3y-5/y 2. 4b/2x(6x-5) 3. 7/x(7-2x) 4. x2-32/(x+5)(x-6) 5. If y=5x+4/2x+3 a. For what value of x is y not defined? b. Determine the value of x for which y Vanishes c. For what range of values of x is y>3?

Here's how to solve the problem: We are given the function y = (5x+4)/(2x+3). a) For what value of x is y not defined? Step 1: A rational function is undefined when its denominator is equal to zero. Set the denominator to zero and solve for x. 2x+3 = 0 Step 2: Isolate x. 2x = -3 x = -(3)/(2) The value of x for which y is not defined is -(3)/(2). b) Determine the value of x for which y vanishes: Step 1: The function y vanishes (is equal to zero) when its numerator is equal to zero, provided the denominator is not zero at that same x value. Set the numerator to zero and solve for x. 5x+4 = 0 Step 2: Isolate x. 5x = -4 x = -(4)/(5) The value of x for which y vanishes is -(4)/(5). c) For what range of values of x is y > 3? Step 1: Set up the inequality. (5x+4)/(2x+3) > 3 Step 2: Subtract 3 from both sides to get a single fraction. (5x+4)/(2x+3) - 3 > 0 (5x+4 - 3(2x+3))/(2x+3) > 0 (5x+4 - 6x - 9)/(2x+3) > 0 (-x-5)/(2x+3) > 0 Step 3: Find the critical points where the numerator or denominator is zero. Numerator: -x-5 = 0 x = -5 Denominator: 2x+3 = 0 x = -(3)/(2) Step 4: Use a sign table or test intervals to determine when the expression (-x-5)/(2x+3) is positive. The critical points divide the number line into three intervals: (-, -5), (-5, -(3)/(2)), and (-(3)/(2), ). • For x < -5 (e.g., x=-6): Numerator: -(-6)-5 = 6-5 = 1 (positive) Denominator: 2(-6)+3 = -12+3 = -9 (negative) Fraction: positivenegative = negative. So, (-x-5)/(2x+3) < 0. • For -5 < x < -(3)/(2) (e.g., x=-2): Numerator: -(-2)-5 = 2-5 = -3 (negative) Denominator: 2(-2)+3 = -4+3 = -1 (negative) Fraction: negativenegative = positive. So, (-x-5)/(2x+3) > 0. • For x > -(3)/(2) (e.g., x=0): Numerator: -0-5 = -5 (negative) Denominator: 2(0)+3 = 3 (positive) Fraction: negativepositive = negative. So, (-x-5)/(2x+3) < 0. Step 5: The inequality (-x-5)/(2x+3) > 0 is satisfied when -5 < x < -(3)/(2). The range of values for x for which y > 3 is -5 < x < -(3)/(2). Send me the next one 📸