Question

solve for x.
7x + 39 ≥ 53 and 16x + 15 > 31
choose 1 answer:
a x > 1
b x ≥ 2
c x ≤ 2
d there are no solutions
e all values of x are solutions

Explanation:

Step1: Solve \(7x + 39 \geq 53\)

Subtract 39 from both sides: \(7x \geq 53 - 39\)
\(7x \geq 14\)
Divide both sides by 7: \(x \geq \frac{14}{7}\)
\(x \geq 2\)

Step2: Solve \(16x + 15 > 31\)

Subtract 15 from both sides: \(16x > 31 - 15\)
\(16x > 16\)
Divide both sides by 16: \(x > \frac{16}{16}\)
\(x > 1\)

Step3: Find the intersection of \(x \geq 2\) and \(x > 1\)

The intersection of \(x \geq 2\) (which includes 2 and all numbers greater than 2) and \(x > 1\) (all numbers greater than 1) is \(x \geq 2\) because the values that satisfy both inequalities must satisfy the more restrictive one, which is \(x \geq 2\).

Answer:

B. \(x \geq 2\)