Algebra Help?

Question

"A 2 digit number has its 'tens' number 5 more than 2 times the 'ones/units' number. The number is 1 less than 9 times the sum of it its digits.
Write an equation for this and solve it"

This was a question from one of my maths tests. I was so lost. Can anyone help?

Answer 1

let tens digit be x
and units digit be y
x=2y+5..........................(1)
9(x+y)=(10x+y)+1
9x+9y=10x+y+1
x=8y-1...................................(2)
(1)=(2)
2y+5=8y-1
6y=6
y=1
and x=7
so number is 71

Answer 2

Let x = unit digit, 2x + 5 = tens digit.

Equation:
10(2x + 5) + x = 9(2x + 5 + x) - 1
20x + 50 + x = 9(3x + 5) - 1
21x + 50 = 27x + 45 - 1
21x + 50 = 27x + 44
6x = 6
x = 1 unit digit

2x + 5 = 7 tens digit

Answer: The number is 71.

Proof:
9(7 + 1) - 1 = 71
9(8) - 1 = 71
72 -1 = 71

Answer 3

Let y be the tens digit, and x be the ones digit.

y=5+2x

y*10 + x = 9*(y + x) -1

(5 + 2x) * 10 + x = 9 * ((5 + 2x) + x) - 1

50 + 21x = 44 + 27x

6=6x

x=1

y=7

The number is 71.

The sum of its digits is 8. 9*8=72, which is one more than the number.

Answer 4

Let n be the number, u the units, t the tens.

n = 10t + u

t = 5 + 2u

n= 9(u + t) - 1

------------------------
Plug the third equation into the first:
9(u + t) - 1 = 10t + u
t = 5 + 2u

9u + 9t - 1 = 10t + u // - 10t - 9u
t = 5 + 2u

-t - 1 = -8u
t - 5 = 2u

Add the equations:
-6 = -6u // divide by -6

u=1

Plugging into t-5 = 2u

t - 5 = 2

t=7

n=10t+u=10*7+1=71

Answer 5

♠ equations t=2*u+5, 9*(t+u) = 10*t +u +1;
9*(2u+5 +u) = 10*(2u+5) +u +1;
27u +45 = 21u +51, hence u=1, and t=7; thus 71;
♠ I flee while you are busy with it;