What is 3 5x 2 2 3
Solving equations with an unknown
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Exercises  Repetition: release parentheses
Exercises
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General information on equations
An equation is a mathematical expression: Both sides to the left and right of the equal sign have the same value:
1. Example: 5 * (4  2) = 7 + 3
Variables can also appear in equations, the value of which is not known at first. It is, however, their value so to determine that the equation is "correct" again, i.e. the same value results on the left and right. Letters are used for this unknown quantity (s), usually the x, but any other letter can also be used.
2. Example: 5 * (x  2) = 7 + 3
From the 1st example you can see: If you write the number 4 instead of x, the left side of the equation gives the correct value 10. By the way, there is no other number that you can substitute for x to get the total value on the left 10 to come. So the "correct" solution for x is 4. Write down the solution set: L = {4} or even simpler x = 4.
With simple equations like the one in the example, the solution can still be found out easily by trial and error. With more complicated equations, or if the solution is not an integer, this quickly becomes more difficult:
3rd example: 4 (y  3)  2y = 5 (3y + 1)
However, there are methods of transforming the equation in such a way that the value for the unknown quantity can be read off directly. The prerequisite for these transformations is that they do not change the "equality" of the equation, that is, its "truth content".
Process as a flow chart
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Let's go back to the first example. The first step is always to simplify the expressions right and left as much as possible. This includes breaking up brackets (multiplying and / or minus brackets) and combining similar summands (numbers and variables):
5 (x  2) = 7 + 3  Multiply or calculate 5x  10 = 10
Same with the second example:
4 (y  5)  2y + 8 = 5 (3y + 1)  Multiply out on both sides 4y  20  2y + 8 = 15y + 5  Combining numbers and variables (resorting, application of the commutative law) 4y  2y  20 + 8 = 15y + 5  Calculate 2y  12 = 15y + 5
Help is available here for breaking brackets.
So far so good. The next steps are to transform the equation so that only the variable (x or y) is on one side and only a number on the other. Then you can read off the value of the variable directly.
For this purpose, all "disruptive" elements (summands and factors) can be eliminated, i.e., better said, brought to the other side of the equation by on either side of the equation applies an operation that makes the disruptive summand or factor disappear.
In the equation 5x  10 = 10, the " 10" on the left initially bothers. A minus of 10 can be eliminated with a plus of 10. Caution: The equation only remains correct if you change the same thing on both sides:
5x  10 = 10  Adding 10 5x  10 + 10 = 10 + 10  Calculate 5x = 20
Now the factor 5 in front of the x, which can be eliminated by dividing by 5, still "disturbs". Caution: Always treat both sides of the equation equally!
5x = 20  Divide by 5 5x / 5 = 20/5  Calculate x = 4
Caution: When dividing and multiplying a term (arithmetic expression) you must all Add the number divided by or multiplied by the number!
The intermediate step before the calculation can of course be omitted, because you know that 10 + 10 equals 20.
The respective transformation is noted to the right of the equation by the corresponding mathematical expression. The correct solution to the equation looks like this:
5 (x  2) = 7 + 3  V (Simplify) 5x  10 = 10  + 10 5x = 20  :5 x = 4
The second example goes like this:
4 (y  5)  2y + 8 = 5 (3y + 1)  V 4y  20  2y + 8 = 15y + 5  V 2y  12 = 15y + 5  + 12 2y = 15y + 17  + 15y 17y = 17  : 17 y = 1
In order to check the result, one checks whether the initial equation works if one substitutes the found value for the variable: In the equation 4 (y  5)  2y + 8 = 5 (3y + 1) one writes for all y the number 1: 4 (1  5)  2 · 1 + 8 = 5 (–3 · 1 + 1), calculates both sides and gets:
4·(–4) – 2 + 8 = 5·(–2) –16 + 6 = –10 –10 = –10
This is a true statement, so the solution y = 1 is correct.
Not all transformations are allowed, but all additions and subtractions, as well as all multiplications and divisions with / by numbers other than 0.
A disruptive negative sign in front of the variable at the end of the transformation, e.g. at x = 5, can be reversed by multiplying by (1):
x = 5  ·(1) x = 5
If at the end an equation in which the variable no longer occurs, the solution set is empty ( L = Æ) if this equation is false (e.g. 2 = 3). If the equation is true (e.g .: 1 = 1), then the solution set is equal to the set of real numbers  L =  R.
Exercises
Dissolving parentheses
The method for resolving parentheses depends on the mathematical symbol that precedes the parenthesis.
