac{-18 + 3}{-8} = \frac{-15}{-8} = \frac{15}{8} \) À droite nous avons (-9 + 1)(-9 - 3) = (-8)(-12) = 96 Puisque \( \frac{15}{8} \) n'est pas égal à 96, l'équation n'est pas vraie pour x = -9. Donc, x = -9 n’est pas une solution pour l’équation donnée. Passons au deuxième point de l'exercice : 2. Soit \( p(x) = (x - 2)(x + 5) \) a) Résoudre dans R l'équation p(x) = 0 Pour résoudre cette équation, nous cherchons les valeurs de x pour lesquelles le produit est nul. Un produit de facteurs est nul si au moins l'un des facteurs est nul. Donc, \( x - 2 = 0 \) ou \( x + 5 = 0 \) Cela donne deux solutions : \[ x = 2 \] et \[ x = -5 \] b) Résoudre dans R l'inéquation p(x) > 0 Pour résoudre cette inéquation, nous utilisons les solutions de l'équation trouvées précédemment pour déterminer les intervalles sur lesquels le produit est positif. Les racines de l'équation divisent la ligne numérique en intervalles sur lesquels le signe du produit ne change pas. On a : - Pour \( x < -5 \), le produit (x - 2)(x + 5) est positif car on a un nombre négatif multiplié par un autre nombre négatif. - Pour \( -5 < x < 2 \), le produit est négatif car on a un nombre positif multiplié par un nombre négatif. - Pour \( x > 2 \), le produit est à nouveau positif car on a deux nombres positifs. Donc, les intervalles sur lesquels p(x) > 0 sont : \[ x \in ]-\infty; -5[ \cup ]2; +\infty[ \] Pour le troisième point : 3. Considérons l'équation (E) : x² - 2x + 1 = 0 Nous savons déjà que \( x_1 = \sqrt{2} - 1 \) est une solution de l'équation (E). Pour trouver la deuxième solution, remarquons que l'équation est une équation du second degré qui peut être écrite sous forme factorisée car c'est une équation du type \( (x - a)^2 = 0 \), où \( a \) est le terme racine. L'équation donnée est la forme développée de \( (x - 1)^2 \): \[ (x - 1)^2 = x^2 - 2x + 1 \] Pour que cette équation soit nulle, il est nécessaire que \( x - 1 = 0 \), d'où \( x = 1 \). Ceci est la deuxième solution de l'équation. Pour le quatrième point, l'inéquation à résoudre est : 4. Résoudre dans R l'inéquation \( 4x - 3\sqrt{x} + 1 < 0 \) Pour trouver les valeurs de x qui satisfont cette inéquation, il serait utile de factoriser l'expression ou d'utiliser une autre technique appropriée. Dans ce cas, la résolution exacte pourrait devenir complexe selon le niveau d'études attendu. Si vous avez besoin d'une méthode spécifique pour résoudre cette inéquation, veuillez préciser et je pourrai vous guider en conséquence.
Solving Absolute Value Inequalities
The inequality presented here is \(7 \leq |7y - 9|\). To solve this, we need to consider the absolute value |7y - 9| and split the inequality into two cases because the expression inside the absolute value can be either positive or negative. The absolute value inequality says that the expression inside the absolute value is either greater than or equal to 7 or less than or equal to -7. First, we'll handle the case where the expression inside the absolute value is non-negative: 1. \(7y - 9 \geq 7\) 2. \(7y \geq 16\) 3. \(y \geq \frac{16}{7}\) Next, we handle the case where the expression inside the absolute value is non-positive: 1. \(7y - 9 \leq -7\) 2. \(7y \leq 2\) 3. \(y \leq \frac{2}{7}\) Combining both inequalities, we get the solution set for y: \(y \leq \frac{2}{7}\) or \(y \geq \frac{16}{7}\). Graphically, this means y is either in the interval \((-\infty, \frac{2}{7}]\) or in the interval \([\frac{16}{7}, \infty)\).
