word problems

4:33 AM |


WORD PROBLEMS fall into distinct types. Below are examples. The only difficulty will be translating verbal language into algebraic language. See Lesson 1, Problem 7.  And practice.
Example 1.   ax ± b = c.  All problems like the following lead eventually to an equation in that simple form.
Jane spent $42 for shoes.  This was $14 less than twice what she spent for a blouse.  How much was the blouse?
Solution.   Every word problem has an unknown number. In this problem, it is the price of the blouse.  Always let x represent the unknown number.  That is, let x answer the question.
Let x, then, be how much she spent for the blouse.  The problem states that "This" -- that is, $42 -- was $14 less  than two times x.
Here is the equation:
2x − 14 = 42.
 
2x = 42 + 14   (Lesson 9)
 
  = 56.
 
x = 56
 2
 
  = 28.
The blouse cost $28.
Example 2.   There are b boys in the class.  This is three more than four times the number of girls.  How many girls are in the class?
 Solution.   Again, let x represent the unknown number that you are asked to find:  Let x be the number of girls.
(Although b is not known, it is not what you are asked to find.)
The problem states that "This" -- b -- is three more than four times x:
 4x + 3=b. 
 Therefore,
 4x=b − 3 
 
x=b − 3
   4		.
The solution here is not a number, because it will depend on the value of b.  This is a type of "literal" equation, which is very common in algebra.
Example 3.  The whole is equal to the sum of the parts.
The sum of two numbers is 84, and one of them is 12 more than the other.  What are the two numbers?
 Solution.  In this problem, we are asked to find two numbers. Therefore, we must let x be one of them.  Let x, then, be the first number.
We are told that the other number is 12 more, x + 12.
The problem states that their sum is 84:
  = 84
The line over x + 12 is a grouping symbol called a vinculum.  It saves us writing parentheses.
We have:
 
2x= 84 − 12
 
  = 72.
 
x = 72
 2
 
  = 36.
This is the first number.  Therefore the other number is
x + 12 = 36 + 12 = 48.
The sum of 36 + 48 is 84.
Example 4.   The sum of two consecutive numbers is 37.  What are they?
Solution.   Two consecutive numbers are like 8 and 9, or 51 and 52.
Let x, then, be the first number.  Then the number after it is x + 1.
The problem states that their sum is 37:
  = 37
2x=37 − 1
 
 =36.
 
x=36
 2
 
 =18.
The two numbers are 18 and 19.
Example 5.  One number is 10 more than another.  The sum of twice the smaller plus three times the larger, is 55.  What are the two numbers?
 Solution.  Let x be the smaller number.
Then the larger number is 10 more:  x + 10.
The problem states:
2x + 3(x + 10)=55.
        That implies
2x + 3x + 30=55.  Lesson 14.
 
5x=55 − 30 = 25.
 
x=5.
That's the smaller number.  The larger number is 10 more:  15.
Example 6.   Divide $80 among three people so that the second will have twice as much as the first, and the third will have $5 less than the second.
Solution.   Again, we are asked to find more than one number.  We must begin by letting x be how much the first person gets.
Then the second gets twice as much, 2x.
And the third gets $5 less than that, 2x − 5.
Their sum is $80:
 
5x=80 + 5
 
x=85
 5
 
 =17.
This is how much the first person gets.  Therefore the second gets
2x=34.
 
        And the third gets
 
2x − 5=29.
The sum of 17, 34, and 29 is in fact 80.
Example 7.  Odd numbers.   The sum of two consecutive odd numbers is 52.  What are the two odd numbers?
Solution.   First, an even number is a multiple of 2:  2, 4, 6, 8, and so on. It is conventional in algebra to represent an even number as 2n, where, by calling the variable 'n,' it is understood that n will take whole number values:  n = 1, 2, 3, 4, and so on.
An odd number is 1 more (or 1 less) than an even number. And so we represent an odd number as 2n + 1.
Let 2n + 1, then, be the first odd number. Then the next one will be 2 more -- it will be 2n + 3.  The problem states that their sum is 52:
2n + 1  +  2n + 3=52.
We will now solve that equation for n, and then replace the solution in 2n + 1 to find the first odd number.  We have:
4n + 4=52
 
