**D ^{efinition}**

Last time we learnt how to answer simple algebraic equations. In this part, we'll learn how to answer more difficult algebra, including simultaneous equations&&equations with brackets.

**S ^{imultaneous} E^{quations};;**

*Simply 2 equations;

-with 2 unknowns

-usually

*x*and

*y*

*To SOLVE the equations

means we find values of

*x* and *y* that;

-satisfy BOTH equations [work in]

-at the same time [simultaneously]

**E ^{quations}**

I^{ntermediate} A^{lgebra} (1);;

A-> 2x-y=1

B-> 3x+y=9

[[We have the same number of *y's* between each]]

If we add the equations, the *y's* disappear;

5x=10….so…..x=2

Now, substitute x=2 into equations *A*;

2x2-y=1

4-y=1

y=3

So the ANSWER is… x=2, y=3

I^{ntermediate} A^{lgebra} (2);;

A-> 4x-2y=8

B-> 3x+6y=21

This is slightly more tricky. We have no "like" values! Oh heavens above! No problremo though! If we multiply A by 3, then we will have the same number of *y's* in each equation;

12x-6y=24

3x+6y=21

Now doesn't that look familiar?!?

We again add the similar values;

15x=45…so….x=3

And again we substitute the x value into the equation;

3x3+6y=21

9+6y=21

6y=12

y=2

So the ANSWER is…x=3, y=2

**E ^{xpanding} B^{rackets}**;;

Sometimes we need to solve an equation and it involves removing 2 brackets that are multiplied together. There are afew methods for doing this. Here are some of them;;

F^{oil};;

(x+7)(x-5)

**F**-First=unknown values multiplied together

**O**-Outer=first unknown value multiplied with last known value

**I**-Inner=last value in the first bracket multiplied with first value in the second bracket

**L**-Last=known values multiplied together

The above equation is now identical to; x^{2}-5x+7x-35

Which is; x^{2}+2x-35

This technique is also known as the *smiley face* because if you draw in lines attached to the values you're multiplying, it resembles a smiley face =D

G^{rid} M^{ethod};;

(3x-1)(4x-7)

X | 4x | -7 |

3x | 12x^{2} |
-21x |

-1 | -4x | 7 |

Multiply along the columns and rows and write in your answer in the provided boxes;

12x^{2}-21x-4x+7

ANSWER=12x^{2}-25x+7

**SUPER MEGA DOOPER IMPORTANT EXAMPLEE!!**;;

(x-4)^{2} means; (x-4)(x-4)

**REMEMBER**;;

- To solve simultaneous equations, you must find values that satisfy BOTH equations
- Simplify your answers! Makes life a lot easier
- In simultaneous equations, chose the simpler value to multiply that suits you
- Chose a technique that suits you. Chose the one you feel most comfortable with

Well this is the end of Algebra For The Intermediate!

The next stage hasn't yet been made, but when it is made, click Algebra for the Advanced.

Later!