Question

The sum of two numbers is 41. The larger number is one less than twice the smaller number. Find the numbers.

Answer 1

Answer:

14 and 17

Step-by-step explanation:

x+y=41

y=2x-1

x+(2x-1)=41

3x-1=41

3x=42

x=14

y=2(14)-1=27

Answer 2

Step-by-step explanation:

Step 1:  Make a system of equation

The sum of two numbers is 41

    x        +        y        =        41

The larger number is one less than twice the smaller number.

    x                =                2y               -                1

Step 2:  Substitute x from the second equation into the first

$x + y = 41$

$(2y - 1) + y = 41$

$3y - 1+ 1 = 41 + 1$

$3y/3 = 42 / 3$

$y = 14$

Step 3:  Substitute y into the second equation

$x = 2y - 1$

$x = 2(14) - 1$

$x = 28 - 1$

$x = 27$

Answer:  $x = 27, y = 14$