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Question
From the sum of 5x + 7y – 12 and 3x – 5y + 2, subtract the sum of 2x – 7y – 1 and – 6x + 3y + 9
Solution
Sum of 5x + 7y – 12 and 3x – 5y + 2.
= 5x + 7y – 12 + 3x – 5y + 2
= (5 + 3)x + (7 – 5)y + ((– 12) + 2)
= 8x + 2y – 10.
Again Sum of 2x – 7y – 1 and – 6x + 3y + 9
= 2x – 7y – 1 + (– 6x + 3y + 9)
= 2x – 7y – 1 – 6x + 3y + 9
= (2 – 6)x + (– 7 + 3)y + (– 1 + 9)
= – 4x – 4y + 8
Now 8x + 2y – 10 – (– 4x – 4y + 8)
= 8x + 2y – 10 + (4x + 4y – 8)
= 8x + 2y – 10 + 4x + 4y – 8
= (8 + 4)x + (2 + 4)y + ((– 10) + (– 8))
= 12x + 6y – 18
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| Column I | Column II |
| (i) The number of corners of a quadrilateral | (A) = |
| (ii) The variable in the equation 2p + 3 = 5 | (B) constant |
| (iii) The solution of the equation x + 2 = 3 | (C) +1 |
| (iv) solution of the equation 2p + 3 = 5 | (D) –1 |
| (v) A sign used in an equation | (E) p |
| (F) x |