Solving Inequalities Involving Absolute Value
To solve the inequality \(-9|c| < -50\), let's start by isolating the absolute value term. Firstly, divide both sides by -9 to get the absolute value by itself. It's important to remember that when we divide or multiply both sides of an inequality by a negative number, the direction of the inequality symbol changes. So the inequality will reverse in this case. \[ -9|c| < -50 \quad / :(-9) \] \[ |c| > \frac{-50}{-9} \] \[ |c| > \frac{50}{9} \] Since the absolute value of a number is always nonnegative, we can write this as two separate inequalities: \[ c > \frac{50}{9} \quad or \quad c < -\frac{50}{9} \] So the solution to the inequality is that \(c\) is either greater than \(\frac{50}{9}\) or less than \(-\frac{50}{9}\).
Solving Absolute Value Inequalities
To solve the inequality \( 8 - 7|1 - 6s| < -6 \), follow these steps: 1. First, isolate the absolute value term on one side of the inequality: \( 8 - 7|1 - 6s| + 6 < 0 \) \( 14 - 7|1 - 6s| < 0 \) \( -7|1 - 6s| < -14 \) Now divide by -7, remembering that dividing by a negative number will flip the inequality sign: \( |1 - 6s| > 2 \) 2. Next, we need to consider the two cases due to the absolute value, where \(1 - 6s\) can be either greater than 2 or less than -2. **Case 1 - Positive case:** \( 1 - 6s > 2 \) Subtract 1 from both sides: \( -6s > 1 \) Divide by -6 and flip the inequality sign: \( s < \frac{1}{6} \) **Case 2 - Negative case:** \( 1 - 6s < -2 \) Subtract 1 from both sides: \( -6s < -3 \) Divide by -6 and flip the inequality sign: \( s > \frac{1}{2} \) 3. Combine both cases to get the final solution: \( s < \frac{1}{6} \) or \( s > \frac{1}{2} \) The solution to the inequality is all \( s \) that satisfy either \( s < \frac{1}{6} \) or \( s > \frac{1}{2} \).
Solving an Inequality Involving Absolute Value
To solve the given inequality, \( 8 - 7|-6s| < -6 \), let's first simplify the expression inside the absolute value sign by multiplying 7 and the absolute value of \(-6s\): \[ 8 - 7 \cdot |-6s| < -6 \] Let \( A = |-6s| \), then we have: \[ 8 - 7A < -6 \] Now, let's solve for \( A \) by isolating it on one side: \[ -7A < -6 - 8 \] \[ -7A < -14 \] Divide both sides by -7, and remember to reverse the inequality sign since we are dividing by a negative number: \[ A > 2 \] But \( A \) was defined as \( |-6s| \), so we substitute back in: \[ |-6s| > 2 \] Now we have to consider the two cases due to the absolute value. This inequality splits into two separate inequalities: 1. When the expression inside the absolute value is positive or zero: \[ -6s > 2 \] \[ s < -\frac{1}{3} \] 2. When the expression inside the absolute value is negative: \[ -6s < -2 \] \[ s > \frac{1}{3} \] Since no values of \( s \) can be simultaneously less than \(-\frac{1}{3}\) and greater than \(\frac{1}{3}\), this inequality has no solution. It means there are no values of \( s \) that can satisfy the original inequality \( 8 - 7|-6s| < -6 \).
Solving Absolute Value Inequalities
Sure, let's solve the inequality presented in the image: \[ |2p| - 8 \geq 13 \] To solve this, you first need to isolate the absolute value expression on one side of the inequality. Let's move the -8 to the other side by adding 8 to both sides: \[ |2p| \geq 13 + 8 \] \[ |2p| \geq 21 \] Now that the absolute value is isolated, we can split the inequality into two separate cases, because the absolute value of a number is either equal to the number itself if the number is positive or equal to the negative of the number if it is negative. So we have: \[ 2p \geq 21 \quad \text{or} \quad 2p \leq -21 \] Now, solve each inequality for \( p \): For the first inequality: \[ 2p \geq 21 \] \[ p \geq \frac{21}{2} \] \[ p \geq 10.5 \] For the second inequality: \[ 2p \leq -21 \] \[ p \leq \frac{-21}{2} \] \[ p \leq -10.5 \] Therefore, the solution to the inequality is \( p \leq -10.5 \) or \( p \geq 10.5 \). This means \( p \) can be any number less than or equal to -10.5 or any number greater than or equal to 10.5.
