2. Algebra Flashcards
rule of opposites
addition gets rid of subtraction
subtraction gets rid of addition
multiplication gets rid of division
division gets rid of multiplication
the “Denominator Trick”
fractions
To simplify an equation that contains a fraction, multiply the equation by the denominator of that fraction!
Upon distribution, the denominator of your fraction will cancel out!
When multiple den -> use the LCD (Lowest Common Denominator)
Equations With Multiple Variables: Isolating Variables
The key to determining which variable to solve for lies in the phrasing of the question
Always solve for the variable that the question asks about or that the question tells you to solve for!
The phrase “Solve for x in terms of y and z” tells you to solve for x .
The phrase “In terms of x and y, what is z?” asks about z
Evaluation : Plug ‘n Chug
Any question that supplies a question and a value that can be plugged into that question is commonly known an evaluation.
» plug the supplied value into the question!
In many cases, evaluations supply an equation instead of a value. To solve such problems, simplify solve the equation and plug the resulting value into the question.
Never multiply or divide an inequality by a variable because
Unless you know whether that variable represents a positive or negative number, you do not know whether to flip the sign!
Which method is used to combine inequalities?
Stacking method
What do you need to be able to stack/combine inequalities?
At least one common term
What do you do if your inequalities do not contain a common term
You can give them one by multiplying or dividing the inequalities by any factors necessary
What do you do if your inequalities do not contain a common term
You can give them one by multiplying or dividing the inequalities by any factors necessary
Absolute value (definition)
non-negative value of a number or expression within
two vertical bars
Is the absolute value of any number positive?
No, since |0| = 0, and 0 is not a positive number
How many solutions do equations with absolute value (AV) brackets have?
two solutions, since the value of any term or expression inside AV brackets can be positive or negative (or zero).
How to solve equations that involve absolute value
1) isolate the AV brackets algebraically.
2) rewrite the equation as:
|Expression| = Answer ⇒ Expression = ± Answer
3) solve
Are both AV solutions always valid?
No.
Depends if match equation limitations
How to solve inequalities that involve absolute valu?
In almost exactly the same manner as equations that involve absolute value.
Difference: flip the sign of the negative equation
|x – 2| ≤ 8
x – 2 ≤ ±8
EQUATION #1
x – 2 ≤ 8
x ≤ 10
EQUATION #2
x – 2 ≥ –8 (Flip the Sign!)
x ≥ –6
How to translate graphs?
Use the midpoint
How to translate graphs?
Use the midpoint (represents the middle point of a shaded line segment)
|x – (Midpoint)| ≤ Distance From Endpoints to Midpoint
‘What is the algebraic expression of the number line below?’
According to the graph above, the line segment extends from –3 to 5 and has a midpoint of 1, since the average of –3 and 5 is 1:
(−3)+5 / 2 = 1
Likewise, the distance from the endpoints to the midpoint is ≤ 4, since the distance of every point from –3 to 1 is 4 spaces or less from the midpoint, as is the distance of every point from 5 to 1.
Thus, the algebraic expression for shaded portion of the number line above is |x – 1| ≤ 4, since the graph has a midpoint of 1 and the distance of every point from the endpoints to the midpoint is 4 or less.
How to solve Absolute Value on Both Sides of an Equation (Advanced)
Consider one scenario in which both brackets are positive and one scenario in which the first AV bracket is positive and the second AV bracket is negative.
‘If |4x + 7| = |2x – 1|, what is the value of x?’
SCENARIO #1: +/+
4x+7=2x–1
2x = –8
x = –4
SCENARIO #2: +/- 4x + 7 = -(2x-1) 4x + 7 = -2x + 1 6x = -6 x = -1
! one positive and one
negative EQ
How to solve Absolute Value on Both Sides of an Equation (Advanced)
Consider one scenario in which both brackets are positive and one scenario in which the first AV bracket is positive and the second AV bracket is negative.
‘If |4x + 7| = |2x – 1|, what is the value of x?’
SCENARIO #1: +/+
4x+7=2x–1
2x = –8
x = –4
SCENARIO #2: +/- 4x + 7 = -(2x-1) 4x + 7 = -2x + 1 6x = -6 x = -1
! one positive and one
negative EQ
How to solve Absolute Value on Both Sides of an INequality (Advanced)
just like equations.
Just be sure to flip the sign of the negative inequality!
How to solve multiple AV Brackets (Advanced)?
Test numbers.
Most difficult AV questions, since involves testing positivity ad negativity (difficult to solve algebraically)
Chart and test 4 scenarios (if 2 variables)
1) Positive/Positive
2) Positive/Negative
3) Negative/Negative
4) Negative/Positive
What is particular about equations with Roots & Exponents compared to normal EQ?
