We are given this system of functions:
[tex]2x + 3y = 18[/tex]
[tex]3x -4y \ \textgreater \ 16[/tex]
They are written in standard form, [tex]ax+by=c[/tex], so our first step is to convert them into slope-intercept form, [tex]y=mx+b[/tex], where m is the slope and b is the y-intercept.
We rearrange to basically solve for y in each function.
1) [tex]2x + 3y = 18 \\ 3y = -2y+18 \\ y= \frac{-2}{3}y+6[/tex]
2) [tex] -4y\ \textgreater \ -3x+16[/tex]
We divide by –4 to isolate y, but remember when multiplying or dividing with a negative number in an inequality, we flip the sign of the inequality.
[tex]y \ \textless \ \frac{3}{4}x-4[/tex]
This gives us our slope-intercept forms for each.
For the domain, each function is defined for all real x. This is a property of linear (and polynomial) functions. The inequality does not affect this.
The range, however, is a bit tricky. The range of a linear or polynomial function is always all real numbers y. It might be tempting to restrict it for an inequality, however, the range is still all reals.
In set-builder notation, we write for the domains of both functions: {[tex]x \ | \ all \ real \ numbers[/tex]}. For the ranges, we write {[tex]y \ | \ all \ real \ numbers[/tex]}.
In interval notation, we write [tex](- \infty, \infty)[/tex] for both the domain and range.
To obtain the slope and y-intercept, we look back to the functions in slope-intercept form, [tex]y=mx+b[/tex], again where m is the slope and b is the y-intercept.
For the equation, we have
[tex]m= \dfrac{-2}{3} [/tex]
and [tex]b=6[/tex].
For the inequality, we have
[tex]m= \dfrac{3}{4} [/tex]
and [tex]b=-4[/tex].
We can use the slope and y-intercept to graph each function. Remember that an inequality with less than or greater than (i.e., no "equal to" part) is represented with a dotted line.
To find where you shade the inequality, think about its sign. Since y is less than the expression, we shade below the line. We can also verify this by plugging in a test-point above or below the line and shading the region where the test-point is a valid solution.
See the graphed solution set attached.
