Rational Equations

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Rational Equations · Grade 9 · Worksheet 3

1. (x + 3)/(x - 2) = (2x - 1)/(x + 4) Answer: x = -5 Solution: Rational equations contain fractions with variables in denominators. The key strategy is to multiply through by the LCD to eliminate fractions, creating a simpler equation to solve.
   Full step-by-step solution

   Rational equations contain fractions with variables in denominators. The key strategy is to multiply through by the LCD to eliminate fractions, creating a simpler equation to solve. For example, when solving (a)/(b) = (c)/(d), cross-multiplying gives ad = bc. After finding potential solutions, you must check that they don't make any original denominator equal to zero, as these would be extraneous solutions.

2. Aroha is analyzing a geometric diagram on a coordinate grid. She sees a large rectangle with vertices at (0, 0), (12, 0), (12, 9), and (0, 9). Inside the rectangle, a diagonal is drawn from (0, 0) to (12, 9). A smaller rectangle is formed by the origin, a point P on the diagonal, and the lines through P parallel to the axes. If the ratio of the area of the smaller rectangle to the area of the large rectangle is 4/9, what are the coordinates of P? Answer: (8, 6) Solution: Find the area of the large rectangle. Length = 12, width = 9. Area = 12 * 9 = 108 square units.
   Full step-by-step solution

   Step 1: Find the area of the large rectangle. Length = 12, width = 9. Area = 12 * 9 = 108 square units. Step 2: Let the coordinates of P be (x, y). Since P lies on the diagonal from (0,0) to (12,9), the slope is 9/12 = 3/4. So y = (3/4)x. Step 3: The smaller rectangle has corners at (0,0), (x,0), (x,y), (0,y). Its area = x * y = x * (3/4)x = (3/4)x^2. Step 4: The ratio of areas is (area of small rectangle)/(area of large rectangle) = ((3/4)x^2)/108 = 4/9. Step 5: Simplify: (3x^2)/(4 * 108) = 4/9. Multiply both sides by 4*108: 3x^2 = (4/9)*432 = 192. So x^2 = 64, thus x = 8 (positive since P is in first quadrant). Step 6: y = (3/4)*8 = 6. Step 7: The coordinates of P are (8, 6). The answer is (8, 6).

3. Liam is designing a rectangular garden with an area of 48 square meters. He wants the length to be 4 meters more than the width. What are the dimensions of Liam's garden? Answer: width = 6 meters, length = 10 meters Solution: These can be modeled using quadratic equations, where you set up an equation based on the area formula and the given relationship, then solve for the variable representing one dimension.
   Full step-by-step solution

   Many real-world geometry problems involve finding dimensions when given area and a relationship between length and width. These can be modeled using quadratic equations, where you set up an equation based on the area formula and the given relationship, then solve for the variable representing one dimension.

4. A chemical reaction proceeds at a rate modeled by the function r(t) = (3t^2 - 12) / (t^2 - 4), where r is the reaction rate in moles per minute and t is time in minutes. At what time does the reaction rate become undefined due to a mathematical discontinuity in the model? Answer: 2 Solution: Step 1: Identify when the reaction rate function r(t) = (3t^2 - 12)/(t^2 - 4) becomes undefined Step 2: A rational function is undefined when its denominator equals zero Step 3: Set the denominator equal to zero: t^2 - 4 = 0 Step 4: Solve for t: t^2 = 4 Step 5: t = ±2 Step 6: Since time cannot…
   Full step-by-step solution

   Step 1: Identify when the reaction rate function r(t) = (3t^2 - 12)/(t^2 - 4) becomes undefined Step 2: A rational function is undefined when its denominator equals zero Step 3: Set the denominator equal to zero: t^2 - 4 = 0 Step 4: Solve for t: t^2 = 4 Step 5: t = ±2 Step 6: Since time cannot be negative in this context, the relevant solution is t = 2 Step 7: Verify: At t = 2, denominator = (2)^2 - 4 = 4 - 4 = 0 Step 8: The reaction rate becomes undefined at t = 2 minutes The answer is 2.

5. A chemical reaction follows the equation y = 5/(x - 2) where y is the reaction rate and x is the temperature in Celsius. The reaction becomes unstable when the rate exceeds 2.5 units. At what temperature does the reaction first become unstable? Answer: 4 Solution: Set up the inequality for when the reaction becomes unstable: 5/(x - 2) > 2.5 Solve the inequality: 5/(x - 2) > 2.5 Multiply both sides by (x - 2): 5 > 2.5(x - 2) Distribute on the right side: 5 > 2.5x - 5 Add 5 to both sides: 10 > 2.5x Divide both sides by 2.5: 4 > x This means x < 4, but we…
   Full step-by-step solution

   Step 1: Set up the inequality for when the reaction becomes unstable: 5/(x - 2) > 2.5 Step 2: Solve the inequality: 5/(x - 2) > 2.5 Step 3: Multiply both sides by (x - 2): 5 > 2.5(x - 2) Step 4: Distribute on the right side: 5 > 2.5x - 5 Step 5: Add 5 to both sides: 10 > 2.5x Step 6: Divide both sides by 2.5: 4 > x Step 7: This means x < 4, but we must also consider the domain restriction x ≠ 2 Step 8: The reaction first becomes unstable as we approach x = 4 from below The answer is 4.

