Question

five times the sum of the digits of a two digit number is 9 less than the number formed by reversing its digits . if four times the value of the digit at ones place is equal to half of the place value of the digit at tens place, find the numbers​

Answer 1

9514 1404 393

Answer:

  45

Step-by-step explanation:

Let x and y represent the tens and ones digits, respectively. The 5 times the sum of digits is 5(x+y). The value of the digit-reversed number is (10y+x), so the required relation is ...

  5(x +y) = (10y +x) -9

The other relationship between the digits is given as ...

  4y = 1/2(10x)

A graphing calculator shows the solution to these equations is (x, y) = (4, 5).

The two-digit number is 45.

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Additional comment

You can solve these equations in any of a variety of ways. Using a graphing calculator to find integer solutions is fast and easy, so is one of my favorites. Here, the coefficients on one equation are not easy multiples of those in the other, so substitution and/or elimination can get messy. In this situation, I like to use the "cross-multiplication" method, which starts with the equations in general form:

• 4x -5y +9 = 0
• 5x -4y +0 = 0

From the coefficients of these equations, differences of cross products are formed:

• d1 = 4(-4) -(5)(-5) = 9
• d2 = -5(0) -(-4)(9) = 36
• d3 = 9(5) -0(4) = 45

Then the solutions are the solutions to 1/d1 = x/d2 = y/d3:

• x = d2/d1 = 36/9 = 4
• y= d3/d1 = 45/9 = 5

(x, y) = (4, 5)   ⇒   the number is 45.

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This method is one of several variations of Cramer's Rule, the general solution of systems of linear equations using matrix methods.