EQUATIONS

 

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Rules: An equation is a statement or mathematical expression which says one side is equal to the other side, eg. 122 = 3 + 3. Usually equations are combined with algebra and contain at least one unknown, eg. 12a = 3 + 3, and to solve the equation the unknown value has to be determined.

The basic rule to be observed is any operation done one one side of the equals sign has to be copied on the other side. In doing this it may be necessary to include brackets around expressions which have to be done first in the order of operations.

Consider the example above: 12a = 3 + 3.

Multiplying both sides by 'a' removes the denominator: 12aa = 3 + 3 x a.
The left side of the statement is now 12, but the right side of the statement looks wrong.
The order of operations states multiply before sum, but this would give an incorrect answer.
In this case the 3 + 3 has to be preserved and is bracketed: (3 + 3)a.
Now to make 'a' the subject of the formula: divide both sides by (3 + 3), leaving:
12(3 + 3) = a, or a = 2. Check by putting into the original equation: 122 = 3 + 3, correct.

Without brackets it would have been 12 = 3 + 3a, minus 3 from both sides giving 9 = 3a, dividing both sides by 3 leaves: a = 3, which when put into the original equation gives: 123 = 3 + 3, which is incorrect.

Always check your result by substitution.

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Simple Equations: Using the rules stated above:

Example: Solve the equation, 4x - 8 = 32.

Answer: Add 8 to both sides: 4x = 40.
Divide both sides by 4: x = 10.
Check: 4 x 10 -8 = 32. Correct.

Example: Solve the equation, 6 (8x + 4) = 48.

Answer: Divide both sides by 6: 8x + 4 = 8.
Subtract 4 from both sides: 8x = 4.
Divide both sides by 8: x = .
Check: 6(8 x + 4) = 48 or 6(4 + 4) = 48. Correct.

Example: Solve the equation, 6x - 4 = 18x - 12.

Answer: Put the x's on one side, the numbers on the other: -4 + 12 = 18x - 6x.
Sum both sides: 8 = 12x.
Divide both sides by 12: x = .
Check: -4 + 12 = 18 x - 6 x or 8 = 13 - 4. Correct.

Example: Solve the equation: (p - 1)p = p(p - 3).

Answer: Remove both denominators by multiplying through by each: (p - 1)(p - 3) = p x p.
Expand the left hand side: p- 4p + 3 = p.
Subtract pfrom both sides: - 4p + 3 = 0.
Subtract 3 from both sides: - 4p = -3.
Change the signs: 4p = 3.
Divide both sides by 4: p = .

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Exercise: Solve the equation for ' x '.

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Simultaneous Equations: A simultaneous equation is one where there is more than one unknown and more than one equation presented. These are normally done on two unknowns and two lines of equations and normally solved by elimination, but an example is done solving by substitution.

Solution by Elimination: To solve by elimination, one of the unknowns has to be made equal to the other and eliminated by addition or subtracting, leaving a simple equaltion to solve for the other unknown.

Example: 4x - y = 8
2x + y = 10
This has two unknowns 'x' and 'y' and two lines of equations.
By adding the two equations the 'y' is eliminated leaving: 6x = 18.
'x' is solved: x = 186 or x = 3.
Going back to the original equation, replace 'x' with 3:
12 - y = 8
6 + y = 10
Taking either equation; 12 - y = 8, solving for y gives: -y = -4; or y = 4.
The last task is to insert the y = 4 in both equations and check your answer is correct.
12 - 4 = 8.
6 + 4 = 10.
It is normal to number each equation, which will be done in the next example where the co-efficients have to be made equal.

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Example: 3x + 6y = 30   1
6x - 2y = 18                      2
This has two unknowns 'x' and 'y' and two lines of equations.
Before adding or subtracting to eliminate an unknown, the co-efficients have to be equal.
Rule: with equations what is done to one side of the equals has to be done to the other side.
To eliminate the 'y' co-efficient simply multiply the 2 equation through by 3.
3x + 6y = 30      1
18x - 6y = 54     3
Add the equations: 21x = 84. Solving for 'x'; x = 8421, x = 4.
Substitute x = 4 back into any equation eg.     3.
72 - 6y = 54. Solving for 'y'; -6y = -18; or y = 3.
Finally substitute numbers in the original equation and check your answer is correct.
12 + 18 = 30.    1
24 - 6 = 18.       2


Example: 4x = 2y - 12      1
-9x +7y = 32                      2

1. Make the equation look normal:
4x - 2y = -12                      1
-9x + 7y = 32                     2

2. Eliminate the 'y' co-efficient (cross multiply: equation 1 by 7; equation 2 by 2).
28x - 14y = -84                  3
-18x + 14y = 64                 4

3. Add the equations:
10x = -20                           5

4.Solve for 'x':
x = -2010. x = -2.


