Math Question... Word problem?

Let x = number of rooms with 2 beds

let y = number of rooms with 3 beds

x + y = 60 (because that's how many rooms there are)

2x + 3y = 140 (because that's how many beds there are)

Multiply the first equation by - 2

-2x -2y = - 120

2x + 3y = 140

Add the columns.

y = 20, so x = 40

So there should be 12 rooms left with 3 beds.

Let x be the number of rooms with 2 beds

Let y be the number of rooms with 3 beds

Since there are only 60 rooms, x + y = 60

Since there are 140 beds: 2x + 3y = 140

You now have two equations for two unknowns.

Solve for x:

x + y = 60

x = 60 -y

ok now use the other equation, subsituting our new x in:

2x + 3y = 140

2( 60 - y ) + 3y = 140

120 - 2y + 3y = 140

120 + y = 140

y = 20

ok so we know y... now just go back to one of the equations and find x:

x = 60 -y

x = 60 - ( 20 )

x = 40

20 - 8 = 12 rooms that have 3 beds remain.

x + y = 60 (let x = 2 beds and y = 3 beds)

2x + 3y = 140

Solve for x and y

x + y = 60 (subtract x from both sides)

y = 60 -x

use the above equation and substitute into the second equation for y

2x + 3(60-x) = 140

2x + 180 - 3x = 140

-x + 180 = 140 (subtract 180 from both sides)

-x = -40 (divide both sides by -1)

x = 40

y = 60 -x

y = 60 -40

y = 20 (number of 3-bed rooms)

20 -8 = 12 3-bed rooms left.