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lidiya [134]
3 years ago
13

How do I solve this problem

Mathematics
2 answers:
Lubov Fominskaja [6]3 years ago
5 0
I'm assuming you want to solve for x.

When solving algebraic equations, there is one principle that you must learn.

The principle is as follows: doing any operation on one side of the equation must be done the same on the other side. For example, if you add 10 to one side of the equation, you also add 10 on the opposite side. This is done to keep the balance or equality of the equation.

We don't just do an operation on both sides of an equation for the sake of it. Every operation we do has a reasoning behind it. We might add 15 to both sides of the equation because one side has a -15 in it and we might want to get rid of it.

Usually, algebraic textbooks will call this process transposing. But, if you want an even fancier and more technical sounding name, you can say, "The Principle Addition of Equality" and "The Multiplication (or Division) of Equality".

For other operations, like taking the square root of an equation, you can simply say, "[name of the operation in present progressive tense] of both sides of the equation".

Now that I showed you one of the most important principles in algebra for solving an equation, let's get back to your original question.

What is the value of x that makes the equation -5 + 3x = 10 true?

We need to get the x variable by itself and everything else on the opposite side. The first operation we could do is add 5 to each side. We do this not only to keep the balance of the equation but to get rid of the -5 on the left-hand side.

-5 + 3x       = 10
-5 + 3x + 5 = 10 + 5
3x               = 15

Great! The -5 is out of the way and we're closer to the answer. We'll have to do one more operation to find the value of x.

When a number and a variable (or a series of variables) are juxtaposed together, that indicates multiplication.

Because the 3 and the variable x are juxtaposed, they are being multiplied and we will have to find an operation to get rid of the 3.

Dividing both sides by 3 will do just this.

3x      = 15
3x / 3 = 15 / 3
x         = 5

We finally arrived at the answer. We did all this by using a simple principle in algebra.

I hope you find my answer immensely helpful and if there is something you don't quite understand, don't be afraid to leave a comment or write a message to me. I'll be glad to be of assistance to you!
zysi [14]3 years ago
4 0
You need to simplify it
lets start with your orgininal equation here:
-5+3x=10
Now lets try and get "x" alone. we will do this by adding 5 to both sides.
-5+3x=10
+5       +5
3x=15
Now we have 3 times "x" equals 15
The next step would be to divide both sides by 3 in order to get "x" alone.
3x=15
x=15/3
x=5


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Ugo [173]

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Let the three sides be represented with A, B, C

Let the angles be represented with \alpha, \beta, \theta

[See Attachment for Triangle]

A = 10cm

C = 12cm

\beta = 30

What the question is to calculate the third length (Side B) and the other 2 angles (\alpha\ and\ \theta)

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When two angles of a triangle are known, the third side is calculated as thus;

B^2 = A^2 + C^2 - 2ABCos\beta

Substitute: A = 10,  C =12; \beta = 30

B^2 = 10^2 + 12^2 - 2 * 10 * 12 *Cos30

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\sqrt{B^2} = \sqrt{36.153903093}

B = \sqrt{36.153903093}

B = 6.0128115797

B = 6.0 <em>(Approximated)</em>

Calculating Angle \alpha

A^2 = B^2 + C^2 - 2BCCos\alpha

Substitute: A = 10,  C =12; B = 6

10^2 = 6^2 + 12^2 - 2 * 6 * 12 *Cos\alpha

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100 = 36 + 144 - 144 *Cos\alpha

100 = 180 - 144 *Cos\alpha

Subtract 180 from both sides

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Divide both sides by -144

\frac{-80}{-144} = \frac{- 144 *Cos\alpha}{-144}

\frac{-80}{-144} = Cos\alpha

0.5555556 = Cos\alpha

Take arccos of both sides

Cos^{-1}(0.5555556) = Cos^{-1}(Cos\alpha)

Cos^{-1}(0.5555556) = \alpha

56.25098078 = \alpha

\alpha = 56.3 <em>(Approximated)</em>

Calculating \theta

Sum of angles in a triangle = 180

Hence;

\alpha + \beta + \theta = 180

30 + 56.3 + \theta = 180

86.3 + \theta = 180

Make \theta the subject of formula

\theta = 180 - 86.3

\theta = 93.7

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