Algebraic Expressions and Equations

Algebraic Expressions: Saying it with Algebra

Don and John the quarrelsome twins were at it again. This time, they were fighting about their marbles. It seems both of them collected red ones and both collections had been dumped on the rug and had rolled together. So, they were fighting about how many marbles each one owned. Don said " I'm not sure how many marbles were mine, but I know that I had eight more than you did!" Then John suggested they count all the marbles on the rug.

" How will that help?" asked Don. "We still won't know how many marbles belong to each one of us. " Well maybe we can figure it out" said John. So, they counted all the marbles and got 44.

"Now," said John, "we know that there are 44 marbles in all, and you had eight more than I did. Let's write it down and think about it." So here is what they wrote:

John's marbles + Don's marbles = 44, and
Don's marbles =
John's marbles + 8
so,
John's marbles + (John's marbles + 8) = 44

Then they had a brilliant idea – at exactly the same time. They decided to remove 8 marbles from the group, so that half of the remaining marbles were John's, and half were Don's.
If they separated the remaining marbles into two equal parts, they would know how many each one had owned originally.So now they wrote this:

John's marbles + John's marbles = 36 (44 – 8)
or 2 ×
John's marbles = 36

and from this statement they knew that
John must have had 18 marbles originally,
and Don, with 8 more -- had 26.

What the twins did was solve their mathematical problem with an algebraic approach.
Let's write an algebraic expression and equation to show what they did:

Since algebra uses symbols (letters or variables) to represent values, words etc.,

We'll let J represent the number of marbles John had originally
So,
J + 8 represents the number of marbles Don had originally
therefore
J + (J + 8) represents the sum total of both marble collections.
We know the total is 44
So,
J + J + 8 = 44 which is the same as 2J + 8 = 44
Then they removed 8 marbles to get
(2J + 8) – 8 = (44) – 8
And since 8 – 8 = 0, this statement is the same as 2J = 36
therefore, the value of J must be ½ of 36 which is 18.
The twins concluded that John had 18 marbles originally,
and Don, with 8 more -- had 26.
Since 18 + 26 = 44, they knew they had figured it out correctly.

Algebraic Expressions

Unless we're newbies, when our online chat buddy types LOL in response to a joke, we know exactly what it means. It tells us that he or she is "laughing out loud". Though we may not be aware of it, we're using algebra to communicate with each other, since we assign or name a recognizable symbol to represent something -- a concept, situation, or a value.

In our solution to the twins' problem, we used J to represent the number of marbles John had,
and J + 8 for the number of marbles Don had. These are algebraic expressions.

Then we wrote an equation that said the sum of all the marbles is equal to 44. From that information, we were able to solve for the number of marbles each twin owned.

The first step in creating an algebraic expression is to define the variable. This is sometimes done with a diagram but most often with a let statement. In our solution above, we let J represent John's marbles. Once we defined what J stood for, we knew J + 8 represented the number of Don's marbles. And once we equated the sum to 44, we could solve the problem.

It's a good idea to choose variables that remind us of the quantity we're describing. We usually set length = l, width = w, height = h etc. Notice how we used J for John's marbles.

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Example:
Say we own a very exclusive toy store that sells only 2 items –
teddy bears and hula hoops.
Bears sell for
\$7 each, and hoops cost \$2 each. Now, say a customer buys a few of each item.
If we let
b represent the number of bears and h be the symbol for the number of hoops they buy, we would express their total bill (before taxes) with this algebraic expression:

7b + 2h

Remember that in algebra, we indicate multiplication by writing the terms together,
so 6k means 6 times k. Also remember that "a" means 1 × a – we don't write the 1 in front.

Once we know how many of each item our customer wants to buy, we can evaluate the expression to find the cost of the purchase (before taxes), since we now have assigned values for b and h.
A customer who buys 3 bears and 5 hoops will pay (7 × 3) + (2 × 5) or \$31 -- plus tax, of course! If he buys 1 bear and 4 hoops, he'll pay 7 + (2 × 4) or \$15 before taxes, and so on.
Notice how we don't bother to write 7 × 1 – we just write 7.

An algebraic expression is a SYMBOLIC representation of a situation.

Examples:

1) If a represents the price of an apple and b represents the price of a banana, write
an algebraic expression for the price of:

 a dozen apples12a 3 apples and 7 bananas3a + 7b 5 bananas5b half dozen of each.6a + 6b

2) Evaluate each of the expressions in #1 if the prices are per apple and per banana.

 a dozen apples12a 12(5¢) = 60¢ 3 apples and 7 bananas3a + 7b 3(5¢) + 7(8¢) = 71¢ 5 bananas5b 5(8¢) = 40¢ half dozen of each.6a + 6b 6(5¢) + 6(8¢) = 78¢

Though we see equal signs in the statements above, they are not really equations. Yes, they're statements of equality but there are no variables (letters) in them.

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3) An algebraic expression for the Perimeter of a rectangle with base = 5 and height = h is
P = 2(5 + h). The Area is expressed as A = 5h

Formulas such as A = l × w for the area of a rectangle, are also statements of equality, but, they too are not equations because there's nothing to solve for. We use them as templates or stencils to evaluate certain measures.

Now get a pencil, an eraser and a note book, copy the questions,
do the practice exercise(s),then check your work with the solutions.
If you get stuck, review the examples in the lesson, then try again.

Practice Exercise 1: algebraic expressions

1) Write an algebraic expression for the Perimeter and Area of each figure. 2) Evaluate (find a number value for) each algebraic expression in #1 given that:
b = 7 inches, s = 4 feet, and h = 9 inches. Algebraic Equations: Solving it with Algebra

this is an equation An Equation is an algebraic statement of Equality
between knowns (constants) and unknowns (variables).

