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Lesson 2B: Linear Equations cont...
Lesson Outline:
1. Linear equations
2. Quadratic equations
a. Background
b. FOIL
3. Helpful tips
Plugging in
Coming up with your own equations
LINEAR EQUATIONS
Problem:
Which of the following ordered pairs graph onto the equation
y = ½x  6 ?
A) (6, 0)
B) (3,1/2)
C) (8,2)
D) (2,1)
E) (1,5/2)
Ordered pairs are of the form (x,y) so all you have to remember is that x comes before y, like in the alphabet. Then try each of the ordered pairs in the linear equation to see which one holds true.
 First try A) (6, 0). You plug 6 into the equation for x
y = ½(6) 6
= 3 6
=  3
Y = 3, not 0 so A) is not the answer
y = ½(3) 6
= 1.5  6
= 4.5
y = 4.5, not ½ (or .5) so B) is not the answer
y = ½(8) 6
= 4  6
= 2
Y = 2, which matches, so the answer is C)
QUADRATIC EQUATIONS
Another type of equation you may work with is the quadratic equation
It has a standard form of Ax 2 + Bx + C = 0
For example,
x 2 + 5x + 6 =0
(For our purposes, the ‘A’ in front of the x 2 term will always equal 1.)
On problems with a quadratic equation, you may need to find solutions by factoring.
When you learned that 3 and 5 were the factors of 15, you first needed to know that 3 and 5 multiplied out to 15. Similarly we must look at what multiplies out to form a quadratic before we can understand the factoring process.
 In this case, the factors happen to be
(x + 3) and (x +2)
 Multiplying them together uses a method often called “FOIL”, which stands for:
F irst terms
O utside terms
I nside terms
L ast terms
F + O + I + L
When we multiply (x + 3) and (x +2)
 We start with the first terms, the two x’s
( x + 3) ( x +2) which gives us
x2 + __ + __ + __
 then the outside terms x and 2
( x + 3)(x + 2 ) which gives us
x2 + 2x + __ + __
 then the inside terms 3 and x
(x + 3 )( x +2) which gives us
x2 + 2x + 3x + __
 then the last terms, 3 and 2
(x + 3 )(x + 2 ) which gives us
x2 + 2x + 3x + 6
 Finally, notice that the two middle terms in our result, 2x and 3x, are like terms. That is, they have the same variable, x. This means they can be combined by adding the numbers in front of them (called the “coefficients") like so:
x2 + ( 2x + 3x ) + 6 =
x2 + 5x + 6
And we have the original quadratic equation we started with in standard form.
 Now, to factor it, we do the reverse of what we just did.
x 2 + 5x + 6 = 0
 Let us first set up how it will look when we find the factors
(x + ?)(x +?) = 0
 For factoring any simple quadratic equation we look at the coefficients in the 2nd and 3rd term
x 2 + 5 x + 6 = 0
The coefficients are 5 and 6. We then ask “what two numbers add up to 5 and multiply out to six?” The answer, as you may know, are the numbers 2 and 3
So in the factor form, we have (x + 3)(x +2) = 0
 Now, finally, we are ready to solve
Notice that the equation is set equal to zero.
Let us, for the moment, restate it as
A* B = 0 , where A is (x + 3) and B is(x +2)
In such a situation the equation is true if either A or B is equal to zero (because zero times any number or quantity equals zero.)
 So, the equation is true if either A or B equals zero.
A = 0 or B = 0
 This can be worked out by solving for x in each expression.
(x + 3) = 0 or (x +2) = 0
x + 3 = 0 or x + 2 = 0
x = 3 or x = 2
We have found the solutions for the quadratic equation through factoring.
Problem:
What is the solution to x 2 5x 14 = 0 ?
A) 5 or 14
B) 5 or 15
C) 2 or 5
D) 2 or 7
E) 2 or 7
 To solve this problem, first we must factor the equation
x 2 5x 14
 Remember how to set up for factoring a quadratic equation
(x + ?)(x + ?) = 0
 Look at the coefficients of the 2nd and 3rd terms
x 2 5x  14
The numbers are  5 and  14
 Then, find two numbers that add up to the first number and multiply out to the second. In this case, the factors are 2 and 7
Now that you have your two factor numbers, you can put them into factor form
(x + 2 )(x + 7 ) = 0
 The answers for x that make the equation true are then 2 and 7 because
When x = 2 (2 + 2)(2 + 7) = 0
(0)(9) = 0
0 = 0
 And when x = 7 (7 + 2)(7 + 7) = 0
(9)(0) = 0
0 = 0
So the answer is choice D) 2 or 7.
Helpful tips:
On some multiplechoice questions it may be faster to simply plug the answer values into the problem to find the right answer.
In a previous problem you could choose to do that instead of factoring.
The problem was:
What is the solution to: x 25x 14 = 0 ?
F) 5 or 14
G) 5 or 15
H) 2 or 5
I) 2 or 7
J) 2 or 7
Instead of factoring the quadratic equation you could simply start plugging the numbers from the answer choices in for x until you get two that work. It depends on which method is faster for you and can also depend on the problem.
Algebra is a useful tool for solving a variety of problems
There are times when you may have to come up with your own algebra equations to solve a problem
Problem:
A bottle of 250 vitamin tablets sells for $3.75. At this rate, what should a bottle of 40 tablets cost? (Choose the answer, marked correctly, from the standard grids.)
Recognize that you have a price that goes with a certain amount ($3.75 and 250) and you have another amount and a price that goes with it that follows the same pricing rate. You can then set up an equation.
$3.75 $x
 = 
250 pills 40 pills
Crossmultiply to get
250 * x = 30 * 3.75
250 * x = 30 * 3.75
Then isolate x
250 * x 30 * 3.75
 = 
250 250
40 * 3.75 112.5
x =  =  = 0.6
250 250
The answer is .6
You then enter your answer at the top of the standard grid for this problem and bubble in the corresponding values.
Click on the link below to move on to lesson 3A.
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