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uranmaximum [27]
3 years ago
11

Please help me understand how to solve equations like these, I don’t understand them:

Mathematics
1 answer:
algol [13]3 years ago
4 0

Answer:

  • subtract the unwanted b term
  • divide by the coefficient of b
  • b = 60

Step-by-step explanation:

The numbers 4/5 and 9/10 are fractions. They multiply the variable 'b' in each case.

<u>How to solve an equation like this</u>

An example of an equation like this without fractions is ...

  6 + 3b = 8b

You solve this sort of equation by rearranging the equation so the terms having 'b' in them are all on one side of the equal sign. You want all of the terms not containing 'b' to be on the other side of the equal sign.

Here, the terms are mixed on the left, but the right has only a 'b' term. So, if we can remove the 'b' term from the left side, we will have reached the goal

In the example equation, we do that by adding -3b to both sides of the equation.

  6 +3b -3b = 8b -3b

Simplifying, we have ...

  6 = 5b

Now, we divide by the coefficient of 'b', which is 5. We divide both sides of the equation by that:

  6/5 = (5b)/5

  6/5 = b . . . . after simplifying

_____

<u>How to solve this equation</u>

The given equation is solved exactly the same way. We add the opposite of the 'b' term that we don't want (which is 4/5b):

  6 + 4/5b -4/5b = 9/10b -4/5b

Simplifying, this is ...

  6 = (9/10 -4/5)b = (9/10 -8/10)b

  6 = 1/10b

We can make the coefficient of 'b' be 1 by multiplying it by 10. We have to do that to both sides of the equation:

  (10)(6) = (10)(1/10)b

  <u>60 = b</u> . . . . . the solution

_____

<em>Comment on equations in general</em>

The one absolute rule in Algebra is that the equal sign must remain valid. You keep that commandment by observing the properties of equality. Essentially, you can do anything you like to one side of an equation, as long as you do exactly the same thing to the other side.

Here, when we say "subtract 4/5b", the way we keep this commandment is that we do it to both sides of the equation. The same is true for "multiply by 10". This is emphasized by the bold highlighting in the solution shown above.

<em>Comment on arithmetic</em>

In solving this equation, we did the arithmetic using the given fractions. Often, for an equation like this, you will be told the first step is to "eliminate fractions". You can do that here by multiplying the equation by 10:

  10(6 +4/5b) = 10(9/10b)

  60 +8b = 9b

Now, the unwanted term is 8b, so subtracting that (from both sides) gives ...

  60 = b

This is essentially a "2-step" equation, either way you do it.

It is helpful if you are comfortable doing arithmetic with fractions, mixed numbers, decimals, percentages, and numbers in scientific notation, as well as integers.

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Round £168.51 to the nearest pound
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The required round of  £168.51 to the nearest pound is £169.

Given that,
To round £168.51 to the nearest pound.

<h3>What is rounding of values?</h3>

The rounding of values is superseding a number with an inexact value that has a more ephemeral, more uncomplicated, or more direct representation.

Here,
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The expressions that are equivalent when m = 1 and m = 4 is;

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For option B; 3m + 4 and m + 4 + 2m

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<h3>How to find an angle?</h3>

A point where two or more line segments meet is called a vertex.

The vertex point also forms an angle.

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Substitute the given values into the expression above:

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