Final answer:
To ensure that the system of equations 12x-20y=108 and 3x+ky=27 graphs as a single line, we need to find a value of k that makes their slopes identical. By transforming both equations to slope-intercept form and equating slopes, we determine that k must equal -5.
Step-by-step explanation:
To find the value of k that will result in the system of linear equations 12x-20y=108 and 3x+ky=27 having a graph which is a single line, we need to set the two equations to have the same slope, since the slope is constant along a straight line. We can use the slope-intercept form, y=mx+b, where m is the slope and b is the y-intercept.
First, let's write the first equation in the slope-intercept form:
- Divide every term by 4: 3x-5y=27
- Isolate y: y = \(\frac{3}{5}x - \frac{27}{5}\)
The slope of this line is 3/5 and the y-intercept is not relevant for this problem as we are only concerned with the slope for equating two lines.
Now we'll rewrite the second equation to find its slope:
- Isolate y: y = \(-\frac{3}{k}x + \frac{27}{k}\)
For the two lines to coincide, their slopes must be equal, which means:
\(\frac{3}{5} = -\frac{3}{k}\)
By solving the above equation, we get k = -5. Thus, when k is -5, both equations represent the same line in the coordinate plane, meeting the condition provided.