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18) solve the inequalities a) $x - 9 > 28$ b) $3x + 2 < 8$

Question

  1. solve the inequalities

a) $x - 9 > 28$
b) $3x + 2 < 8$

Explanation:

Part (a)

Step1: Isolate \( x \) by adding 9 to both sides.

To solve \( x - 9 > 28 \), we add 9 to both sides of the inequality to isolate \( x \). This is based on the addition property of inequalities, which states that if \( a > b \), then \( a + c > b + c \) for any real number \( c \).

$$ x - 9 + 9 > 28 + 9 $$

Step2: Simplify both sides.

Simplifying the left - hand side, \( x-9 + 9=x \). Simplifying the right - hand side, \( 28 + 9 = 37 \). So we get:

$$ x>37 $$
Part (b)

Step1: Subtract 2 from both sides.

To solve \( 3x+2 < 8 \), we first use the subtraction property of inequalities. The subtraction property of inequalities states that if \( a < b \), then \( a - c < b - c \) for any real number \( c \). Subtract 2 from both sides:

$$ 3x+2 - 2<8 - 2 $$

Step2: Simplify both sides.

Simplifying the left - hand side, \( 3x+2 - 2 = 3x \). Simplifying the right - hand side, \( 8 - 2=6 \). So we have \( 3x<6 \).

Step3: Divide both sides by 3.

We use the division property of inequalities. The division property of inequalities states that if \( a < b \) and \( c>0 \), then \( \frac{a}{c}<\frac{b}{c} \). Divide both sides of \( 3x < 6 \) by 3:

$$ \frac{3x}{3}<\frac{6}{3} $$

Step4: Simplify both sides.

Simplifying the left - hand side, \( \frac{3x}{3}=x \). Simplifying the right - hand side, \( \frac{6}{3} = 2 \). So we get:

$$ x < 2 $$

Answer:

a) \( x>37 \)

b) \( x < 2 \)