Solving Absolute Value Inequalities
Certainly! We have an inequality involving an absolute value: |u - 7| > 29. Recall that |x| > a, where a is a positive number, implies two situations: 1. x > a 2. x < -a Applying this to our inequality, we split it into two separate inequalities: u - 7 > 29 and u - 7 < -29 For the first inequality (u - 7 > 29), we solve for u by adding 7 to both sides: u > 29 + 7 u > 36 For the second inequality (u - 7 < -29), we again solve for u by adding 7 to both sides: u < -29 + 7 u < -22 Therefore, the solutions to the inequality |u - 7| > 29 are: u > 36 or u < -22 These inequalities represent the range of values that u can take to satisfy the original inequality.
Solving Absolute Value Inequalities
To solve the inequality |6 - w| < 59, you will need to consider two cases due to the absolute value. Case 1: When 6 - w is positive or zero, 6 - w < 59 Now, solve for w: -w < 59 - 6 -w < 53 Multiply both sides by -1 (and remember to reverse the inequality when you multiply or divide by a negative number): w > -53 Case 2: When 6 - w is negative, -(6 - w) < 59 -w + 6 < 59 Now, solve for w: -w < 59 - 6 -w < 53 Again, multiply both sides by -1 (and reverse the inequality): w > -53 Since both cases lead to the same inequality, w > -53, the solution to the inequality |6 - w| < 59 is: w > -53 This means that w can be any number greater than -53.
Solving Absolute Value Inequalities
To solve the absolute value inequality \( |4q - 1| \leq 85 \), we must consider the two cases of the absolute value definition: 1. The expression inside the absolute value, \( 4q - 1 \), is non-negative. 2. The expression inside the absolute value, \( 4q - 1 \), is negative. For the first case, when \( 4q - 1 \) is non-negative: \[ 4q - 1 \leq 85 \] To solve this inequality, add 1 to both sides: \[ 4q \leq 86 \] Now, divide both sides by 4: \[ q \leq 21.5 \] For the second case, when \( 4q - 1 \) is negative, we consider the opposite of the expression, which gives us: \[ -(4q - 1) \leq 85 \] Simplify the left side by distributing the negative sign: \[ -4q + 1 \leq 85 \] Now, to isolate the q term, subtract 1 from both sides: \[ -4q \leq 84 \] Divide both sides by -4 and remember to reverse the inequality sign because we are dividing by a negative number: \[ q \geq -21 \] Putting both cases together gives us the compound inequality: \[ -21 \leq q \leq 21.5 \] This is the solution set for the inequality \( |4q - 1| \leq 85 \).
Solving Absolute Value Inequalities
To solve the inequality |4q - 1| ≤ 85, you need to consider the definition of the absolute value. The absolute value of a number is the distance of that number from 0 on the number line, which is always nonnegative. Therefore, |x| ≤ a means that x is within the distance of a from 0. This leads to two scenarios: x ≤ a and x ≥ -a. Applying this idea to |4q - 1| ≤ 85, we get two inequalities: 1. 4q - 1 ≤ 85 2. 4q - 1 ≥ -85 Now, let's solve each inequality separately. For the first inequality: 4q - 1 ≤ 85 Add 1 to both sides: 4q ≤ 86 Divide both sides by 4: q ≤ 86 / 4 q ≤ 21.5 For the second inequality: 4q - 1 ≥ -85 Add 1 to both sides: 4q ≥ -84 Divide both sides by 4: q ≥ -84 / 4 q ≥ -21 Combining these two inequalities gives us the compound inequality: -21 ≤ q ≤ 21.5 This is the solution to the original inequality, which means that q must be greater than or equal to -21 and less than or equal to 21.5.
