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Sagot :
To graph the solution of the inequality [tex]\(350 \geq 125 + 15x\)[/tex], let's follow these steps carefully:
1. Isolate the term with [tex]\(x\)[/tex]:
- Start with the given inequality:
[tex]\[ 350 \geq 125 + 15x \][/tex]
- To isolate [tex]\(15x\)[/tex] on one side, subtract 125 from both sides of the inequality:
[tex]\[ 350 - 125 \geq 15x \][/tex]
- Simplify the left side:
[tex]\[ 225 \geq 15x \][/tex]
2. Solve for [tex]\(x\)[/tex]:
- Divide both sides of the inequality by 15 to solve for [tex]\(x\)[/tex]:
[tex]\[ \frac{225}{15} \geq x \][/tex]
- Simplify the fraction:
[tex]\[ 15 \geq x \][/tex]
- This can be rewritten as:
[tex]\[ x \leq 15 \][/tex]
3. Graph the solution on a number line:
- Draw a horizontal number line.
- Identify and mark the critical point [tex]\(x = 15\)[/tex] on the number line.
- Since the inequality is [tex]\(x \leq 15\)[/tex], we include the value [tex]\(x = 15\)[/tex] in our solution, so we place a solid point or closed circle at [tex]\(x = 15\)[/tex].
- Shade the portion of the number line that includes all values less than or equal to 15, extending towards negative infinity.
Here's a representation of the number line:
[tex]\[ \begin{array}{cccccccccccccc} \text{Negative Infinity} \leftarrow & \cdots & -2 & -1 & 0 & 1 & 2 & \cdots & \cdots & 14 & 15 & 16 & 17 & \cdots \rightarrow \text{Positive Infinity} \end{array} \][/tex]
[tex]\[ \begin{array}{ccccccccccccccccccc} \leftarrow & & & & & & & & & \bullet & \text{~~~~~~~~~~~~~}\cdots\\ & & & & & & & & 14 & \ 15 \ & \text{~~~~~~~~}\cdots \\ \end{array} \][/tex]
- The shaded part shows all values for [tex]\(x\)[/tex] that satisfy [tex]\(x \leq 15\)[/tex].
Thus, the solution [tex]\([ -\infty, 15 ]\)[/tex] has been represented on the number line with a closed circle at [tex]\( x = 15 \)[/tex] and shading to the left.
1. Isolate the term with [tex]\(x\)[/tex]:
- Start with the given inequality:
[tex]\[ 350 \geq 125 + 15x \][/tex]
- To isolate [tex]\(15x\)[/tex] on one side, subtract 125 from both sides of the inequality:
[tex]\[ 350 - 125 \geq 15x \][/tex]
- Simplify the left side:
[tex]\[ 225 \geq 15x \][/tex]
2. Solve for [tex]\(x\)[/tex]:
- Divide both sides of the inequality by 15 to solve for [tex]\(x\)[/tex]:
[tex]\[ \frac{225}{15} \geq x \][/tex]
- Simplify the fraction:
[tex]\[ 15 \geq x \][/tex]
- This can be rewritten as:
[tex]\[ x \leq 15 \][/tex]
3. Graph the solution on a number line:
- Draw a horizontal number line.
- Identify and mark the critical point [tex]\(x = 15\)[/tex] on the number line.
- Since the inequality is [tex]\(x \leq 15\)[/tex], we include the value [tex]\(x = 15\)[/tex] in our solution, so we place a solid point or closed circle at [tex]\(x = 15\)[/tex].
- Shade the portion of the number line that includes all values less than or equal to 15, extending towards negative infinity.
Here's a representation of the number line:
[tex]\[ \begin{array}{cccccccccccccc} \text{Negative Infinity} \leftarrow & \cdots & -2 & -1 & 0 & 1 & 2 & \cdots & \cdots & 14 & 15 & 16 & 17 & \cdots \rightarrow \text{Positive Infinity} \end{array} \][/tex]
[tex]\[ \begin{array}{ccccccccccccccccccc} \leftarrow & & & & & & & & & \bullet & \text{~~~~~~~~~~~~~}\cdots\\ & & & & & & & & 14 & \ 15 \ & \text{~~~~~~~~}\cdots \\ \end{array} \][/tex]
- The shaded part shows all values for [tex]\(x\)[/tex] that satisfy [tex]\(x \leq 15\)[/tex].
Thus, the solution [tex]\([ -\infty, 15 ]\)[/tex] has been represented on the number line with a closed circle at [tex]\( x = 15 \)[/tex] and shading to the left.
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