5. Substitute 'x' for -2:
-8 - 2y = -12                      1
18 +7y = 32                      2

6. Solve for 'y' from any equation:
-8 -2y = -12; -2y = - 4; y =- 4-2 ; y = 2.

7. Check by substituting both letters:
-8 - 4 = -12                       1
18 + 14 = 32                    2

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Simultaneous equations may be solved using a different method, called solving by substitution. The same conditions must apply as solving by elimination; two unknowns to two lines of equations.

Solution by Substitution: Done by making one of the unknowns the subject of one equation.

Example: 4x - 2y = -12       1
-9x + 7y = 32                       2

1. Make 'y' the subject of equation 1 (this is easier than equation 2):
-2y = -4x - 12; y = 2x + 6.

2. Substitute '2x + 6' for 'y' in equation    2:
-9x + 7(2x + 6) = 32           3

3. Remove brackets and collect like terms:
-9x + 14x + 42 = 32;
5x = -10;


4. Solve for 'x';
x = -105; x = -2.

5. Substite x = -2 in equation 1 or 2 and solve for 'y' :
-8 - 2y = -12                             1
-2y = - 4; y = - 4-2; y = 2.

6. Complete by checking.

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Exercise: Simultaneous Equation. Solve for each unknown.

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Quadratic Equations: There are different methods to solving the unknown co-efficient in a quadratic equation. The simplest method is factorisation, as shown in Algebra. This does not always work when the unknown is not a whole number. The unknown in a quadratic equation may then be solved by completing the square or by using the quadratic formula. Another way of solving is by drawing a graph of the equation.

The general form for a quadratic is: ax+ bx + c = 0. where a, b and c are real numbers. A quadratic has two solutions for (in this case) ' x '. The two solutions may be equal. The solutions are called roots. At GCSE level only solutions with real roots, which are real numbers are studied.

Solution by Factors: This means putting the equation into two brackets.
Using the ' x ' as the unknown, this method is used: (x )(x ) = 0.
Start by using the standard format: ax+ bx + c = 0.
x is squared therefore it is present in both brackets: (x )(x ) = 0.
Find two numbers that multiply together to give the ' c ', and add or subtract together to give the ' b '.
Enter these within the brackets with the correct signs.
Always check that the answer given is correct.
Finally the solution is easily found from the factorised forms.
To make the equation equal zero make each ' x ' the same value as the number within it's brackets, but with the opposite sign.

Example: Solve for ' x ': x+ x - 6 = 0.

1. The equation is in its standard format.
2. Create brackets enter an ' x ' in each: (x )(x ) = 0.
3. Pair of numbers that are factors of 6: 3 and 2; and 6 and 1.
    Eliminate the 6 and 1 as adding or subtracting cannot give ' b ' = +1x.
    The factors of 3 and 2 if subtracted will give ' b ' = +1x.
4. Check the signs: 3 is plus and 2 is minus.
     Enter in the brackets: (x + 3)(x - 2) = 0.
5. Check: x- 2x + 3x - 6 = 0. OK.
6. If (x + 3)(x - 2) = 0 Then when x = -3, the equation balances, and when x = +2, the equation balances.
Therefore the solution is: x = -3 and x = +2.

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Factorising is not quite so straight forward when the ' a ' co-efficient has a numerical value that is not 1.

Example: Solve for ' x ': 4x+ 26x = 48.