4n=48
 
n=12.
Therefore the first odd number is  2· 12 + 1 = 25.  And so the next one is 27.  Their sum is 52.
Problems
Problem 1.   Julie has $50, which is eight dollars more than twice what John has.  How much has John?  (Compare Example 1.)
First, what will you let x represent?
To see the answer, pass your mouse over the colored area.
To cover the answer again, click "Refresh" ("Reload").
Do the problem yourself first!
The unknown number -- which is how much that John has.
What is the equation?
2x + 8 = 50.
Here is the solution:
x = $21
Problem 2.   Carlotta spent $35 at the market.  This was seven dollars less than three times what she spent at the bookstore; how much did she spend there?
Here is the equation.
3x − 7 = 35
Here is the solution:
x = $14
Problem 3.   There are b black marbles.  This is four more than twice the number of red marbles.  How many red marbles are there?  (Compare Example 2.)
Here is the equation.
2x + 4 = b
Here is the solution:
x = b − 4
   2
Problem 4.    Janet spent $100 on books.  This was k dollars less than five times what she spent on lunch.  How much did she spend on lunch?
Here is the equation.
5x − k = 100
Here is the solution:
x = 100 + k
     5
Problem 5.  The whole is equal to the sum of the parts.
The sum of two numbers is 99, and one of them is 17 more than the other.  What are the two numbers?  (Compare Example 3.)
Here is the equation.
Here is the solution:
x =41
 
x + 17=58
Problem 6.   A class of 50 students is divided into two groups; one group has eight less than the other; how many are in each group?
Here is the equation.
Here is the solution:
x =29
 
x − 8=21
Problem 7.   The sum of two numbers is 72, and one of them is five times the other; what are the two numbers?
Here is the equation.
x + 5x = 72.
Here is the solution:
x = 12.   5x = 60.
Problem 8.   The sum of three consecutive numbers is 87; what are they?  (Compare Example 4.)
Here is the equation.
Here is the solution:
28, 29, 30.
Problem 9.   A group of 266 persons consists of men, women, and children.  There are four times as many men as children, and twice as many women as children.  How many of each are there?
(What will you let x equal -- the number of men, women, or children?)
Let x=The number of children.  Then
 
4x=The number of men. And
 
2x=The number of women.
 
Here is the equation:
x + 4x + 2x = 266
Here is the solution:
x = 38.  4x = 152.  2x = 76.
Problem 10.   Divide $79 among three people so that the second will have three times more than the first, and the third will have two dollars more than the second.   (Compare Example 6.)
Here is the equation.
Here is the solution:
$11,  $33,  $35.
Problem 11.   Divide $15.20 among three people so that the second will have one dollar more than the first, and the third will have $2.70 more than the second.
Here is the equation.
Here is the solution:
$3.50,  $4.50,  $7.20.
Problem 12.   Two consecutive odd numbers are such that three times the first is 5 more than twice the second.  What are those two odd numbers?
(See Example 7, where we represent an odd number as 2n + 1.)
Solution. Let the first odd number be 2n + 1.
Then the next one is 2n + 3 -- because it will be 2 more.
The problem states, that is, the equation is:
 3(2n + 1)=2(2n + 3) + 5.
       That implies:
 6n + 3=4n + 6 + 5.
 2n=8.
 n=4.
Therefore the first odd number is 2· 4 + 1 = 9.  The next one is 11.
And that is the true solution, because according to the problem:
3· 9 = 2· 11 + 5.

