When would you use Substitution to solve Simultaneous Equations?

When one variable can be easily expressed in terms of the other

When would you solve 2 simultaneous equations by Combination?

Whenever it's easy to manipulate the equations so that the coefficients for one variable are the SAME or OPPOSITE

If you plan to ADD 2 simultaneous equations to solve by combination, what should you do?

If you plan to SUBTRACT 2 simultaneous equations by combination, what should you do?

How would you solve 3 simultaneous equations with 3 variables in each?

Using substituion or combination, or both

What's the sum of x, y and z  ?:

x + y = 8

x + z = 11

y + z = 7 

x + y         = 8

x +   +  z = 11

+ y +  z = 7

--------------------------------

2x + 2y + 2z = 26

thus, x + y + z = 13 !

What is the rule for determining whether 2 equations involving 2 variables (say, x and y) will be sufficient to solve for the variables?

1) If both equations are LINEAR (i.e. no squared terms and no xy terms) - the equations are SUFFICIENT, UNLESS the two equations are MATHEMATICALLY IDENTICAL

2) If there are ANY non-linear terms in either of the equations (i.e. x³, xy or x/y), there will USUALLY be 2 more different solutions for each of the variables,and the equations will NOT be sufficient.

What should you do when you see a COMBO problem, i,e, asked to find the value of x + y ?

TRY to manipulate the given equation(s) so that the COMBO is isolated on one side of the equation. ONLY try to solve for the individual variables once you've exhausted all the other avenues.

What are the MADS manipulations in relation to solving most COMBO problems?

M

Multiply/Divide by the WHOLE equation by a single number

A

Add/Subtract a number on BOTH SIDES of the equation

D

Distribute or factor an expression on ONE side of the equation

S

Square/Unsquare both sides of the equation

To solve for a variable combo, what should you do?

Isolate the combo on one side of the equation

In DS problems, when you detect that it may involve a combo, you should try to manipulate the equation(s) in either the question or the statement so that the combo is isolated on one side of the equation. Then, how do you tell if the equation is SUFFICIENT? What about NOT SUFFICIENT?

Sufficient: The other side of an equation from a statement contains a VALUE.

NOT Sufficient: The other side of the equation contains a VARIABLE EXPRESSION

What are the 3 steps for solving ABSOLUTE VALUE EQUATIONS?

1. ISOLATE the Absolute Value expression

2. Once you have an equation of the form |x| = a and a>0 , you know that x = (+ - ) a... Remove the absolute value brackets and solve the RHS of the equation for 2 DIFFERENT CASES.

3. Check to see whether each solution is valid by putting each one back into the original equation and verify that the LHS = RHS of the equation. Sometimes one solution may fail!

Once we have an equation of the form |x| = a, and x>0, what do we know about x ?

x = (+-) a

What rule is essential to follow when solving ABSOLUTE VALUE EQUATIONS? 

To make sure to solve for BOTH cases.

Why are EVEN EXPONENTS dangerous?

Because they hide the sign of the base, and can have a POSITIVE and a NEGATIVE solution!

x² = 25

|x| = 5

What do these have in common?

What rule to explain this?

In both cases, x = (+ -)5

RULE: for any x, sqrt.(x) = |x|

Solve:

x² + 9 = 0

x² = -9

Therefore x has NO SOLUTION (squaring can NEVER product a negative number!)

How many solutions does an equation with an odd exponent have?

1 only

How would you solve problems that involve exponential expressions on BOTH sides of the equation ?

REWRITE the bases so that either the same base, or the same exponent, appears on both sides of the exponential equation.

THEN you can usually eliminate the bases or the exponents, writing what's left over as an equation...

0^x = 0^y

so, x=y...

True or false? Why?

FALSE

Because for example 0²=0⁵=0¹¹ etc.

So, we can't claim that x = y

1^a = 1^d

so, a=d... 

True or false? Why?

FALSE

Because for example 1²=1⁵=1¹¹ etc.

So, we can't claim that a = d

When are you allowed to divide by a variable, (or ANY expression) ?

When you are absolutely sure the variable or expression <> 0

Be careful not to assume that a quadratic equation always has _____ _____. Always _____ quadratic equations to determine their solutions. This will enable you to see whether a quadratic equation has ____ or ____ solutions.

x² - y² = ?

