Formulate the following problems as a pair of equations, and hence find their solutions:
A two digit number is seven times the sum of its digits. The number formed by reversing the digits is 18 less than the given number. Find the given number.
Let x denote the digit in the tenth place and y denote the digit in the unit place. So the number may be written as 10x + y.
When digits are reversed, the number becomes 10y + x.
First condition is
10x + y = 7 (x + y)
⇒ 10x + y – 7x – 7y = 0
⇒ 3x – 6y = 0
⇒ x – 2y = 0
Second condition is
10y + x = (10x + y) – 18
⇒ 10y + x – 10x – y + 18 = 0
⇒ – 9x + 9y + 18 = 0
⇒ – x + y + 2 = 0
For cross multiplication method, we write the coefficients as

Hence, we get
=
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⇒
=
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⇒
=
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⇒ x =
= 4
⇒ y =
= 2
∴ The number is 10x + y = 10 (4) + 2 = 40 + 2 = 42.
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