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Posted by Sumon Sengupta

Linear equation

4:29 AM |


AN EQUATION is an algebraic statement in which the verb is "equals" = .  An equation involves an unknown number, typically called x.  Here is a simple example:
x + 64 = 100.
"Some number, plus 64, equals 100."
We say that an equation has two sides:  the left side, x + 64, and the right side, 100.
In what we call a linear equation, x appears only to the first power, as in the equation above.  A linear equation is also called an equation of thefirst degree.
The degree of any equation is the highest exponent that appears on the unknown number. An equation of the first degree is called linear because, as we will see much later, its graph is a straight line.
Now, the statement -- the equation -- will become true only when the unknown has a certain value, which we call the solution to the equation.
We can find the solution to that equation simply by subtracting:
x = 100 − 64
 
  = 36.
36 is the only value for which the statement "x + 64 = 100" will be true.  We say that x = 36 satisfies the equation.
Algebra depends on how things look.  As far as how things look, then, we will know that we have solved an equation when we have isolated x on the left.
Why the left? Because that's how we read, from left to right. "x equals . . ."
In the standard form of a linear equation -- ax + b = 0 -- x appears on the left, not the right.
In fact, we are about to see that for any equation that looks like this:
x + a = b,
 
  the solution will look like this:
x = b − a.
Inverse operations
There are two pairs of inverse operations.  Addition and subtraction, multiplication and division.
Formally, to solve an equation we must isolate the unknown(typically x) on the left.
ax − b + c = d.
To solve that equation, we must get a, b, c over to the right, so that xalone is on the left.
The question is:
Answer:
By writing it on the other side with the inverse operation.
For, that preserves the arithmetical relationship on the one hand between addition and subtraction:
100 − 64 = 36   implies   100 = 36 + 64;
and on the other, between multiplication and division:
10
 2		=5 implies 10 = 2· 5.
Algebra is, after all, abstracted -- drawn from -- arithmetic.
And so, to solve this equation:
ax − b + c=d
 
then since b is subtracted on the left, we will add it on the right:
 
ax + c=d + b.
 
 Since c is added on the left, we will subtract it on the right:
 
ax=d + b − c.
 
 And finally, since a multiplies on the left, we will divide it on the right:
 
x=d + b − c
     a
We have solved the equation.

The four forms of equations
Solving any linear equation, then, will fall into four forms, corresponding to the four operations of arithmetic.  The following are the basic rules for solving any linear equation.  In each case, we will shift a to the other side.
1.    If  x + a  = b,  then  x  =  b − a.
2.    If  x  − a  = b,  then  x  =  b + a.

3.    If  ax  = b,  then  x  =  b
a		. 
4.    If   x
a		 = b,  then  x  =  ab.

In every case, we shifted a to the other side by means of the inverseoperation.  Every linear equation can be solved by combining those fourformal rules.
Transposing
When the operations are addition or subtraction (Forms 1 and 2 above), we call that transposing.
We may shift a term to the other side of an equation
by changing its sign.

+ a goes to the other side as − a.
− a goes to the other side as + a.
Transposing is one of the most characteristic operations of algebra, and it is thought to be the meaning of the word algebra, which is of Arabic origin.  (Arabic mathematicians learned algebra in India, from where they introduced it into Europe.)  Transposing is the technique of those who actually use algebra in science and mathematics -- because it is skillful. And as we are about to see, it maintains the clear, logical sequence of statements.  Moreover, it emphasizes that we do algebra with our eyes. When you see
x + a = b,
 
 then you immediately see that +a goes to the other side as −a:
 
x = b − a.
The way that is often taught these days, is to add −ato both sides, draw a line, and add:
(Lesson 6.) While that is logically correct, it's clumsy, 
it disregards the logical sequence of statements and 
therefore offers no preparation to read a calculus text,
for example.

What, after all, is the purpose of it? The purpose is to 
transpose +a to the other side as −a. Therefore the 
student should simply learn to transpose

A logical sequence of statements
In an algebraic sentence, the verb is typically the equal sign = .
ax − b + c = d.
That sentence -- that statement -- will logically imply other statements.  Let us follow the logical sequence that leads to the final statement, which is the solution.
 (1)  ax − b + c = d
 
implies   (2)  ax = d + b − c
 
implies   (3)  x =  d + b − c .
      a
The original equation (1) is "transformed" by first transposing theterms (Lesson 1).  Statement (1) implies statement (2).
That statement is then transformed by dividing by a.  Statement (2) implies statement (3), which is the solution.
Thus we solve an equation by transforming it -- changing its form -- statement by statement, line by line  according to the rules of algebra, until x finally is isolated on the left.  That is how books on mathematics are written (but unfortunately not books that teach algebra!).  Each line is its own readable statement that follows from the line above -- with no crossings out
In other words, What is a calculation?  It is a discrete transformation of symbols.  In arithmetic we transform "19 + 5" into "24".  In algebra we transform  "x + a = b"  into  "x = b − a."
Problem 1.   Write the logical sequence of statements that will solve this equation for x :
abcx − d + e − f  =  0
To see the answer, pass your mouse from left to right 
over the colored area. 
To cover the answer again, click "Refresh" ("Reload").
Do the problem yourself first!