Plus sign: + (...)
Brackets directly in front of a plus sign can simply be left out:
5x + (11  3x) = 5x + 11  3x Minus sign:  (...)
Parentheses preceded by a minus are treated as follows:
The minus sign and the brackets are omitted, all signs in the brackets are reversed.
1. Example: 4x  (5 + 3x  7y) = 4x  5  3x + 7y = x + 7y  5
2nd example: 3x  36  (–x^{2} + 23  71x) = 3x  36 + x^{2}  23 + 71x = x^{2} + 74x  59
3. Ex .:  (4x  4)  (3x  5) = 4x + 4 + 3x + 5 = x + 9Multiplication sign: · (...) or just a factor
If there is a factor in front of the brackets, each summand in the brackets is multiplied by this factor when the brackets are resolved. Sign rules are:
(+)·(+) = (+)
(+)·(–) = (–)
(–)·(+) = (–)
(–)·(–) = (+)
1. Example: 5 (x  2) = 5x  10 (the multiplication point can be omitted)
2. Ex .: 3 (5x + 2y) = 15x  6y
3. Ex .: 4x (2 + 3x) = 8x + 12x^{2}
4. Ex .: 17a (2b + 3c1) = 34ab51ac + 17aBrackets times brackets: (...) (...)
When multiplying out two brackets, all the summands in the first bracket must be multiplied by all the summands in the second bracket. Note the sign!
1. Ex .: (a + b) (c + d) = ac + ad + bc + bd
2. Example: (2  3x) (5x + 7) = 10x + 14  15x^{2}  21x
3rd example: (3a  11b + 2) (5x  7) = 15ax  21a  55bx + 77b + 10x  14
4. Ex .: (a + b)^{2} = (a + b) (a + b) = a^{2} + from + from + b^{2} = a^{2} + 2ab + b^{2} (1st binomial formula)
5. Ex .: (a  b)^{2} = (a  b) (a  b) = a^{2}  from  from + b^{2} = a^{2}  2ab + b^{2} (2nd binomial formula)
6. Ex .: (a + b) (a  b) = a^{2}  from + from  b^{2} = a^{2}  b^{2} (3rd binomial formula)Minus bracket times factor or minus bracket times bracket
It is recommended to use the rule "point before line calculation", i.e. it is first multiplied and only then subtracted. To do this, however, the entire multiplication expression must be put in brackets, because the scope of the minus sign must be retained:
1. Ex .:  (3 + x) * 2 =  [(3 + x) * 2)] =  [6 + 2x] = 6  2x
2. Example: 2x  (3x  1) (2 + y) = 2x  [(3x  1) (2 + y)] = 2x  (6x + 3xy  2  y) = 2x  6x  3xy + 2 + y = 4x  3xy + y + 2Binomial formulas
1. Binomial formula (a + b) ² = a² + 2ab + b²
Expressions of the form (a + b) ² can be solved without writing out the square: (a + b) (a + b), to be multiplied out: a² + ab + ab + b² and to summarize: a² + 2ab + b², if you have that knows the summarized result a² + 2ab + b² and applies it to the summands in brackets: (a + b) ² = a² + 2ab + b²
a stands for the first summand in brackets and b for the second.
1. Example: (x + 3) ² = x² + 2 * x * 3 + 9 = x² + 6x + 9
2. Ex .: (2a + 5bx) ² = 4a² + 20abx + 25b²x²2. Binomial formula (a  b) ² = a²  2ab + b²
Here multiplying (a  b) results in (a  b) = a²  ab  ab + b² = a²  2ab + b².
This can also be used as a blueprint for resolving the square bracket directly.
1. Ex .: (3p  q) ² = 9p²  6pq + q²
2. Ex: (7x³  3xyz) ² = 49x^{6}  42x^{4}yz + 9x²y²z²3. Binomial formula (a + b) (a  b) = a²  b²
When multiplying and combining, it turns out that in this case the nonsquare summand vanishes: (a + b) (a  b) = a²  ab + ab  b² = a²  b².
1. Example: (3x + 0.5) (3x  0.5) = 9x²  0.25
2. Ex .: (6a  2b) (2b + 6a) = 36a²  4b²
In the 2nd example, the commutative law was applied twice: The bracket with the minus is in front (multiplication is commutative) and the summands in the second bracket are interchanged (addition is commutative). Note that the order in the result is based on the bracket with the minus!
More exercises
Source: Helmut Postel: Collection of exercises for exercise and repetition. Hanover: Schroedel, 1998. ISBN 3507732211
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