Problems solved with linear equations
by Math Student
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
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
Follow the same procedure for your second problem.
Best of luck!
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