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

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Q. A line with equation

(3k + 2)y + 5x = 7

has the same gradient as the line

y + (3 - 2k)x = 8

. Find the value of

k

where

k > 0

Rearrange equation for gradient: To find the gradient of the first line, we need to rearrange the equation $(3k + 2)y + 5x = 7$ into the slope-intercept form $y = mx + b$ , where $m$ is the gradient.
Isolate y-term: First, we isolate the y-term on one side of the equation: $\newline$ $(3k + 2)y = -5x + 7$
Find gradient of first line: Next, we divide both sides by $(3k + 2)$ to solve for $y$ : $y = \frac{-5x + 7}{3k + 2}$
Rearrange second line equation: The gradient of the first line is the coefficient of $x$ , which is $-\frac{5}{3k + 2}$ .
Isolate y-term: Now, we find the gradient of the second line by rearranging $y + (3 - 2k)x = 8$ into the slope-intercept form.
Find gradient of second line: We isolate the $y$ -term on one side of the equation: $\newline$ $y = - (3 - 2k)x + 8$
Set gradients equal: The gradient of the second line is the coefficient of $x$ , which is $-(3 - 2k)$ .
Clear fraction: Since the two lines have the same gradient, we set the gradients equal to each other: $\newline$ $-\frac{5}{3k + 2} = -\frac{3 - 2k}{1}$
Expand equation: We can remove the negative signs from both sides of the equation: $\frac{5}{3k + 2} = 3 - 2k$
Combine like terms: Next, we multiply both sides by $(3k + 2)$ to clear the fraction: $\newline$ $5 = (3 - 2k)(3k + 2)$
Move terms to one side: We expand the right side of the equation: $5 = 9k + 6 - 6k^2 - 4k$
Solve quadratic equation: Combine like terms on the right side: $5 = -6k^2 + 5k + 6$
Calculate discriminant: We move all terms to one side to set the equation to zero: $\newline$ $0 = -6k^2 + 5k + 6 - 5$
Calculate possible values for $k$ : Simplify the equation: $\newline$ $0 = -6k^2 + 5k + 1$
Calculate possible values for k: Now, we solve the quadratic equation for k. Since we are looking for k > $0$ , we can use the quadratic formula: $\newline$ k = $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $\newline$ where a = $-6$ , b = $5$ , and c = $1$ .
Identify valid solution: Calculate the discriminant $(b^2 - 4ac)$ : $\newline$ Discriminant = $(5)^2 - 4(-6)(1) = 25 + 24 = 49$
Identify valid solution: Calculate the discriminant $(b^2 - 4ac)$ : $\text{Discriminant} = (5)^2 - 4(-6)(1) = 25 + 24 = 49$ Calculate the two possible values for $k$ : $k = \frac{-5 \pm \sqrt{49}}{2 \cdot -6}$
Identify valid solution: Calculate the discriminant $(b^2 - 4ac)$ : $\text{Discriminant} = (5)^2 - 4(-6)(1) = 25 + 24 = 49$ Calculate the two possible values for $k$ : $k = \frac{-5 \pm \sqrt{49}}{2 \cdot -6}$ Simplify the square root of the discriminant: $k = \frac{-5 \pm 7}{-12}$
Identify valid solution: Calculate the discriminant $(b^2 - 4ac)$ : $\text{Discriminant} = (5)^2 - 4(-6)(1) = 25 + 24 = 49$ Calculate the two possible values for $k$ : $k = \frac{-5 \pm \sqrt{49}}{2 \times -6}$ Simplify the square root of the discriminant: $k = \frac{-5 \pm 7}{-12}$ Calculate the two possible values for $k$ : $k_1 = \frac{-5 + 7}{-12} = \frac{2}{-12} = -\frac{1}{6}$ $k_2 = \frac{-5 - 7}{-12} = \frac{-12}{-12} = 1$
Identify valid solution: Calculate the discriminant $(b^2 - 4ac)$ : $\text{Discriminant} = (5)^2 - 4(-6)(1) = 25 + 24 = 49$ Calculate the two possible values for $k$ : $k = \frac{-5 \pm \sqrt{49}}{2 \cdot -6}$ Simplify the square root of the discriminant: $k = \frac{-5 \pm 7}{-12}$ Calculate the two possible values for $k$ : $k_1 = \frac{-5 + 7}{-12} = \frac{2}{-12} = -\frac{1}{6}$ $k_2 = \frac{-5 - 7}{-12} = \frac{-12}{-12} = 1$ Since we are looking for $k > 0$ , the only valid solution is $k = 1$ .