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發問:
(3y-8)／(2y-1)＞(y-4)／(y+1) find the value of y 更新: i mean the range of values of y
最佳解答:
Method 1: (Splitting different cases) If y＞1/2, then 2y-1 and y+1 are both ＞0 (3y-8)/(2y-1)＞(y-4)/(y+1) (3y-8)(y+1)＞(y-4)(2y-1) [Since we are multiplying two positive numbers, we do not need to change the inequality sign] 3y2-5y-8＞2y2-9y+4 y2+4y-12＞0 (y+6)(y-2)＞0 y＜-6 or y＞2 Combining with y＞1/2, the solution is y＞2 for this case. If 1/2＞y＞-1, then 2y-1＜0 and y+1＞0 (3y-8)/(2y-1)＞(y-4)/(y+1) (3y-8)(y+1)＜(y-4)(2y-1) [Since we are multiplying 1 negative number and 1 positive number, we need to change the inequality sign] 3y2-5y-8＜2y2-9y+4 y2+4y-12＜0 (y+6)(y-2)＜0 -6＜y＜2 Combining with 1/2＞y＞-1, the solution is 1/2＞y＞-1 for this case. If -1＞y, then 2y-1 and y+1 are both ＜0 (3y-8)/(2y-1)＞(y-4)/(y+1) (3y-8)(y+1)＞(y-4)(2y-1) [Since we are multiplying two negative numbers, we need to change the inequality sign 2 times, which means no change XD] 3y2-5y-8＞2y2-9y+4 y2+4y-12＞0 (y+6)(y-2)＞0 y＜-6 or y＞2 Combining with -1＞y, the solution is y＜-6 for this case. Combining all the cases, the solution is y＞2 or 1/2＞y＞-1 or y＜-6 Method 2: (Factorization) (3y-8)/(2y-1)＞(y-4)/(y+1) (3y-8)/(2y-1)*(2y-1)2 *(y+1)2＞(y-4)/(y+1)*(2y-1)2 *(y+1)2 [We are sure that (2y-1)2 *(y+1)2 is not a negative number, so we do not need to change the inequality sign.] (3y-8)(2y-1)(y+1)2＞(y-4)(y+1)(2y-1)2 (3y-8)(2y-1)(y+1)2 - (y-4)(y+1)(2y-1)2＞0 (y+1)(2y-1)[(3y-8)(y+1) - (y-4)(2y-1)]＞0 (y+1)(2y-1)[(3y2-5y-8) - (2y2-9y+4)]＞0 (y+1)(2y-1)(y2+4y-12)＞0 (y+1)(2y-1)(y+6)(y-2)＞0 Consider the equation (x+1)(2x-1)(x+6)(x-2)=0, since it is a quartic equation (degree 4) and the coefficient of x^4 is positive, the graph is W-shaped and crosses the x-axis at -6, -1, 1/2, 2. Therefore, when y＜-6 or -1＜y＜1/2 or y＞2, (y+1)(2y-1)(y+6)(y-2)＞０. I have shown two methods here in order to show the common ways (Splitting cases and Multiplying non-negative factors) to handle fractions in solving inequalities. Hope it helps! ^^
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Inequality發問:
(3y-8)／(2y-1)＞(y-4)／(y+1) find the value of y 更新: i mean the range of values of y
最佳解答:
Method 1: (Splitting different cases) If y＞1/2, then 2y-1 and y+1 are both ＞0 (3y-8)/(2y-1)＞(y-4)/(y+1) (3y-8)(y+1)＞(y-4)(2y-1) [Since we are multiplying two positive numbers, we do not need to change the inequality sign] 3y2-5y-8＞2y2-9y+4 y2+4y-12＞0 (y+6)(y-2)＞0 y＜-6 or y＞2 Combining with y＞1/2, the solution is y＞2 for this case. If 1/2＞y＞-1, then 2y-1＜0 and y+1＞0 (3y-8)/(2y-1)＞(y-4)/(y+1) (3y-8)(y+1)＜(y-4)(2y-1) [Since we are multiplying 1 negative number and 1 positive number, we need to change the inequality sign] 3y2-5y-8＜2y2-9y+4 y2+4y-12＜0 (y+6)(y-2)＜0 -6＜y＜2 Combining with 1/2＞y＞-1, the solution is 1/2＞y＞-1 for this case. If -1＞y, then 2y-1 and y+1 are both ＜0 (3y-8)/(2y-1)＞(y-4)/(y+1) (3y-8)(y+1)＞(y-4)(2y-1) [Since we are multiplying two negative numbers, we need to change the inequality sign 2 times, which means no change XD] 3y2-5y-8＞2y2-9y+4 y2+4y-12＞0 (y+6)(y-2)＞0 y＜-6 or y＞2 Combining with -1＞y, the solution is y＜-6 for this case. Combining all the cases, the solution is y＞2 or 1/2＞y＞-1 or y＜-6 Method 2: (Factorization) (3y-8)/(2y-1)＞(y-4)/(y+1) (3y-8)/(2y-1)*(2y-1)2 *(y+1)2＞(y-4)/(y+1)*(2y-1)2 *(y+1)2 [We are sure that (2y-1)2 *(y+1)2 is not a negative number, so we do not need to change the inequality sign.] (3y-8)(2y-1)(y+1)2＞(y-4)(y+1)(2y-1)2 (3y-8)(2y-1)(y+1)2 - (y-4)(y+1)(2y-1)2＞0 (y+1)(2y-1)[(3y-8)(y+1) - (y-4)(2y-1)]＞0 (y+1)(2y-1)[(3y2-5y-8) - (2y2-9y+4)]＞0 (y+1)(2y-1)(y2+4y-12)＞0 (y+1)(2y-1)(y+6)(y-2)＞0 Consider the equation (x+1)(2x-1)(x+6)(x-2)=0, since it is a quartic equation (degree 4) and the coefficient of x^4 is positive, the graph is W-shaped and crosses the x-axis at -6, -1, 1/2, 2. Therefore, when y＜-6 or -1＜y＜1/2 or y＞2, (y+1)(2y-1)(y+6)(y-2)＞０. I have shown two methods here in order to show the common ways (Splitting cases and Multiplying non-negative factors) to handle fractions in solving inequalities. Hope it helps! ^^
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