To solve the inequality \( 8 + 2x \le 20 \), we first isolate \( x \).
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Subtract 8 from both sides: \[ 2x \le 20 - 8 \] \[ 2x \le 12 \]
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Divide both sides by 2: \[ x \le 6 \]
Now we have \( x \le 6 \). Any value less than or equal to 6 will satisfy the inequality. Here are two examples of such values:
- \( x = 5 \)
- \( x = 6 \)
Both of these values make \( 8 + 2x \le 20 \) true:
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For \( x = 5 \): \[ 8 + 2(5) = 8 + 10 = 18 \quad (\text{which is } \le 20) \]
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For \( x = 6 \): \[ 8 + 2(6) = 8 + 12 = 20 \quad (\text{which is } \le 20) \]
To summarize, two values for \( x \) that satisfy the inequality are \( 5 \) and \( 6 \).