WebQuest
Beginning Algebra
Process
Day 1:
In algebra, a simple function will be given with a missing number. To replace this number we use a variable such as x. With the given equation it is the students job to solve the equation for x, meaning what number would replace x.
For Example:
We are given the equation x - 4 = 2. We want to know what number x represents. To solve for that number the easiest way is to put both numbers on one side of the equation. We add 4 to both sides to cancel the one out on the x side. So, x - 4 (+4) = 2 (+4) becomes x = 6. Therefore, the number that replaces x is 6 and we can check our work by ensuring that 6 - 4 does in fact equal 2.
Quiz 1:
1. x - 5 = 3
2. x + 4 = 6
3. x - 12 = 5
4. x + 16 = 20
5. x + 5 = 13
Day 2:
Once the concept of solving simple addition and subtraction equations for x we can begin to add a number in front of x. This number simply means that you simply multiply x by that number. For example, 2x is the same as 2 times x. To solve algebraic equations with these factors we must first move any addition or subtraction numbers to one side and then deal with the x. To get rid of the number that is multiplying by x, we simply divide both sides by that factor.
For Example:
We are given the equation 2x - 5 = 1. Remember to solve for x we must get rid of the 5 on the x side by adding it to both sides. By doing this, we get 2x - 5 (+5) = 1 (+5) which is 2x = 6. To get the x by its self we divide both sides by 2 to get 2x(/2) = 6 (/2) which gives us the answer x = 3.
Quiz 2:
1. 2x - 4 = 8
2. 3x + 5 = 14
3. 5x - 6 = 14
4. 4x + 8 = 12
5. 2x + 5 = 11
Day 3:
Now that we understand how to solve for equations that have a number multiplying by x, we can do the same process for numbers dividing by x. However, instead we will multiply by the number x is dividing by to cancel the numbers and solve for x.
For Example:
We are given the equation (x/2) + 5 = 9. We must first cancel the 5 by subtracting it from both sides and get (x/2) + 5 (-5) = 9 (-5) to get x/2 = 4. We will now multiply both sides by 2 to get x by its self to get (x/2) (x2) = 4(x2) to get x = 8.
Quiz 3:
1. (x/2) - 4 = 3
2. (x/4) + 5 = 6
3. (x/3) - 2 = 2
4. (x/2) + 6 = 10
5. (x/5) + 6 = 8
Day 4:
Once solving equations has completely been mastered, we can now use these steps to solve for functions. The first function we will solve for is y = mx + b where m and b are both constants, or numbers. The same technique used in the previous lessons is used, however, instead of getting x by its self on one side we only want y by its self.
For example if we are given the function 2x + 3 = y - 4, we can get y by adding 4 to both sides. Now when we add 4 to the side with x, we add what we call "like terms" that is anything that looks the same. For instance we can not add 4 to the 2x because the 4 does not have the x in front of it. Therefore solving the equation would go as follows:
2x + 3 (+4) = y - 4 (+4) to get 2x + 7 = y. Are answer is then y = 2x + 7.
Quiz 4:
1. 2x + 5 = y - 4
2. 3x + 3 = y - 4x - 4
3. 2x + 1 = 4 - 3x + y
4. 5x = y - 6 + 2x
5. 6 = y + 3x - 2
Day 5:
The final step would be just having a number divided or multiplied by y. To solve for y in this case we do the same process as we did with the x. If y is multiplied by a number you divide both sides, if it is divided by a number you multiply both sides. Note: the entire equation must be multiplied or divided by the factor of y.
For example, if we're given 4x + 5 = 2y - 1 we first cancel the 1 from both sides to get y by its self. Doing this we get 4x + 6 = 2y. Now we divide the entire equation by 2 to get y alone. Doing this we get (4x + 6) /2 = 2y/2. and get 2x + 3 = y. We get this because 4x/2 = 2x and 6/2 = 3. Your answer is then y = 2x + 3
Quiz 5:
1. 2x + 3 = 2y - 1
2. 3x + 1 = y/3
3. 4x - 5 = 2y - 9
4. 3x - 1 = 3y + 4
5. 10x + 5 = 5y - 5
Attachments
Solving For X
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