 (1)  abcx − d + e − f  = 0
 
implies   (2)  abcx  =d − e + f
 
implies   (3)  x  =  d − e + f .
    abc
First, transpose the terms. Line (2).
It is not necessary to write the term 0 on the right.
Then divide by the coefficient of x.
Problem 2.   Write the logical sequence of statements that will solve this equation for x :
 (1)  2x + 5 = 27
 
implies   (2)  2x = 27 − 5 = 22
 
implies   (3)  x =  22
 2
 
implies   (4)  x =  11.
Problem 3.   Solve for x :   (p − q)x + r = s
x = s − r
p − q
Problem 4.   Solve for x :   ab(c + d)x − e + f = 0
x =    e − f   
ab(c + d)
Problem 5.   Solve for x :   2x + 1= 0
x = −½
That equation, incidentally, is in the standard form, namelyax + b = 0.
   Problem 6 .   Solve:  ax + b = 0.
 
 x = −b
a
Each of these Problems illustrates doing algebra with your eyes.  That is, the student should see the solution immediately.  In the example above, the student should see that b will go to the other side as −b, and that a will divide.
That is skill in algebra.
Problem 7.   Solve for x :   ax = 0  (a0).
Now, when the product of two numbers is 0, then at least one of them must be 0 (Lesson 5).  Therefore, any equation with that form has the solution,
x = 0.
We could solve that formally, of course, by dividing by a.
x =  0
a		 = 0.
 
(Lesson 5.)
Problem 8.   Solve for x :
4x − 2 = −2
 
4x = −2 + 2 = 0
 
x = 0.
Problem 9.   Write the sequence of statements that will solve this equation:
(1)   6 − x = 9
 
(2)   −x = 9 − 6
 
(3)   −x = 3
 
(4)   x =  −3.
When we go from line (1) to line (2),  −x remains on the left.  For, the terms in line (1) are 6 and −x.
We have "solved" the equation when we have isolated x -- not −x -- on the left.  Therefore we go from line (3) to line (4) by changing the signs on both sides. (Lesson 6.)
Alternatively, we could have eliminated −x on the left by changing all the signs immediately:
(1)   6 − x = 9
 
(2)   −6 + x = −9
 
(3)   x = −9 + 6 = −3
   Problem 10.   Solve for x :    3 − x  =  −5
 
 x = 8
Problem 11.   Solve for x :
5 − 2x  = −11
 
−2x  = −11 − 5
 
2x  = 16  Lesson 6
 
x  = 8
Problem 12.   Solve for x:
3x − 15
2x + 1 		 = 0
(Hint:  Compare Lesson 5, Problem 18.)
x = 5.
Transposing versus exchanging sides
   Example 1. a + b = c − x
We can easily solve this -- in one line -- simply by transposing x to the left, and what is on the left, to the right:
x  =  c −a − b.
   Example 2. a + b = c + x
In this Example, +x is on the right.  Since we want +x on the left, we can achieve that by exchanging sides:
c + x = a + b     
Note:  When we exchange sides, no signs change.
The solution easily follows:
c + x = a + b − c
In summary, when −x is on the right, it is skillful simply to transpose it.  But when +x is on the right, we may exchange the sides.
Problem 13.   Solve for x :
 p + q = r − x − s
 