An algebraic equation behaves exactly the same way a seesaw does or balanced scales do, with the equal sign replacing the pivot point of the seesaw or scales. Whatever we do to one side of the equation we do to the other side in order to maintain equality or balance. Those are the operational words. We must maintain equality at all times -- why do you think we call it an EQUATION?? We would call it something else if it was something else -- but it's not.

In the twins question, the equation was J + J + 8 = 44
when we removed 8 marbles, we subtracted 8 from both sides of the statement
and got 2J = 36.
Then we divided by 2 or took ½ of 36 to get 18.

What we did was to perform the inverse operation for each one indicated in the equation. We eliminate the additional 8 marbles when we subtract (or remove) 8 from both sides of the equation and then, since we know twice J but want just one J, we divide both sides of the equation by 2.

Now let's solve a problem with an equation from the example about the apples and bananas.
Say we know an apple costs 5¢ but we don't know the price of a banana.
At the check out, we find that 3 apples and 7 bananas cost 71¢.
So, we let b represent the price of a banana. Our equation for this situation is:

3(5¢) + 7b = 71¢ which simplifies to 15 + 7b = 71¢

Now, we'll be brilliant like the twins and take away the 15¢ we pay for the 3 apples.
We get 15 + 7b – 15 = (71 – 15)¢ which becomes 7b = 56¢

This says that the price of 7 bananas is 56¢ so we know that each banana costs , which is one seventh of 56. Again, we solved this equation with 2 operations -- first subtraction and then division because the operations shown in it were addition and multiplication.

When we solve an equation, our job is to find the number value of the unknown or variable. In this example, the variable was b, the price of a banana. For the twins, it was J, the number of John's marbles. This diagram depicts the marbles problem. Here's the apples and bananas picture. If we see subtraction in the equation, we will add. If we see division, we will multiply. The only rule is no prejudice!! What we do to one side, we do to the other -- ALWAYS!!

Examples

1) Solve this equation: x – 7 = 33
We see subtraction, so we will add 7 to both sides to get: (x – 7) + 7 = 33 + 7. So x = 40.
To check our work, we verify if the original statement is true. 40 – 7 does = 33 so we're done.

2) Solve this equation: 3x – 4 = 29
We see subtraction, so we will add 4 to both sides to get: (3x – 4) + 4 = 29 + 4
Which, when we add + 4 and – 4 to get 0 on the left gives us: 3x = 33
Now we see multiplication, so we divide both sides by 3 to get x = 11.
When we check our answer in the original statement we see it is true. 3(11) – 4 does = 29.

3) Solve this equation: Here we see addition and division (remember, fractions mean division), so we'll first subtract 9 from both sides and then we will multiply by 5 to find x. We'll get: This says that when we divide x by 5, we get 6, so x must be 5 × 6 or 30. Now get a pencil, an eraser and a note book, copy the questions,
do the practice exercise(s),then check your work with the solutions.
If you get stuck, review the examples in the lesson, then try again.

Practice Exercise 2: algebraic equations

1) Solve these equations by addition and division:
 a) 5x – 19 = 31 b) 2x – 10 = 12 c) 7x – 1 = 139 d) 4x – 13 = 35

2) Solve these equations by subtraction and division:
 a) 5x + 19 = 74 b) 2x + 10 = 48 c) 7x + 17 = 73 d) 4x + 13 = 45

3) Solve these equations by addition or subtraction and multiplication:
 a) b) c) d) hint: In d) you need addition, multiplication and division, in that order.

4) Harry wants to by new shoes that cost \$78. He states that if he had \$6 more,
he would have half the price of the shoes.
a) write an equation that states in algebra what the question states in words.
b) Solve the equation to find how much money Harry has right now? .

Solutions

Practice Exercise 1: algebraic expressions

1) Write an algebraic expression for the Perimeter and Area of each figure. Practice Exercise 2: algebraic equations

1) Solve these equations by addition and division:
 a) 5x – 19 = 31 5x – 19 + 19 = 31 + 19 5x = 50 so x = 10 add 19 divide by 5 b) 2x – 10 = 12 2x – 10 + 10 = 12 + 10 2x = 22 so x = 11 add 10 to both sides divide both sides by 2. c) 7x – 1 = 139 7x – 1 + 1 = 139 + 1 7x = 140 so x = 20 add 1 divide by 7 d) 4x – 13 = 35 4x – 13 + 13 = 35 + 13 4x = 48 so x = 12 add 13 to both sides divide both sides by 4.

2) Solve these equations by subtraction and division:
 a) 5x + 19 = 74 5x + 19 – 19 = 74 – 19 5x = 55 so x = 11 subtract 19 divide by 5. b) 2x + 10 = 48 2x + 10 – 10 = 48 – 10 2x = 38 so x = 19 subtract 10 divide by 2. c) 7x + 17 = 73 7x + 17 – 17 = 73 – 17 7x = 56 so x = 8 subtract 17 divide by 7. d) 4x + 13 = 45 4x + 13 – 13 = 45 – 13 4x = 32 so x = 8 subtract 13 divide by 4.

3) Solve these equations by addition or subtraction and multiplication:
 a)   so x = 18 add 1 to both sides multiply both sides by 3 b)   so x = 42 subtract 3multiply by 7 c)   so x = 54 add 4 to both sidesmultiply both sides by 6 d)   so 2x = 18 and x = 9 add 1 to both sidesmultiply by 3divide by 2.

4)

a) Let a = the amount of money Harry has, so the equation is:

a + 6 = ½ (78)

b) How much money does Harry have right now?
Since ½(78) = 39, our equation is: a + 6 = 39 -- so we subtract 6 from both sides
to find that Harry has 39 – 6 or \$33 right now. (all content © MathRoom Learning Service; 2004 - ).