Sample Problem: Solving Algebraic Equations

Example:

• a + 1 = 3
• a = 3 - 1
• If you transpose a certain value to the other side of the equation, it "becomes the opposite of the original," so plus becomes minus, minus becomes plus, multiply becomes divide and vice versa.
• a = 2
• 2 + 1 = 3
• 3 = 3
• a = 2

Sample Problems:

• Solve for a:
• 5 + a = 7
• a - 13 = 4
• 3a  = 9
• a/2 = 10
• Solve for b:
•  9 + b = 15
• b - 2 = 1
• 2b = 2
• b/3 = 4
• Solve for c:
• 6 + c = 13
• c - 3 = 2
• 2c = 20
• 6/c = 3
• Solve for x: 
• 2 + x = 24
• x - 3 = 20
• 8x = 16
• x/7 = 5
• Solve for y:
• 8 + y = 28
• y - 4 = 12
• 3y = 15
• y/8 = 2

Sample Problem: Substituting Variables

Example Problem:

• What is the value of a + 1 if:
• a = 1
• a = 2
• a = 3
• a = 4
• a = 5

Answer to Example Problem:

• a + 1 =
• 2
• 3
• 4
• 5
• 6

Sample Problems:

• What is j + 7 if...
• j = 5
• j = 9
• j = 6
• What is u + 5 if...
• u = 4
• u = 2
• u = 8
• What is d + 9 if...
• d = 5
• d = 2
• d = 11
• What is b + 6 if...
• b = 1
• b = 4
• b = 3
• What is x + 8 if...
• x = 1
• x = 2
• x = 3

Simple Words to Magical Algebra

In case most of you are wondering what to do when your Algebra teacher gives you a quiz and decides to dictate the questions in sentences... well, don't fret, LAMEFHM is here to help! With this post, we'll show you how to turn simple words to algebraic expressions and/or equations and solve problems!

Don't worry, your teacher is not crazy by dictating stuff instead of writing the number problems on the board during your quiz on Algebra. That's normal. And there's a way to solve it! Mathematical word problems are most commonly solved by translating them into expressions and equations or, as I like to call them, "algebraic sentences."

Here's an example for you:

• Your teacher says... "I bought a coffee maker for $50 after they deducted $20 from its original price because it was on sale and had a discount. How much was the original price of the coffee maker I bought?" (Credits to TutoringMaths for the cool image and to Basic-Mathematics for their awesome word problems!)

To solve this, you have to write it as an algebraic sentence (meaning write it as an expression or equation). To do that, you have to:

1. Distinguish the quantity missing or unknown. Know what you're trying to find. (This is usually what your teacher asks you to find.)
2. Get variables to represent the quantity you're looking for. (Usually, the most common variables used are a, b, c, x, y, and z.)
3. Find out what kind of operations you have to use in solving the given problem.

Following the steps mentioned lets you know that:

1. The original price of the coffee maker my teacher bought is unknown. I must try to find out how much that coffee maker originally cost!
2. I'll get the letter x (since the letter x is so cool) and use it as the variable to my equation (you'll be using an equation instead of an expression with this problem... I'll explain why later) and name the unknown quantity x, making x = original coffee maker price.
3. I'll be using subtraction with this problem ... and let me show you why subtraction is the operation I'll be using... later.

Ready to know the answer? Let's solve!

• x = original price
• x - 20 = 50
• We subtracted 20 from x since x is our original price, and the teacher said he bought his coffee maker with a $20-discount for $50. Subtract $20 from the original price and we get the amount the teacher paid for his coffee maker which was $50.
• x = 50 + 20
• Since $50 is the price of the coffee maker once the $20-discount has already been deducted from its original price, we therefore transpose 20 (since we only need x to be the only one remaining on one particular side of the equation in order to come up with its value) to the other side of the equation (and since we transposed it to the other side, its sign changes, hence the plus sign instead of the minus) and add 20 to 50 to come up with the original price of the product before the $20-discount has already been deducted.
• x = 70
• Here we now have the answer, which is $70 as the coffee maker's original price before its $20-discount has already been deducted.

We here at LAMEFHM hope our posts help--especially this one, since this one was originally created to help--you... so, if you've got questions or more ideas/suggestions/opinions/comments/messages regarding how you think we can be more helpful to you and your mathematical needs... just leave a comment and we'll answer as soon as we possibly can! Thanks! :)

Properties of Operations and Identities

• Commutative Property of Addition
• a + b = b + a
• Example:
• a + b = b + a
• a = 7
• b = 5
• (7) + (5) = (5) + (7)
• 12 = 12
• Commutative Property of Multiplication
• ab = ba
• Example:
• ab = ba
• a = 9
• b = 6
• 9 · 6 = 6 · 9
• 54 = 54
• Associative Property of Addition
• (a + b) + c = a + (b + c)
• Example:
• (a + b) + c = a + (b + c)
• a = 4
• b = 1
• c = 2
• (4 + 1) + 2 = 4 + (1 + 2)
• (5) + 2 = 4 + (3)
• 7 = 7
• Associative Property of Multiplication
• (ab)c = a(bc)
• Example:
• a = 9
• b = 8
• c = 5
• (9 · 8) · 5 = 9 · (8 · 5)
• (72) · 5 = 9 · (40)
• 360 = 360
• Distributive Property of Multiplication Over Addition
• a (b + c) = ab + ac
• Example:
• a (b + c) = ab + ac
• a = 12
• b = 7
• c = 5
• 12 · (7 + 5) = (12 · 7) + (12 · 5)
• 12 · (12) = (84) · (60)
• 144 = 144
• Additive Identity Property
• a + 0 = 0 + a = a
• Example:
• a + 0 = 0 + a = a
• a = 5
• (5) + 0 = 0 + (5) = 5
• 5 = 5 = 5
• Multiplicative Identity Property
• 1(a) = a(1) = a
• Example:
• 1(a) = a(1) = a
• a = 3
• 1(3) = 3(1) = 3
• 3 = 3 = 3
• Quotient Property
• a (1/b) = a/b
• Example:
• a (1/b) = a/b
• a = 1
• b = 2
• 1 (1/2) = 1/2
• ½ = ½
• Multiplicative Inverse Property
• a (1/a) = 1
• Example:
• a (1/a) = 1
• a = 3
• 3 (1/3) = 1
• 1 = 1
• Multiplication Property of Zero
• 0(a) = a(0) = 0
• Example:
• 0(a) = a(0) = 0
• a = 1
• 0(1) = 1(0) = 0
• 0 = 0 = 0