Transpose: 
 
 x = r − s − p − q
Problem 14.   Solve for x :
 p − q + r = s + x
 
Exchange sides: 
 
 s + x = p − q + r
 
 x = p − q + r − s
Problem 15.   Solve for x :
0  = px + q
 
px + q = 0 
 
px = −q
 
x = −q
p
Problem 16.   Solve for x :
−2  = −5x + 1
 
5x = 1 + 2 = 3
 
x = 3
5
Canceling
x + b + d  = c + d.
d appears on both sides.  Therefore, we may cancel them.
x + b  = c.
Theoretically, we can say that we subtracted d from both sides.
Finally, on solving for x:
x  = c − b.
Problem 17.   Solve for x :
x² + x − 5  =  x² − 3

Cancel the x²'s:
x − 5 = −3
 
x = −3 + 5 = 2.
Problem 18.   Solve for x :
x − a + b  =  a + b + c
Cancel the b's but not the a's.  On the left is −a, but on the right is +a.  They are not equal.
Transpose −a:
x = a + a + c
 
x = 2a + c.

The unknown on both sides
Example 3.   Solve for x :
4x − 3=2x − 11.
 
      1.  Transpose the x's to the left and the numbers to the right:
 
4x − 2x=−11 + 3.
 
      2.  Collect like terms, and solve:
 
2x=−8
 
x=−4.
This is another example of doing algebra with your eyes.  You shouldsee that 2x goes to the left as −2x, and that −3 goes to the right as +3.
As a general rule for solving any linear equation, we can now state the following:
Problem 19.   Solve for x :
15 + x=7 + 5x
 
x − 5x=7 − 15
 
−4x=−8
 
x=2.
Problem 20.   Solve for x :
1.25x − 6=x 
 
1.25x −x=6 
 
(1.25 − 1)x=6 On combining the like terms
on the left.
.25 x=6 
 
x=24. On multiplying both sides
by 4.
Problem 21.   Remove parentheses, add like terms, and solve for x :
(8x − 2) + (3 − 5x)=(2x − 1) − (x − 3)
 
8x − 2 + 3 − 5x=2x − 1 − x + 3
 
3x + 1=x + 2
 
3x − x=2 − 1
 
2x=1
 
x=1
2

Simple fractional equations
   Example 4.      x
2		  =  4.
Since 2 divides on the left, it will multiply on the right:
x=2· 4    
 
 =8.
Problem 22.   Solve for x :
−x
 5 		=3
 
−x=15
 
x=−15.
  Problem 23. x
4		=1
2
 
  x=4· 1
2		= 2.
Example 5.   Solve for x:
4
x		  =  5.
Solution.   In the standard form of a simple fractional equation, x is in the numerator.  But we can easiy make that standard form by taking thereciprocal of both sides.
 x
4		  =  1
5		 
This implies
 x  =  4 ·  1
5
  =  4
5		.
  Problem 24. 2
x		=1
3
 
  x
2		=3
 
  x=6.
Example 6.  Fractional coefficient.
3x
4  		=y
 
     Since 4 divides on the left, it will multiply on the right:
 
3x=4y.
 
     And since 3 multiplies on the left, it will divide on the right:
 
x=4y
 3
In other words, 3
4		 goes to the other side as its reciprocal, 4
3		.
Note that 3
4		 is the coefficient of x :
3x
4		= 3
4		x.
Coefficients go to the other side as their reciprocals!
Problem 25. 2x
3 		=a
 
  x=3a
 2
Problem 26. 5
8		x=a − b
 
   x=8
5		(a − b)
  Problem 27. 4
5		x  + 6=14
 
   4
5		x=14 − 6 = 8
 
    x=5
4		· 8 = 10.
  Problem 28. A=½xB
 
         Exchange sides: 
 
 ½xB=A
 
 x=2A
 B
The reciprocal of ½ is 2.
Problem 29.   The Celsius temperature C is related to the Fahrenheit temperature F  by this formula,
F  =  9
5		C + 32
F=9
5		· 10 + 32
 
 =18 + 32
 
 =50°
b)  Solve the formula for C.
9
5		C + 32=F
 
 9
5		C=F − 32
 
  C=5
9		(F − 32)
c)  What is the Celsius temperature when the Fahrenheit temperature
c)   is 68°?

C=5
9		(68 − 32)
 
 =5
9		· 36
 
 =5· 4
 
 =20°

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