Q2) Linear, quadratic and other equations Flashcards
What are linear equations vs what are not linear equations?
y = 4x + 1
xy - 2x = 3y
5x - 6y = 30
Characteristics of linear equations:
> variables are raised only to 1 or -1 AND
> variables are NOT multiplied together
Essentially any equation that can be REARRANGED into y = mx + b or x = #
> constants and coefficients can be integers or fractions
**What is the general rule for solving linear equations with unknowns?
SOMETIMES if you have n number of variables, you need at least n number of different equations (for LINEAR equations)
> Be ware of trap: two equations that are the SAME are NOT sufficient
HOWEVER, there are EXCEPTIONS:
EXCEPTION 1) if you are presented with another CONSTRAINT/RESTRICTIONS on variables (e.g., variables must be POSTIIVE INTEGERS, think about REALISTIC APPLICATIONS like # of balls must be NON NEGATIVE integers, >=0)
> one equation, two variables = SUFFICIENT
e.g., 15a + 29b = 440
e.g., 23a + 21b = 130
> usually unknowns are QUANTITIES (not prices)
> Typically also a good sign when you see multiples of 5 and/or 10 as coefficients (units digit must be 0 or 5)
Strategy:
> get rid of decimals by multiplying both sides of equation by 10s
> First try isolate one of the variables (x) and see what must the other variable (y) be in order for x to be an integer
> OR test / write out the multiples of each term and see if more than one valid combo works —> rearrange so that you can find the easiest way to test!
e.g., 5x + 8y = 55 —> 5 and 55 share factors, so isolate for y
y = (5*(11-x))/8 —-> 11-x must be divisible by 8, so x = 3
——> set 11-x equal to multiples of the DIVISOR (8)
11-x = 8(1) –> x = 3
11-x = 8(2) –> x = negative
Therefore x = 3 is the ONLY solution
EXCEPTION 2) Asked to solve for a COMBO of variables (as opposed to the value of a single variable)
> see whether it is possible to ISOLATE the expression in question
> might be given 4 variables, but not enough unique equations –> SIGN to REARRANGE for the combo
EXCEPTION 3) One equation may be sufficient to determine unique values for two variables
> occurs when one of the variables cancels out
ALSO remember that a PRODUCT of variables is NOT a linear equation (could result in quadratic equation)
Tips for solving linear equations with fractions
e.g., 5/n + 1/2 = 2/5
Get rid of the fractions at the start by multiplying every term on both sides by the LCM of the DENOMINATORS (including n since it is in the denom)
> generally makes the equation easier to solve
e.g., multiply both sides by n (what you do the one side you must do to the other side)
**The product of integers = 1, what does this mean about the integers?
Integers must be mix of 1 and -1 only (|1|)
> CANNOT be equal to 0
> CANNOT BE DECIMALS
> CANNOT be non +/- 1 integers
+/- 1 is unique for PRODUCTS
Beware of:
> DS questions asking about knowing what the value of unknown integers are (could be +1 or -1)
> Could show up as product of unknowns (e.g., y^2*(1 + x) = 1)
Tips:
> To determine whether unknowns are +1 or -1, start by understanding the SIGNS of the unknowns
e.g., x^2 * y = 1 —> x^2 is always positive, so y must = 1
The product of integers =/ 0, what does this mean about the integers?
NONE of the integers can be 0
0 is unique for PRODUCTS (if product of integers = 0, it means that one OR both of the factors must equal 0)
What characteristics do quadratic equations have?
The highest power (exponent) of a variable (unknown) is 2
General form is ax^2 + bx + c
How do you factor a quadratic equation?
Step 1) Make sure the equation is in the GENERAL form: ax^2 + bx + c = 0
Step 2) Need to think of two numbers that can ADD to “b” and MULTIPLY to “ac”
> list out factors of “a” and list out factors of “c”
> figure out the criss cross multiplication that works (equals “b”)
Step 3) Present factors as multiplication (e.g., (x+#)*(x+#) = 0
RECALL that to get from factored form back into general form, we use FOIL method OR use short cut by memorizing common quadratic “identities”
What are ROOTS of a quadratic equation?