6. (9x - 12)/(x + 8) = 6 = ? Answer: 20 Solution: Start with (9x - 12)/(x + 8) = 6 Multiply both sides by (x + 8): 9x - 12 = 6(x + 8) Distribute the 6 on the right: 9x - 12 = 6x + 48 Subtract 6x from both sides: 3x - 12 = 48 Add 12 to both sides: 3x = 60 Divide both sides by 3: x = 20 Check: (9*20 - 12)/(20 + 8) = (180 - 12)/28 = 168/28 = 6.
   Full step-by-step solution

   Step 1: Start with (9x - 12)/(x + 8) = 6 Step 2: Multiply both sides by (x + 8): 9x - 12 = 6(x + 8) Step 3: Distribute the 6 on the right: 9x - 12 = 6x + 48 Step 4: Subtract 6x from both sides: 3x - 12 = 48 Step 5: Add 12 to both sides: 3x = 60 Step 6: Divide both sides by 3: x = 20 Step 7: Check: (9*20 - 12)/(20 + 8) = (180 - 12)/28 = 168/28 = 6. The denominator is not zero. The answer is 20.

7. (5x - 3)/(x + 7) = 3 = ? Answer: 12 Solution: Start with (5x - 3)/(x + 7) = 3 Multiply both sides by (x + 7): 5x - 3 = 3(x + 7) Distribute the 3 on the right: 5x - 3 = 3x + 21 Subtract 3x from both sides: 2x - 3 = 21 Add 3 to both sides: 2x = 24 Divide both sides by 2: x = 12 Check: (5*12 - 3)/(12 + 7) = (60 - 3)/19 = 57/19 = 3.
   Full step-by-step solution

   Step 1: Start with (5x - 3)/(x + 7) = 3 Step 2: Multiply both sides by (x + 7): 5x - 3 = 3(x + 7) Step 3: Distribute the 3 on the right: 5x - 3 = 3x + 21 Step 4: Subtract 3x from both sides: 2x - 3 = 21 Step 5: Add 3 to both sides: 2x = 24 Step 6: Divide both sides by 2: x = 12 Step 7: Check: (5*12 - 3)/(12 + 7) = (60 - 3)/19 = 57/19 = 3. The denominator is not zero. The answer is 12.

8. A chemical reaction proceeds at a rate modeled by the rational function r(t) = (3t^2 - 12) / (t^2 - 4), where t is time in minutes and r(t) is the reaction rate in moles per minute. At what time does the reaction rate become undefined due to a vertical asymptote? Answer: 2 Solution: Let me reconsider the problem. The original function is r(t) = (3t^2 - 12)/(t^2 - 4) Factor numerator: 3t^2 - 12 = 3(t^2 - 4) Factor denominator: t^2 - 4 = (t - 2)(t + 2) So r(t) = 3(t^2 - 4)/((t - 2)(t + 2)) = 3(t - 2)(t + 2)/((t - 2)(t + 2)) Cancel common factors: r(t) = 3 for t ≠ 2 and t ≠ -2…
   Full step-by-step solution

   Step 1: Identify when the reaction rate becomes undefined by setting the denominator equal to zero: t^2 - 4 = 0 Step 2: Solve for t: t^2 = 4, so t = 2 or t = -2 Step 3: Since time cannot be negative, we only consider t = 2 Step 4: Verify this is a vertical asymptote by checking if the numerator is also zero at t = 2: 3(2)^2 - 12 = 3(4) - 12 = 12 - 12 = 0 Step 5: Since both numerator and denominator equal zero at t = 2, we need to check if this is a removable discontinuity or vertical asymptote Step 6: Factor both numerator and denominator: r(t) = 3(t^2 - 4)/(t^2 - 4) = 3 for t ≠ 2 Step 7: The function simplifies to r(t) = 3, meaning the discontinuity at t = 2 is removable, not a vertical asymptote Step 8: Wait, I made an error in my initial reasoning. Let me reconsider the problem. Step 9: The original function is r(t) = (3t^2 - 12)/(t^2 - 4) Step 10: Factor numerator: 3t^2 - 12 = 3(t^2 - 4) Step 11: Factor denominator: t^2 - 4 = (t - 2)(t + 2) Step 12: So r(t) = 3(t^2 - 4)/((t - 2)(t + 2)) = 3(t - 2)(t + 2)/((t - 2)(t + 2)) Step 13: Cancel common factors: r(t) = 3 for t ≠ 2 and t ≠ -2 Step 14: The function has removable discontinuities at both t = 2 and t = -2, not vertical asymptotes Step 15: I need to re-read the problem statement more carefully Step 16: The problem asks when the reaction rate becomes undefined, which occurs when the denominator equals zero: t^2 - 4 = 0 Step 17: Solving t^2 - 4 = 0 gives t = 2 or t = -2 Step 18: Since time cannot be negative, the reaction rate becomes undefined at t = 2 minutes Step 19: Even though this is a removable discontinuity in the mathematical function, in the physical context the reaction rate would be undefined at this exact moment Step 20: Therefore, the answer is t = 2 minutes The answer is 2.