Normal EQ = 1 solution
Roots / exp. = 2 possible solutions
Exponents and roots are opposite operations
They cancel each other out
How to solve for a variable underneath a radical
(1 Isolate the radical.
(2 Multiply both sides of the equation by the corresponding exponent.
3) Cancel out
How to solve for a variable that has an exponent
1) Isolate the variable with the exponent.
2) Take the corresponding root of both sides of the equation.
3) Cancel out
Even exponents ; absolute value
Taking the root of a term with an even exponent always results in the absolute value of that term!
What needs to be considered when calculating quadratic inequalities?
Whether the factors have a positive or negative product.
» look at inEQ sign
+ aka. > 0
++ : (x+a)(x+b) > 0
– : (x-a)(x-b) > 0
- aka. < 0
+- : (x+a) < 0
-+ : (x-a)(x+b) < 0
> > always consider both scenarios
When is the discriminant used?
To tell you the number of solutions that a quadratic has:
1) Discriminant > 0 = 2 solutions
2) Discriminant = 0 = 1 solution
3) Discriminant < 0 = no solution
Discriminant (formula)
b^2 - 4ac
Discriminant (formula)
b^2 - 4ac
Quadratic formula
x = –b ± (square root b^2 –4ac) / 2a
When do we use the Quadratic Formula?
If a quadratic equation cannot be solved through factoring
Extremely rare
If the factors of a problem are already set equal to 0, FOIL or not?
No.
How are quadratic equations solved?
By factoring.
(1) Set the equation equal to zero.
(2) Make sure the x2 term in positive.
(3) Set up brackets (x )(x ).
(4) Choose two numbers that multiply to the last term and add to the co-efficient o f the middle term.
How are quadratic equations solved?
By factoring.
(1) Set the equation equal to zero.
(2) Make sure the x2 term in positive.
(3) Set up brackets (x )(x ).
(4) Choose two numbers that multiply to the last term and add to the co-efficient o f the middle term.
More Variables than Equations
The oft-taught maxim that the same number of variables and equations are needed to solve for one (or more) of the variables is a myth!
having an equal number of equations and variables does not guarantee that you can solve for a variable. If two equations provide the same information, it is impossible to solve for either variable!
Combining Inequalities & Equations
If given a question that contains an equation and an inequality, always substitute the equation into the inequality!
(1) Isolate the variable in common to both systems.
(2) Substitute the equation into the inequality.
Adding inEQ (adv)
As long as their inEQ signs point in the same direction
> > elimination if needed
What to do if inEQ can’t be added?
Combine them by giving a common term
> > multiply / divide either inEQ or both by any factors necessary
Minimisation/Maximisation problem (def)
Any problem that asks you to identify the greatest or smallest value for a particular variable or equation
2 types:
1) range provided
2) determine inputs
2 types of minimisation / maximisation
1 ) provides a range of possible inputs
» need to determine greatest/smallest output for a certain expression given those inputs
2) asks to determine inputs
» that will lead to the greatest/smallest output for a particular expression
Min / max
Range of possible inputs provided:
1) Identify the greatest and smallest values that are possible for each input.
2) Test out every combination of those inputs.
Min / max
Determine the inputs
Understand the relationship between the input and the output
x^2y – xy^2
xy(x – y)
3^5 +3^5 +3^5
3^5(1+1+1)
= 3^5 (3)
= 3^5 (3^1)
= 3^6
x^3 – 2x^2 +x
x (x^2 – 2x + 1)
= x(x–1)(x–1)
15^x + 15^(x+1)
15^x + 15^x(15^1)
= 15x(1 + 15^1)
= 15x(16)
9 – x
–1(–9 + x)
= –1(x – 9)
= –(x – 9)
2^x +2^x +2^x +2^x
2^x (1+1+1+1)
=2^x(4)
=2^x(2^2)
=2^(x+2)
2^29 – 2^28
2^28(21 – 1)
= 2^28(1)
= 2^28
63(36) + 64(63)
63(36 + 64)
= 63(100)
= 6,300
8(4)^12 – 7(4)^11
4^11(8 * 4 – 7 * 1)
= 4^11 (32 – 7)
= 4^11 (25)
ac + bc + ad + bd
c(a + b) + d(a + b) = (a + b )(c + d)
3^x − 3^(x-1)
3^x - (3^x / 3^1)
= 3^x (1-⅓)
= 3^x (⅔)
q^3 –q
q(q^2 –1)
= q(q–1)(q+1)
2^6 +2^7 +2^8
2^6(1+2^1 +2^2)
=2^6 (1+2+4)
=2^6(7)
