Simplify: 7x²(3x – 9) + 3 and find its values for x=4 and x=6.
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Let his salary be x and his fixed increment be y.
After 4 years of service → x + 4y = 1500 → (1)
After 10 years of service → x + 10y = 1800 → (2)
Subtracting, - 6y = - 300
Thus, y = \frac{-300}{-6}−6−300 =50
Putting y = 50 in x + 4y = 1500
We get, x = 1500 - (4 \times 50)1500−(4×50) = 1500 - 200 = 1300
Hence, his starting salary = Rs.1300
Fixed Increment = Rs.50
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