Solving Absolute Value Inequalities
To solve the inequality \( |8r| + 2 > 42 \), we will first isolate the absolute value part by subtracting 2 from both sides of the inequality: \[ |8r| + 2 - 2 > 42 - 2 \] \[ |8r| > 40 \] Now, by the definition of absolute value, \( |8r| > 40 \) means that 8r is either greater than 40 or less than -40. We can split this into two separate inequalities to represent the compound inequality: 1) \( 8r > 40 \) 2) \( 8r < -40 \) Next, we can solve each inequality for r. Starting with the first inequality: \[ 8r > 40 \] \[ r > \frac{40}{8} \] \[ r > 5 \] For the second inequality: \[ 8r < -40 \] \[ r < \frac{-40}{8} \] \[ r < -5 \] Now, we present these two solutions as a compound inequality: \[ r < -5 \text{ or } r > 5 \] This is the solution to the original inequality \( |8r| + 2 > 42 \).
Solving Absolute Value Inequalities
The inequality provided in the image is \( |u + 6| \geq 46 \). Let's solve the inequality step by step. The absolute value inequality \( |u + 6| \geq 46 \) means that whatever expression is inside the absolute value signs must be either greater than or equal to 46 or less than or equal to -46. We can split this into two separate inequalities: 1. \( u + 6 \geq 46 \) 2. \( u + 6 \leq -46 \) For the first inequality, \( u + 6 \geq 46 \): Subtract 6 from both sides to get \( u \geq 40 \). For the second inequality, \( u + 6 \leq -46 \): Subtract 6 from both sides to get \( u \leq -52 \). Therefore, the solution to the inequality \( |u + 6| \geq 46 \) is \( u \geq 40 \) or \( u \leq -52 \). These solutions represent all the values of \( u \) that make the original inequality true.
Solving Absolute Value Inequalities
To solve the inequality |u + 6| ≥ 46, we have to consider the definition of absolute value. The expression |u + 6| represents the distance of u + 6 from zero on the number line, and this distance is greater than or equal to 46. We can break this into two separate cases, one for u + 6 being positive and one for u + 6 being negative. Case 1 (u + 6 is positive or 0): u + 6 ≥ 46 u ≥ 46 - 6 u ≥ 40 Case 2 (u + 6 is negative): -(u + 6) ≥ 46 -u - 6 ≥ 46 -u ≥ 46 + 6 -u ≥ 52 Multiply both sides by -1 and reverse the inequality sign (because multiplying by a negative number reverses the inequality): u ≤ -52 So the solution to the inequality |u + 6| ≥ 46 is: u ≥ 40 or u ≤ -52
Solving System of Inequalities
好的，這個問題是關於不等式系統的解法。我們需要求解下面的不等式系統： \( \frac{22+3x}{7} < \frac{8+3x}{3} \) 并且 \( \frac{5x - 4}{3} > 2x + 3 \) 首先，我們解第一個不等式： \( \frac{22+3x}{7} < \frac{8+3x}{3} \) 通分，得到： \( 3(22+3x) < 7(8+3x) \) 展開後得到： \( 66 + 9x < 56 + 21x \) 這樣我們可以移項： \( 9x - 21x < 56 - 66 \) 簡化得到： \( -12x < -10 \) 我們再除以 -12 記得要反轉不等號： \( x > \frac{5}{6} \) 所以第一個不等式的解集是 \( x > \frac{5}{6} \)。 接著我們解第二個不等式： \( \frac{5x - 4}{3} > 2x + 3 \) 兩邊同乘以3，消分母得到： \( 5x - 4 > 6x + 9 \) 接著移項得到： \( -4 - 9 > 6x - 5x \) 即： \( -x > 13 \) 同樣的，除以 -1 並反轉不等號： \( x < -13 \) 因此第二個不等式的解集是 \( x < -13 \)。 為了找到這兩個不等式系統的共同解，我們需要找到 \( x \) 值同時滿足 \( x > \frac{5}{6} \) 以及 \( x < -13 \) 的區域。仔細觀察可以看出，这两个区间没有交集，因此这个不等式系统没有解集。
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