Solutions of the quadratic equation that EACH make the equation = 0 (solutions)
> if you SUBSTITUTE each root into the equation FOR X AND set the equation equal to 0 —–> can use this information to solve for unknown coefficients
> there is a MAX TWO ROOTS and MIN ZERO ROOTS
ALTERNATIVE question might tell you that “x + #” is a FACTOR of a quadratic expression and ask you to solve for the value of a constant (BE WARE OF SIGNS)
> simply sub in the ROOT based on the FACTOR and solve the quadratic equation when set = 0
3 common quadratic identities (saves you time FOILing or factoring)
Difference of squares: x^2 - y^2 = (x + y)(x - y)
> to help spot the difference of squares, look for the SQUARE of a value minus the SQUARE of another value (fractions, integers, products of variables, EVEN exponents are all FAIR GAME)
> recognize square roots that can be “x”
> x^2 - 1 = (x+1)(x-1)
> x^2y^2 - 16 = (xy - 4)(xy + 4)
> (1/36)x^2 - 25
> x^30 - y^30 = (x^15 + y^15)*(x^15 - y^15)
(x+y)^2 = x^2 + 2xy + y^2
> recognize square roots that can be “x”
(x-y)^2 = x^2 - 2xy + y^2
> recognize square roots that can be “x”
HIDDEN quadratic equations - where do they arise?
e.g., What are the possible values of x?
60/(x+2) + 1 = 60/x
e.g., ab = 55 and a - b = -6
Be ware especially in DATA SUFFICENCY questions (don’t fall for C trap if asked for the single value of a variable and there’s a hidden quadratic equation that could yield TWO solutions = NOT SUFFICIENT)
Case #1) Arise after removing fractions
Upon simplifying the fractions in the equation by multiplying the equation by the LCM (incl. variables in denom), quadratic equations can result
> hidden quadratics often found in FRACTIONS
Case #2) Arise when given two unknowns and two equations (one is a PRODUCT)
Simple math reminders when simplifying equations
y/x = 4/x - 6y
Fractions –> simplify by multiplying BOTH SIDES of the equation by the LCM of the denominators of fractions (incl. variables)
> do NOT need to multiple top and bottom since you are dealing with an EQUATION (not just an expression)
> if simplifying an expression with fractions, multiply top and bottom of fraction by LCM of mini fractions
Whole point is to NOT change the VALUE of the equation or expression
> if you multiply both sides of an equation by the same thing, it is the same as *1
Solving for combos
Involves questions that ask you to solve for the value of a COMBINATION of VARIABLES (expressions) as opposed to single variables
> you DO NOT need to know the individual values of variables
> rearrange smartly e.g., to find x+y
> questions often scare you by containing a large number of variables
—–> don’t fall for equation trap: part of an equation can be substituted into another equation
When would an equation have 2+ solution? How do you solve them?
Generally equations that involve higher order powers, like x^3 or x^4
> cubic equations
> quartic equations
OR you may also face situations with multiple variables and asked to simplify (e.g., (a-b+1)/(ab-b+(1-a^2)))
Recall a SOLUTION is still a value for the unknown that makes the equation = 0
SOLVE:
>by FACTORING out GCF from every term and solving with one side = 0
> OR solve by FACTORING by grouping (groups of terms share GCF but not all, make sure one side = 0) —> after factoring by grouping, a common factor emerges
If x(x+1) = 4(x+1), what is the value of x?
Be ware of the TRAP answer - do NOT divide by an unknown variable OR expression UNLESS you know that it’s not equal to zero
> applies to both variables and unknown expressions e.g., x+1
Instead, solve by MOVING terms to one side and factoring when set to 0
What should you think of when you see DS with these two equations in the question stem:
Q: What is the value of b?
(1) ab = 55
(2) a - b = -6
CONCEPT: Hidden quadratic equation that COULD yield TWO answers
> clue: PRODUCT of VARIABLES
When you look at both equations TOGETHER –> solve for roots of the quadratic equation