(x + y)(x - y)

x² - 2xy + y² = ?

(x - y)(x - y) = (x - y)²

a² - 1 = ?

(a - 1)(a + 1)

(a + b)² = ?

a² + 2ab + b²

Solve:

a² + b² = 9 + 2ab

Step 1. a² + b² - 2ab = 9

Step 2. (a - b)² = 9

Step 3. a - b = (+ - ) 3 ... (important step!)

(x + y)² = x² + y² ?

TRUE or FALSE?

FALSE

(x + y)^{2 =}  x² + 2xy + y² 

(x - y)² = x² - y² ?

TRUE/FALSE?

Always try to ____ a quadratic equation before solving

At first glance, is the following solveable? Why/Why not?

70A + 5B = 63

No - because there are 2 variables and only one equation.

" x percent" = ?

x/100

"y percent less than" = ?

1 - (y/100)

Describe the VIC solving method of Picking Numbers & Calculating a Target... When is this method useful? 

Involves:

1. Picking numbers for all or most of the unknowns in the problem

2. Using those numbers to calculate the ANSWER (i.e. the TARGET) to the problem

3. Plugging in each number you've picked into each answer choice to see which answer choice yields the same value as your target.

What are the rules for picking numbers in VICS?

1. NEVER pick 1 or 0, or 100 for % VICS

2. All numbers you pick must be DIFFERENT

3. Pick SMALL numbers

4. Try to pick PRIME numbers

5. Avoid picking numbers that are COEFFICIENTS in several answer choices

Describe the steps for solving a VICS problem using "Pick Numbers & Calculate a Target"

1. Pick numbers for each variable. Can be helpful to use a chart.

2. Answer the question, walking through the logic with the numbers that we've picked. This answer is the TARGET.

3. Test EACH answer choice, EVEN if you've already found one that equals your target value...

When using a number picking strategy for VICS, you can pick a value for every variable wheter there are explicit or implicit equations in the problem...

TRUE/FALSE?

FALSE

You can NEVER pick a value for EVERY variable when there are explicit or implicit equations in the problem! 

e.g. when the variables are related to each other through an equation

What must you do in a VIC problem, using the Pick Numbers and Calculate a target strategy, when you CANNOT pick a value for each variable?

Pick a value for ALL BUT ONE of the variables and then solve for the value of the remaining variable.

THEN, plug the numbers we've selected into the original expression to get the TARGET value, and TEST EACH ANSWER CHOICE.

When you are trying to figure out the algebraic manipulation method of solving a VIC, but get stuck, what should you do?

IMMEDIATELY switch to a number-picking strategy!

N.B. NEVER give up on a VIC problem before picking numbers... Sometimes very difficult VIC problems are easily solved with test numbers.

List 3 ways to solve an absolute value inequality

1. By shifting the midpoint - and re-compensating... i.e. the midpoint (x) here is -1, so you must add 1 to it to compensate.

2. find the centre of the range (the average of the endpoints) then use that to test the endpoints...

3. test the end-points in the answer choices. when they both produce an equal value, that is the correct answer.

ALWAYS check the solutions you get in the original euqation! Squaring both sides can actually introduce and extraneous solution.

Fill in the missing parts of the following equations:

Total Cost ($) = ?

Total Sales or Revenue = ?

Profit = ?

Unit Profit = ? or Sale Price = ?

Total Earnings ($) = ?

TotalCost($) = UnitPrice ($/unit) x Qty.Purchas'd (units)

Total Sales or Revenue = Unit Price x Qty. Sold

Profit = Revenue ($) - Cost ($)

Unit Profit = Sale Price - Unit Cost

Sale Price = Unit Cost + Markup

Total Earnings ($) = Wage Rate ($ per hr) x Hrs worked

Explain why this is true:

"For any x, √x² = |x|"

"For any x, √x² = |x|" is true because:

For example, x² = 25 and |x| = 5 share the same solution for x... Namely, x = (+/-) 5

When you square a variable x, the result is positive, no matter what the sign of the base.Remember, even exponents hide the sign of the base. Therefore, the square root of the square of the variable x (again regardless of the sign of the base) will always be positive, and therefore is equal to the absolute value of x, which again is always positive no matter whether x is positive or negative.