# Problems solved with linear equations

by Math Student

Problems solved with linear equations, or as commonly known, Algebra word problems.

A bus was driving with 48 passengers, when it reached a stop, x passengers got off and 3 got on. In the next stop, half the passengers got off and 7 got on. There are now 22 passengers, Solve x.

A bus was driving with 52 passengers. When it reached a stop, y passengers got off and 4 got on. In the next stop 1/3 (one-third) of the passengers got off and 3 got on. There are now 25 passengers. Solve y.

- Thank you

Karin from Algebra-Class says:

Problems Solved with Linear Equations
by: Karin

I'll help with the first problem, and then hopefully you'll be able to solve the second problem!

A bus was driving with 48 passengers, when it reached astop, x passengers got off and 3 got on. In the next stop, half the passengers got off and 7 got on. There are now 22 passengers, Solve x.

48 - x + 3 (x got off - subtract - and three got on - add)

48 - x + 3 this simplifies to:
51 - x (48+3 = 51)

So, now we have 51-x.

51-x - .5(51-x) + 7 = 22 (1/2 of the passengers means to multiply 1/2 times the # of passengers which at this point all we know is that there are 51-x passengers) Plus add 7 since 7 got on. This all equals 22 passengers.
Here's our final equation:

51-x - .5(51-x) + 7 = 22

51-x - 25.5 + .5x + 7 = 22

Now combine like terms.

51 - 25.5 + 7 - x+.5x = 22
32.5 -.5x = 22

Now subtract 32.5 from both sides.

32.2 - 32.5 - .5x = 22 - 32.5
-.5x = -10.5

Now divide by -.5 on both sides

-.5x/-.5 = -10.5/-.5

x = 21

X = 21 passengers.

Check by following the story:

48 - 21 +3 - 1/2(30)+7 = 22
22=22

It works!!!