Ch. 2 Linear and Quadratic Equations Flashcards
What are the two methods to solve a system of linear equations of two variables?
- Substitution Method: solve for one variable & plug into other equation
- Combination Method: stack two equations on top of each other & add or subtract one entire equation from the other
4x + 3y = 12 - what are the 4 & 3 called in this equation?
Coefficients - the coefficient of x is 4 and the coefficient of y is 3.
If you’re using the combination method to solve a system of equations (stacking two equations & subtracting/ adding entire equations together), what do you do if the coefficients are all different?
You can find the LCM (least common multiple) of the coefficients and multiply the entire equations, then do combo method.
When to choose substitution or combination method?
Substitution if it’s easy to isolate the variables; combination if not.
How can you solve an equation that contains fractions?
Consider multiplying the entire equation by the least common multiple (LCM) of the denominators to remove the fractions. This generally makes the equation easier to solve.
If the product of two integers is 1, what are the two integers?
1 or -1.
Don’t forget the -1!*
Also, integers can only be whole numbers
Solve for x:
x(x+100) = 0
x = 0
x = -100
Don’t assume that x can’t be zero!!
Some students divide both sides by x and miss that x=0
Zero product property
When the product of two numbers is 0, one or both of those numbers is 0
What is a quadratic equation?
An equation where the highest power of a variable is 2
examples:
x^2 + 3x + 7 = 0
10x^2 = 100x
5c^2 + 10 = 10c - 20
How to factor quadratic equations
- Change to general form: ax^2 + bx + c = 0
- If it’s x^2 + bx + c = 0, then you can factor to (x + p)(x + q) = 0 where p + q = b and pq = c
- To find p & q, list all positive factors of c
- Figure out which two sum to b
In the quadratic equation, x^2 + 9x + 8 = 0, what are the factors & what are the solutions or roots?
x^2 + 9x + 8 = 0
(x + 8)(x + 1) = 0
x = -8, -1
(x + 8) and (x + 1) are the factors
-8 and -1 are the solutions or roots
What’s the difference between an identity and equation?
Equation is true for only one or some values of x
Identity is true for all or almost all values of x
For example, 2x - 3 = 5 is an equation; (x^2)/x = x is an identity.
(x + y) ^2 =
(x + y) ^2
(x + y)(x + y)
x^2 + 2xy + y^2
(x - y) ^2
(x - y) ^2
(x - y)(x - y)
x^2 - 2xy + y^2
(x + y)(x - y)
(x + y)(x - y)
x^2 - y^2
How to spot the difference of squares [(x - y)(x + y) = x^2 - y^2] in a question?
Look for two values subtracted that could be squares!
For example,
x^2 - 1 = (x + 1)(x - 1)
x^2 y^2 - 16 = (xy - 4)(xy + 4)
1/36 x^2 - 25 = (1/6 x - 5)(1/6 x + 5)
3^30 - 2^30 = (3^15)^2 - (2^15)^2 = (3^15 + 2^15)(3^15 - 2^15)
if x != y, then what is (x - y) / (y - x)?
-1
(if you factor out -1 from one of the expressions, it’s easy to see why this = -1)
What is greater:
(x + y)^2 or
(x^2 + y^2) ?
It depends.
(x + y)^2 = x^2 + 2xy + y^2
If xy is positive, then (x + y)^2 is bigger. If xy is negative, then (x^2 + y^2) is bigger.
If you have two equations, and two variables, can you solve the problem?
Not always!!
Sometimes, a trick question will have two equations that are actually just the same equation!
For example,
x + y = 2
10x + 10y = 20
You can’t solve this for x & y!
True or false: you need 3 unique equations to determine the values of 3 unique variables
False!
This is sometimes true, but sometimes we can determine the values in fewer equations!
This usually happens when there are restrictions provided in the problem and b/c of that, those variables can ONLY be certain numbers. For example, x > y and x & y are +ve integers, x + y = 3, what is x? it HAS to be 2 & y = 1; there’s no other number it can be!
What are some examples of restrictions placed on problems that would allow us to solve for two variables w/ one equation?
- x & y are +ve integers
- word problems where the answers have to be whole, +ve numbers (number of baseballs, number of cats, etc.)
When you have one equation, two variables, and it looks like it might be possible to solve for both variables, how do you do it?
Solve for one of the variables (in terms of the other) and then plug in numbers. You may need to look for common factors & factor out.
Try to aim for this format: y = k (j - x) / i where k, j & i are integers & x & y are variables you’re solving for.
For example, x & y are positive integers and 5x + 8y = 55.
8y = 55 - 5x
8y = 5(11 - x)
y = 5(11 - x) / 8
We know y HAS to be a positive integer, so either 5 or (11 - x) has to be divisible by 8; 5 is NOT divisible by 8, so we need to find x such that 11 - x is divisible by 8 (and positive). x can ONLY be = 3
If you have four terms that don’t seem to have a common factor, what can you do?
See if there is a common factor for TWO of the terms & factor that way!
When can you divide by a variable?
ONLY when you know the variable is not equal to 0
If you can’t divide by a variable, what can you do instead?
Subtract!
For example,
x^2 = 100x
x^2 - 100x = 0
x(x-100) = 0
x = 0, 100
If you divided by x, then you would miss the solution x = 0
When can you divide by a variable expression (i.e., x+4)?
ONLY when you know the variable expression is not equal to 0
i.e., x + 4 != 0
so x != -4
(a + b + c)^2 =
a^2 + b^2 + c^2 + 2ab + 2ac + 2bc
