Week 15 – Solving Rational Expressions

This week in math we learned how to actually solve the equation (solving for x)

Here’s how to do it.

for example number one,

I cross multiplied to simplify
I got my variable on one side of the equation
I isolated “x”
REMEMBER TO LIST NON-PERMISSIBLE VALUES

for example number two, it’s a bit more complicated

Get the factor to have a common denominator
Instead of making just the denominator common multiply everything by the common denominator (Numerator and Denominator)
Cancel out zero pairs (when a factor in the numerator is equivalent to the factor on the denominator)
Expand (FOIL)
Get variables on one side of the equation.
Isolate for “x”
LIST NON-PERMISSIBLE VALUES.

Things to remember;

Only cross-multiply when there are two values.
Make sure to cancel out zero pairs
Multiply the whole thing by the common denominator

Week 11 – Graphing Linear Inequalities.

This last week we learned how to graph different types of equations

Grade 10 question

y=x+7

Now we are looking to solve this as an inequality

y>x-4

Right now we have 50% chance of getting it right

But by testing the inequality by putting 2 points in as a check (verify)

I like to use the points (0,0)

y=x-4

0, 0-4

0>-4 (CORRECT)

so the side would shaded in including (0,0)

Week 10 – Completing the Square with Fractions

Week 5 – Factoring Expressions

Last year in grade 10, we learned how to factor simple expression, but this week we added another step

ex

1.) Replace the complicated set of varibles with a simple varible (a)

2.) Factor the expression

3.) Once properly factoring substitue the varibles back in

=( $x^2$ – x) $^2$ + 5 ( $x^2$ – x) $^2$ – 24

= $a^2$ + 5 $a$ – 24

=(a – 3) (a + 8)

=( $x^2$ – x – 3) ( $x^2$ – x + 8)

=(x + 3) (x – 1) (x + 2) (x + 4)

Week 4 – Addition, Subtraction, Multiplication and Division of Radicals

In Grade 9, we looked at the concept of grouping like terms.

Grade 9; 5x + 6y – 3x + 9y + x

x = 3x

y= 15y

Now in Grade 11, we now group radicals.

Grade 11; 4 $\sqrt[3]{9}$ , 6 $\sqrt[3]{9}$ , 15 $\sqrt[3]{9}$

= 25 $\sqrt[3]{9}$ .

Rule

Radicand and index must be the same.

Subtraction

$\sqrt{160}$ – $\sqrt{40}$

= $\sqrt{16\cdot10}$ – $\sqrt{4\cdot10}$

= 4 $\sqrt{10}$ – 2 $\sqrt{10}$

= 2 $\sqrt{10}$

Division and Multiplication Rule

The denominator CANNOT be a radical

Rationalizing the Denominator

$\sqrt{5}$ / x $\sqrt{6}$ = $\sqrt{30}$ / ——-> $\sqrt{30}$ /

$\sqrt{6}$ x $\sqrt{6}$ = $\sqrt{36}$ ——-> 6

Rationalizing the denominator is a way of cancelling out the radical so the denominator is not a radical.

1.) Recognize that the denominator is a radical

2.) Multiply both the numerator and the denominator (what you do the bottom you got to do the top)

3.) Simplify to find the new values

4.) See if there are any perfect squares (if the index is for example 3, then look for perfect cubes)

5.)Simplify the new values

6.) MAKE SURE A RADICAL IS NOT IN THE DENOMINATOR PLACE.