Problems solved with linear equations
by Math Student
(Canada)
Problems solved with linear equations, or as commonly known, Algebra word problems.
Please help me in the following questions with full steps:
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.
When you write your equation, just follow along with the story. You start with 48 passengers.
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
Now start with the distributive property:
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!!!
Follow the same procedure for your second problem.
Best of luck!
Karin