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

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Mar 09, 2013
Rating
starstarstarstarstar 		FIND 3 CONSECUTIVE
by: MERICK

hi!
i have a problem this is my problem, please help me:
FIND THREE CONSECUTIVE NUMBERS SUCH THAT THE SUM OF THE FIRST AND THE THIRD IS 23 MORE TAN THE SECOND,
I HOPE FOR YOUR KINDLY CONSIDERATION TO ANSWER THIS PROBLEM.
THANK U!


Karin from Algebra Class says:

If you have 3 consecutive integers, we can identify those integers as: x, x+1, and x+2.

The problem states that the sum of the first and third is 23 more than the second. This means:

first (x) + third (x+2) = second (x+1) + 23 (23 more).

Or: x + x+2 = x+1 +23

Now, solve for x.

x+x+2 = x+1 + 23
2x+2 = x+24 (Combine like terms)

2x-x +2 = x-x +24 (Subtract x from both sides)

x+2 = 24

x+2-2 = 24 -2 (Subtract 2 from both sides)

x = 22

If x = 22, then the other two numbers are 23 and 24 since the numbers are consecutive.

Check:
22 + 24 = 46 Is this 23 more than the second number (23)?
46 - 23 = 23. yes!!

So, the three numbers are: 22, 23, and 24.

Hope this helps!

Karin
Jun 29, 2011
Rating
starstarstarstarstar 		thanks
by: Anonymous

Thank you. This was very nice to help

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