Question

Solve the inequality both algebraically and graphically. Draw a number line graph of the solution and give interval notation.

Answer 1

Finding the solution algebraically

To answer this inequality, we can follow the next steps:

1. Multiply by 7 both sides of the inequality:

$7\cdot\frac{(x-7)}{2}<7\cdot\frac{41}{7}\Rightarrow7\cdot\frac{(x-7)}{2}<41$

2. Multiply by 2 both sides of the inequality:

$7\cdot2\cdot^{\cdot}\frac{(x-7)}{2}<2\cdot41\Rightarrow7\cdot(x-7)<82$

3. Apply the distributive property at the left side of the inequality:

$7\cdot x-7\cdot7<82\Rightarrow7x-49<82$

4. Add 49 to both sides of the inequality:

$7x-49+49<82+49\Rightarrow7x<131$

5. Finally, divide both sides of the inequality by 7:

$\frac{7x}{7}<\frac{131}{7}\Rightarrow x<\frac{131}{7}$

We can graph this inequality in the number line as follows:

Notice the parenthesis indicating that the solution is the number below 131/7 (but not equal to 131/7). In interval notation the solution is:

$(-\infty,\frac{131}{7})$$(-\infty,18\frac{5}{7})$

Or, approximately:

$(-\infty,18.7142857143)$

The number 131/7 in decimal is equivalent to 18.7142857143, so the graph of the solution is given by graph A (we can see that there are seven divisions between 18 and 19; since we have that the shaded division is in the 5th division, then, we have 5/7 = 0.714285714286, that is, the decimal part of the above number).

We can express the number 131/7 as a mixed number as follows:

$\frac{131}{7}=\frac{126}{7}+\frac{5}{7}=18+\frac{5}{7}=18\frac{5}{7}$

Again, notice also the symbol for the left part of the interval notation is a parenthesis since the interval is open at the point 131/7 = 18 + 5/7.

Finding the solution graphically

To find the solution graphically, we can represent the inequality as two lines as follows:

$y=\frac{x-7}{2},y=\frac{41}{7}$

Then, if we graph the first line, we can find the x- and the y-intercepts to find two points to graph the line. We have that the x- and the y-intercepts are:

The x-intercept is (that is, when y = 0):

$0=\frac{x-7}{2}\Rightarrow x-7=0\Rightarrow x=7$

Then, the x-intercept is (7, 0), and the y-intercept (the point on the graph when x = 0) is:

$y=\frac{x-7}{2}\Rightarrow y=\frac{0-7}{2}\Rightarrow y=-\frac{7}{2}$

Then, the y-intercept is (0, -7/2).

The other line is given by:

$y=\frac{41}{7}=\frac{35}{7}+\frac{6}{7}=5\frac{6}{7}$

With this information, we can graph both lines:

And we can see that the point where the two lines coincide is:

$(\frac{131}{7},\frac{41}{7})$

Then, the values for x of the line (x-7)/2 [that is, the values of y = (x-7)/2] that are less than y = 41/7, represented as:

$\frac{(x-7)}{2}<\frac{41}{7}$

Are those values of x less than 131/7, or the solution is also (we express the solution as a fraction or a mixed number as follows) (the same solution):

$(-\infty,\frac{131}{7})or(-\infty,18\frac{5}{7})$

In summary, we have that the solution to the inequality is:

As an inequality:

$x<\frac{131}{7},x<18\frac{5}{7}$

In interval notation:

$(-\infty,18\frac{5}{7})or(-\infty,\frac{131}{7})$

And the representation of the solution on the number line is (option A):