Question

Which is true about the solution to the system of inequalities shown? y < One-thirdx – 1 y < One-thirdx – 3 On a coordinate plane, 2 solid straight lines are shown. The first line has a positive slope and goes through (0, negative 1) and (2, 0). Everything below the line is shaded. The second line has a positive slope and goes through (0, negative 3) and (3, negative 2). Everything below the line is shaded. All values that satisfy y < One-thirdx – 1 are solutions. All values that satisfy y < One-thirdx – 3 are solutions. All values that satisfy either y < One-thirdx – 1 or y < One-thirdx – 3 are solutions. There are no solutions.

Answer 1

Let

$f(x)=\dfrac{1}{3}x-1,\quad g(x)=\dfrac{1}{3}x-3$

Observe that g(x)=f(x)-2, implying that g(x) is parallel to f(x), and it lies 2 units below.

So, asking that something is below g(x) will make sure that it is below f(x) as well. On the contrary, is something is below f(x), it might be above g(x), i.e. between f(x) and g(x).

In formulas, we have

$y$

So, a solution to the first in automatically a solutions to the second.

Answer 2

Answer:

Option B.

Step-by-step explanation:

The given inequalities are

$y$

$y$

The related equations of given inequalities are

$y=\dfrac{1}{3}x-1$

$y=\dfrac{1}{3}x-3$

The tables of values are

For line 1                    For line 2

x      y                           x      y

0     -1                          0     -3

3     0                           9     0

Plot these ordered pairs on a coordinate plane and connect them by straight lines.

The sign of both inequalities is "<" it means both lines are dotted and everything below the lines is shaded.

From the below graph it is clear that all values that satisfy $y$ are solutions.

Therefore, the correct option is B.