Question

question
solve for ( x ) and graph the solution on the number line below. if possible, resolve your answer to a sing
( 3 x + 6 lt 21 ) or ( 36 geq 3 x + 6 )
answer attempt 1 out of 3
inequality notation:
number line:

Explanation:

Step1: Solve \(3x + 6 < 21\)

Subtract 6 from both sides: \(3x + 6 - 6 < 21 - 6\)
Simplify: \(3x < 15\)
Divide both sides by 3: \(\frac{3x}{3} < \frac{15}{3}\)
Simplify: \(x < 5\)

Step2: Solve \(36 \geq 3x + 6\)

Subtract 6 from both sides: \(36 - 6 \geq 3x + 6 - 6\)
Simplify: \(30 \geq 3x\)
Divide both sides by 3: \(\frac{30}{3} \geq \frac{3x}{3}\)
Simplify: \(10 \geq x\) or \(x \leq 10\)

Step3: Combine the solutions

The first inequality gives \(x < 5\) and the second gives \(x \leq 10\). Since "or" is used, the solution is all real numbers (because the set of \(x < 5\) is a subset of \(x \leq 10\), so the union is \(x \leq 10\) combined with \(x < 5\) which is just all real numbers? Wait, no, wait. Wait, actually, let's check again. Wait, \(3x + 6 < 21\) gives \(x < 5\), and \(36 \geq 3x + 6\) gives \(x \leq 10\). The "or" means the solution is the set of \(x\) that satisfy either inequality. So the solution is \(x < 5\) or \(x \leq 10\). But since \(x < 5\) is already included in \(x \leq 10\) (because if \(x < 5\), then \(x \leq 10\) is also true), the combined solution is \(x \leq 10\)? Wait, no, that's not right. Wait, no: "or" means any \(x\) that satisfies at least one of the inequalities. So \(x < 5\) satisfies the first, and \(x\) between 5 (inclusive) and 10 (inclusive) satisfies the second, and \(x > 10\) satisfies neither. Wait, no, let's solve \(36 \geq 3x + 6\) again. \(36 - 6 \geq 3x\) => \(30 \geq 3x\) => \(x \leq 10\). So the second inequality is \(x \leq 10\). The first is \(x < 5\). So the "or" of \(x < 5\) and \(x \leq 10\) is \(x \leq 10\), because all \(x < 5\) are also \(x \leq 10\). Wait, but that seems odd. Wait, maybe I made a mistake. Wait, let's check with a number. Let's take \(x = 8\). Does \(x = 8\) satisfy either inequality? \(3(8) + 6 = 30\). \(30 < 21\)? No. \(36 \geq 30\)? Yes. So \(x = 8\) is a solution. \(x = 6\): \(3(6) + 6 = 24\). \(24 < 21\)? No. \(36 \geq 24\)? Yes. So \(x = 6\) is a solution. \(x = 5\): \(3(5) + 6 = 21\). \(21 < 21\)? No. \(36 \geq 21\)? Yes. So \(x = 5\) is a solution. \(x = 10\): \(3(10) + 6 = 36\). \(36 < 21\)? No. \(36 \geq 36\)? Yes. So \(x = 10\) is a solution. \(x = 11\): \(3(11) + 6 = 39\). \(39 < 21\)? No. \(36 \geq 39\)? No. So \(x = 11\) is not a solution. So the solution is \(x \leq 10\) (because for \(x \leq 10\), either \(x < 5\) (so satisfies first) or \(5 \leq x \leq 10\) (so satisfies second)). Wait, but actually, the first inequality is \(x < 5\), the second is \(x \leq 10\). So the "or" of these two is \(x \leq 10\), because any \(x \leq 10\) will satisfy at least one of the inequalities (either \(x < 5\) or \(5 \leq x \leq 10\)). Wait, but let's check the original inequalities again. The problem is \(3x + 6 < 21\) or \(36 \geq 3x + 6\). Let's rewrite the second inequality as \(3x + 6 \leq 36\). So we have \(3x + 6 < 21\) or \(3x + 6 \leq 36\). The first inequality is a strict inequality, the second is non-strict. So the solution is all \(x\) such that \(3x + 6 < 21\) or \(3x + 6 \leq 36\). But \(3x + 6 < 21\) implies \(3x + 6 \leq 36\) (because 21 < 36, so if \(3x + 6 < 21\), then \(3x + 6 < 36\), so \(3x + 6 \leq 36\) is true). Therefore, the solution is all real numbers? Wait, no, that can't be. Wait, no, let's plug in \(x = 10\): \(3(10) + 6 = 36\), which satisfies the second inequality. \(x = 11\): \(3(11) + 6 = 39\), which is not less than 21 and not less than or equal to 36 (since 39 > 36), so \(x = 11\) is not a solution. \(x = -10\): \(3(-10) + 6 = -24\), which is less than 21, so satisfies the first inequality. So the solution is \(x \leq 10\), because for \(x > 10\), neither inequality is satisfied. Wait, no, \(x = 10\) satisfies the second, \(x = 9\) satisfies the second, \(x = 5\) satisfies the second (3(5)+6=21, which is ≤36), \(x = 4\) satisfies the first. So the solution is \(x \leq 10\). Wait, but let's solve the inequalities again.

First inequality: \(3x + 6 < 21\) => \(3x < 15\) => \(x < 5\)

Second inequality: \(36 \geq 3x + 6\) => \(30 \geq 3x\) => \(10 \geq x\) => \(x \leq 10\)

So the solution is \(x < 5\) or \(x \leq 10\). But since \(x < 5\) is a subset of \(x \leq 10\) (because if \(x < 5\), then \(x \leq 10\) is true), the union of \(x < 5\) and \(x \leq 10\) is \(x \leq 10\). Wait, no, that's not right. The union of two sets A and B is all elements in A or B. If A is \(x < 5\) and B is \(x \leq 10\), then A is a subset of B, so A ∪ B = B. So the solution is \(x \leq 10\).

Wait, but let's check with \(x = 5\): \(3(5) + 6 = 21\). The first inequality is \(3x + 6 < 21\), which is 21 < 21? No, so \(x = 5\) does not satisfy the first inequality. But \(x = 5\) satisfies the second inequality: \(36 \geq 21\), which is true. So \(x = 5\) is a solution. \(x = 6\): \(3(6) + 6 = 24\), which is not less than 21, but 36 ≥ 24, so it's a solution. \(x = 10\): 36 ≥ 36, solution. \(x = 11\): 36 ≥ 39? No, not a solution. So the solution is \(x \leq 10\).

Wait, but the first inequality is \(x < 5\), the second is \(x \leq 10\). So the "or" means any \(x\) that is in \(x < 5\) or in \(x \leq 10\). But since \(x < 5\) is inside \(x \leq 10\), the combined solution is \(x \leq 10\).

Answer:

Inequality Notation: \(x \leq 10\)

For the number line: Draw a closed circle at 10 (since \(x \leq 10\) includes 10) and shade to the left of 10 (including all numbers less than or equal to 10).