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A line with equation (3k+2)y+5x=7(3k + 2)y + 5x = 7 has the same gradient as the line y+(3−2k)x=8y + (3 - 2k)x = 8. Find the value of kk where k > 0.

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Full solution

Q. A line with equation (3k+2)y+5x=7(3k + 2)y + 5x = 7 has the same gradient as the line y+(3−2k)x=8y + (3 - 2k)x = 8. Find the value of kk where k>0k > 0.
  1. Rearrange equation for gradient: To find the gradient of the first line, we need to rearrange the equation (3k+2)y+5x=7(3k + 2)y + 5x = 7 into the slope-intercept form y=mx+by = mx + b, where mm is the gradient.
  2. Isolate y-term: First, we isolate the y-term on one side of the equation:\newline(3k+2)y=−5x+7(3k + 2)y = -5x + 7
  3. Find gradient of first line: Next, we divide both sides by (3k+2)(3k + 2) to solve for yy:y=−5x+73k+2y = \frac{-5x + 7}{3k + 2}
  4. Rearrange second line equation: The gradient of the first line is the coefficient of xx, which is −53k+2-\frac{5}{3k + 2}.
  5. Isolate y-term: Now, we find the gradient of the second line by rearranging y+(3−2k)x=8y + (3 - 2k)x = 8 into the slope-intercept form.
  6. Find gradient of second line: We isolate the yy-term on one side of the equation:\newliney=−(3−2k)x+8y = - (3 - 2k)x + 8
  7. Set gradients equal: The gradient of the second line is the coefficient of xx, which is −(3−2k)-(3 - 2k).
  8. Clear fraction: Since the two lines have the same gradient, we set the gradients equal to each other:\newline−53k+2=−3−2k1-\frac{5}{3k + 2} = -\frac{3 - 2k}{1}
  9. Expand equation: We can remove the negative signs from both sides of the equation: 53k+2=3−2k\frac{5}{3k + 2} = 3 - 2k
  10. Combine like terms: Next, we multiply both sides by (3k+2)(3k + 2) to clear the fraction:\newline5=(3−2k)(3k+2)5 = (3 - 2k)(3k + 2)
  11. Move terms to one side: We expand the right side of the equation: 5=9k+6−6k2−4k5 = 9k + 6 - 6k^2 - 4k
  12. Solve quadratic equation: Combine like terms on the right side: 5=−6k2+5k+65 = -6k^2 + 5k + 6
  13. Calculate discriminant: We move all terms to one side to set the equation to zero:\newline0=−6k2+5k+6−50 = -6k^2 + 5k + 6 - 5
  14. Calculate possible values for kk: Simplify the equation:\newline0=−6k2+5k+10 = -6k^2 + 5k + 1
  15. Calculate possible values for k: Now, we solve the quadratic equation for k. Since we are looking for k > 00, we can use the quadratic formula:\newlinek = −b±b2−4ac2a\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\newlinewhere a = −6-6, b = 55, and c = 11.
  16. Identify valid solution: Calculate the discriminant (b2−4ac)(b^2 - 4ac):\newlineDiscriminant = (5)2−4(−6)(1)=25+24=49(5)^2 - 4(-6)(1) = 25 + 24 = 49
  17. Identify valid solution: Calculate the discriminant (b2−4ac)(b^2 - 4ac):Discriminant=(5)2−4(−6)(1)=25+24=49\text{Discriminant} = (5)^2 - 4(-6)(1) = 25 + 24 = 49Calculate the two possible values for kk:k=−5±492⋅−6k = \frac{-5 \pm \sqrt{49}}{2 \cdot -6}
  18. Identify valid solution: Calculate the discriminant (b2−4ac)(b^2 - 4ac):Discriminant=(5)2−4(−6)(1)=25+24=49\text{Discriminant} = (5)^2 - 4(-6)(1) = 25 + 24 = 49Calculate the two possible values for kk:k=−5±492⋅−6k = \frac{-5 \pm \sqrt{49}}{2 \cdot -6}Simplify the square root of the discriminant:k=−5±7−12k = \frac{-5 \pm 7}{-12}
  19. Identify valid solution: Calculate the discriminant (b2−4ac)(b^2 - 4ac):Discriminant=(5)2−4(−6)(1)=25+24=49\text{Discriminant} = (5)^2 - 4(-6)(1) = 25 + 24 = 49Calculate the two possible values for kk:k=−5±492×−6k = \frac{-5 \pm \sqrt{49}}{2 \times -6}Simplify the square root of the discriminant:k=−5±7−12k = \frac{-5 \pm 7}{-12}Calculate the two possible values for kk:k1=−5+7−12=2−12=−16k_1 = \frac{-5 + 7}{-12} = \frac{2}{-12} = -\frac{1}{6}k2=−5−7−12=−12−12=1k_2 = \frac{-5 - 7}{-12} = \frac{-12}{-12} = 1
  20. Identify valid solution: Calculate the discriminant (b2−4ac)(b^2 - 4ac):Discriminant=(5)2−4(−6)(1)=25+24=49\text{Discriminant} = (5)^2 - 4(-6)(1) = 25 + 24 = 49Calculate the two possible values for kk:k=−5±492⋅−6k = \frac{-5 \pm \sqrt{49}}{2 \cdot -6}Simplify the square root of the discriminant:k=−5±7−12k = \frac{-5 \pm 7}{-12}Calculate the two possible values for kk:k1=−5+7−12=2−12=−16k_1 = \frac{-5 + 7}{-12} = \frac{2}{-12} = -\frac{1}{6}k2=−5−7−12=−12−12=1k_2 = \frac{-5 - 7}{-12} = \frac{-12}{-12} = 1Since we are looking for k > 0, the only valid solution is k=1k = 1.

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