Question

question 34
solve for t:
\\(\frac{6}{t} = -18\\)
\\(\bigcirc\\) \\(t = -3\\)
\\(\bigcirc\\) \\(t = -\frac{1}{3}\\)
\\(\bigcirc\\) \\(t = 3\\)
\\(\bigcirc\\) \\(t = \frac{1}{3}\\)

question 35
solve for s:
\\(\frac{6}{s - 3} = \frac{7}{s - 3} - 1\\)
\\(\bigcirc\\) \\(s = -2\\)
\\(\bigcirc\\) \\(s = 2\\)
\\(\bigcirc\\) \\(s = -4\\)
\\(\bigcirc\\) \\(s = 4\\)

Explanation:

Question 34

Step1: Multiply both sides by $t$

$6 = -18t$

Step2: Divide by $-18$ to isolate $t$

$t = \frac{6}{-18} = -\frac{1}{3}$

Question 35

Step1: Multiply all terms by $s-3$

$6 = 7 - (s-3)$

Step2: Simplify the right-hand side

$6 = 7 - s + 3$
$6 = 10 - s$

Step3: Isolate $s$

$s = 10 - 6 = 4$

Answer:

Question 34: $\boldsymbol{t=-\frac{1}{3}}$
Question 35: $\boldsymbol{s=4}$