1st Video
Inspiration
2nd Video
Believe
3rd Video
What’s you Know about Math?
4th Video
Solving Differential Equation
5th Video
Solving Algebraic Equation
1. x minus five equals 3
in this case we are looking for the value of variable x.
x – 5 = 3
first we put additional equation on both sides, in this case we adding five (5) as a additional equation.
x – 5 + 5 = 3 + 5
And now we can the value of variable x. It is eight (8).
x = 8
2. seven equals four times x minus one
in this case we are looking for the value of variable a.
7 = 4a – 1
We put the additional equation too on in this case, and in this case we put one (1).
1 + 7 = 4a – 1 + 1
So that we get eight equals four times a.
8 = 4a
Then, we must over both sides with four so that we can get the value of variable a. And the result was two (2).
a = 4
3. two thirth times x equals eight
in this case we should find the value of variable x.
2/3x = 8
Now we must times both sides with the oposites of the constanta of variable and the oposite of two thirth is three second.
3/2 . 2/3 x = 8 . 3/2
we get x on the left side and we must over eight with two then times the result with three. Then we can get the value of variable x.
x = 12
4. five minus two times x equals three times x plus one
we must find the value of variable x.
5 – 2x = 3x + 1
First we must put an additional equation on both sides and I adding minus five (-5), so we can get
(-5) + 5 – 2x = 3x + 1 + (- 5)
then we must adding more additional equation with variable x. And we can add minus three times x (-3x)
- 2x = 3x – 4
-2x + (– 3x) = 3x – 4 + (– 3x)
and then we can get minus five times x equals to minus four.
-5x = -4
Now we must over both sides with minus five.
-5x/(-5) = -4/(-5)
so we can get the value of variable x, it is four fifth.
x = 4/5
5. three minus five open bracket two times m minus five close bracket equals minus two
In this case we are looking for the value of variable m.
3 – 5(2m – 5) = -2
First we must times five with two times m minus five. So we can get the full form of it is three minus ten times m plus twenty five equals minus two.
3 – 10m + 25 = -2
We can adding three with minus twenty five on left sides then we get minus ten times m plus minus twenty eight equals minus two
-10m + 28 = -2
Then we put additional equation minus twenty eight (-28)
– 10m + 28 + (-28) = -2 + (-28)
so that we can minus ten times m equals minus tirty
-10m = -30
and we must over both sides with minus ten (-10) so that we can get the value of variable m, it is three
-10m/(-10) = -30/ (-10)
à m = 3
6. a half times x plus a quarter equals one third times x plus five forth
we must find the value of variable x from
½ x + ¼ = 1/3x + 5/4
first we must adding an additional equation and we put minus a quarter (- ¼)
½ x + ¼ + (- ¼) = 1/3x + 5/4 + (- ¼)
so we can get a half times x equals one third times x plus four forth (four forth equals one)
½ x = 1/3x + 1
then we put another aditional equation, now we put minus one third.
½ x + (-1/3x) = 1/3x + 1+ (-1/3x)
and we get one sixth equals one, the last step is multiply both sides by six
1/6x = 1
1/6x . 6 = 1 . 6
x = 6
so we get the value of variable x and it is six
7. oh point thirty five times x minus oh point two equals oh point fifteen times x plus oh point one
in this case we must get the value of variable x
0,35x – 0,2 = 0,15x + 0,1
first we must adding an additional equation and we can adding minus oh point one
0,35x – 0,2 + (-0,1) = 0,15x + 0,1 + (-0,1)
the we get oh point thirty five times x minus oh point three equals oh point fifteen times x, then we adding another additional equation, but now we adding minus oh point fifteen times x. We get
0,35x – 0,3 + (-0,15x) = 0,15x + (-0,15x)
à 0,20x – 0,3 = 0
and we adding oh point three, then we over both sides by oh point twenty
0,20x – 0,3 + 0,3 = 0 + 0,3
à 0,20x = 0,3
à 0,20x/0,20 = 0,3/0,20
so x equals one point seventy five
x = 1,75
5th Video
Exponen Rule on Logarithm Equation
We already know some exponent rules, like
x to the power of a times x to the power b equals x to the power of a plus b
x to the power of a over x to the power b equals x to the power of a minus b
x to the power of a in bracet to the power of b equals x to the power of a times b
So we can use these exponent rules to proof some logarithm equations.
First we can proof about logarithm basic x of a equals be, so we can find the value of a. Then logarithm basic x of open bracket a to the power of B close bracket, we can change it with B times logarithm basic x of open bracket A. After that we can prove that logarithm basic x of A plus logarithm basic x of B equals logarithm basic x open bracket A times B close bracket is equals. And the last we can proof that logarithm basic x of A minus logarithm basic x of B equals logarithm basic x open bracket A over B close bracket is equals.
From all these videos I can find some mathematics equations, and we can get the way to solve it too. We can get the way to learn mathematics is not only with a serious condition but we can learn it by interesting way.
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