1. Put the equation in its standard format: 4x+26x - 48 = 0.

2. Create brackets enter the ' x ' in each: (4x )(x ) = 0.

3. Now the difficult part: Factors of 48: 48, 24, 16, 12, 8, 6, 4, 2, 1.
Listing these: 48 and 1, 24 and 2; 16 and 3 12 and 4 8 and 6.
Any of them multiply to be ' c ' but what about ' b ' which has to be +26.
Lets try each one:

(4x   48)(x   1)     multiplies to give  4x and 48x which add or subtract to:  52x or 44x.
(4x   1)(x   48)                               192x and  1x                                            193x or 191x.
(4x   24)(x   2)                                    8x and 24x                                             32x or 16x.
(4x   2)(x   24)                                  96x and  2x                                              98x or 94x.
(4x   16)(x   3)                                  12x and 16x                                             28x or  4x.
(4x   3)(x   16)                                  64x and  3x                                              67x or 61x.
(4x   12)(x   4)                                  16x and 12x                                             28x or  4x.
(4x   4)(x   12)                                  48x and  4x                                              52x or  44x.
(4x   8)(x   6)                                    24x and  8x                                              32x or 16x.
(4x   6)(x   8)                                    32x and  6x                                              38x or 26x.


4. Check the signs: 32x - 6x gave the 26x, therefore the factors are +8 and - 6.
Enter into the brackets: (4x - 6)(x + 8) = 0.

5. Check: 4x+32x - 6x -48 = 0. OK.

6. If (4x - 6)(x + 8) = 0. Then when x = 1 the equation balances, and when x = -8 the equation balances.
Therefore the solution is: x = 1 and x = -8.

Sometimes it is easier to solve these problems by an alternative method.

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Exercise: Find the Roots by factorising a quadratic equation.

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Solution by 'Completing the Square': This is not by any means the easiest was to solve a quadratic equation but is however an important method to learn. In some cases it is not possible to find the roots by factorising. There are two methods: 'complete the square' and by 'formula'.

When factorising the method was to find two factors of the non ' x ' number that added or subtracted to give the co-efficient of the ' x '. When this is not possible we can find the solution by 'complete the square'.

Example: Solve x- 6x - 4 = 0.
Here no two factors of 4 can add or subtract to 6.
Now we start rearranging the equation.
Always do the same operation to each side of the equals sign.

1. Check the equation is in it's standard format.      Yes. x- 6x - 4 = 0.
2. The roots cannot be found by factorising.             No.
3. Does the xhave a co-efficient.                              No.    If yes divide through the whole equation by that number.
4. Take the non ' x ' number to the other side: add 4 to both sides: x- 6x = 4.
5. Add to each side the square of half the co-efficient of ' x ' : x- 6x + (-3) = 4 + (-3).
6. Do the addition: x- 6x + 9 = 13.
7. The left hand side can now be squared; therefore factorise: (x - 3)(x - 3) = (x - 3).
leaving the equation: (x - 3) = 13.
8. Square root both sides, remembering that there is always two roots: x - 3 = 13.
9. Solving for ' x ' gives both roots: x = 3 + 13 or x = 3 - 13.
10. The solutions can be completed with a calculator: x = 6606 or -6606.

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Example: Solve 4x- 16x + 3 = 0 by completing the square.

1. Yes, standard format.
2. No not factorise.
3. Yes,     Divide through by 4:                          x- 4x + 075 = 0.
4. Minus both sides by 075:                             x- 4x = -075.
5. Add the square of half co-efficent of ' x ':    x- 4x + (-2) = -075 + (-2).
6. Addition:                                                         x- 4x + 4 = 325.
7. Factorise the left side:                                  (x - 2) = 325.
8. Square root both sides:                                 x - 2 = 325.
9. Solve for ' x ':                                                   x = 2 +325 or x = 2 - 325.
10. Complete using a calculator gives:            x = 3803 or 0197.

 

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Solution by Formula: A formula based on the completing the square method is used. This has to be remembered.

The expression: ax+ bx + c = 0, where
' a ' is the co-efficient of the 'x' ;
' b ' is the co-efficient of the ' x ' ; and
' c ' is the number.

  

Formula:

By substituting the numbers for the letters in the formula and completing the calculation will give the solution.

When factorising the method was to find two factors of the non ' x ' number that added or subtracted to give the co-efficient of the ' x '. When this is not possible we can find the solution by formula.

Note: in the example a dot is used to signify a multiplication.

Example: Solve by formula 2x- 3x - 4 = 0.

1. Check the equation is in it's standard format.      Yes.
2. Can it be done by Factorising.                               No.
3. Substitute the numbers for the letters:     
4. Simplifying the square root:       
5. Completing the calculation with the aid of a calculator:     x = 235 or -085.

 

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Exercise: Find the Roots by Completing the Square or by Formula:

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