(a) Solve the equation : \(\frac{2}{3}(3x - 5) - \frac{3}{5}(2x - 3) = 3\)
(b)
In the diagram, < STQ = m, < TUQ = 80°, < UPQ = r, < PQU = n and < RQT = 88°. Find the value of (m + n).
(b)
In the diagram, < STQ = m, < TUQ = 80°, < UPQ = r, < PQU = n and < RQT = 88°. Find the value of (m + n).
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Correct Answer: Option n
Explanation:
(a) \(\frac{2}{3}(3x - 5) - \frac{3}{5}(2x - 3) = 3\)
\(\frac{10(3x - 5) - 9(2x - 3)}{15} = 3\)
\(30x - 50 - 18x + 27 = 3 \times 15 = 45\)
\(12x = 45 + 23 = 68\)
\(x = \frac{68}{12} = 5\frac{2}{3}\)
(b) \(n + r = 80° .... (1)\)
\(180° - m + r = 88° .... (2)\)
From (1), r = 80° - n.
Putting into (2), we have
\(180 - m + 80 - n = 88 \implies - m - n = 88 - 260 = - 172°\)
\(-(m + n) = - 172 \implies m + n = 172°\).
(a) \(\frac{2}{3}(3x - 5) - \frac{3}{5}(2x - 3) = 3\)
\(\frac{10(3x - 5) - 9(2x - 3)}{15} = 3\)
\(30x - 50 - 18x + 27 = 3 \times 15 = 45\)
\(12x = 45 + 23 = 68\)
\(x = \frac{68}{12} = 5\frac{2}{3}\)
(b) \(n + r = 80° .... (1)\)
\(180° - m + r = 88° .... (2)\)
From (1), r = 80° - n.
Putting into (2), we have
\(180 - m + 80 - n = 88 \implies - m - n = 88 - 260 = - 172°\)
\(-(m + n) = - 172 \implies m + n = 172°\).