If two lines 3x + 5y = 11 and 3x + ky = 7 are parallel to each other, find the value of k.

 Two lines are parallel to each other if their slopes are equal. The slope of a line in the form

$𝐴𝑥+𝐵𝑦=𝐶$ can be found by rearranging the equation to slope-intercept form $𝑦=𝑚𝑥+𝑏$, where $𝑚$ is the slope.

Let's first find the slopes of the given lines:

1. Line 1: $3𝑥+5𝑦=11$ Rearrange the equation to slope-intercept form: $5𝑦=-3𝑥+11$ $𝑦=-\frac{3}{5}𝑥+\frac{11}{5}$ The slope of this line is ${𝑚}_1=-\frac{3}{5}$.

2. Line 2: $3𝑥+𝑘𝑦=7$ Rearrange the equation to slope-intercep

3. t form:

1. ky=−3x+7 $𝑦=-\frac{3}{𝑘}𝑥+\frac{7}{𝑘}$ The slope of this line is ${𝑚}_2=-\frac{3}{𝑘}$.

For the two lines to be parallel, their slopes must be equal. So, we set ${𝑚}_1={𝑚}_2$ and solve for $𝑘$: $-\frac{3}{5}=-\frac{3}{𝑘}$

Cross-multiply to solve for $𝑘$: $-3𝑘=-15$ $𝑘=5$

Therefore, the value of $𝑘$ that makes the lines $3𝑥+5𝑦=11$ and $3𝑥+𝑘𝑦=7$ parallel to each other is $𝑘=5$