$\dfrac{3x}{2}-5 +5 = 4 + 5\\\\\dfrac{3x}{2} = 9 \\\\Now \ multiply \ both \ sides \ by \ 2 \\\\\dfrac{3x}{2}*2 = 9*2\\\\3x = 18\\Now \ divide \ both \ sides \ by \ 3\\\\x =\dfrac{18}{3}\\\\$

x = 6

$b) \dfrac{x-3}{5} + 1 = -2\\\\\\Subtract \ 1 \ from \ both \ sides\\\\\dfrac{x-3}{5} = -2 -1\\\\\dfrac{x-3}{5} = -3\\\\Multiply \ both \ side \ by \ 5\\\\x - 3 = -3*5\\\\x -3 = -15\\\\$

Add 3 to both sides

x = -15 +3

x = -12

7 0

3 years ago

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Sarah took the advertising department from her company on a round trip to meet with a potential client. Including Sarah a total

dimulka [17.4K]

You can set up a system of equations for this problem. x= number of coach tickets and y = number of first class tickets.

$210x + $1200y = $10,230 (cost of coach ticket plus cost of first class tickets is total budget)

x + y = 11 (number of coach tickets plus number of first class tickets is total number of people)

Solve the second equation for y to get y = 11 - x, then plug that into the first equation and solve for x:

$210x + $1200(11 - x) = $10,230

$210x + $13,200 - $1200x = $10,230

-$990x + $13,200 = $10,230

-$990x = $2,970

x = 3

Sarah bought x = 3 coach tickets. Plug that into the second equation and solve for y:

3 + y = 11

y = 8

Sarah bought y = 8 first class tickets.

6 0

3 years ago