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artcher [175]
3 years ago
6

Help? please- Can someone help me solve for x. Like you don't need to tell me the answer just how to. It's just a little hard fo

r me-Thank you to anyone who helps

Mathematics
2 answers:
natulia [17]3 years ago
4 0
X>2.5

Divide both sides by -5.

Two negatives make a positive.

Simplify 25/5 to 5.

Subtract 2.5 from both sides.

Simplify 5 – 2.5 to 2.5.

Then switch the sides
trasher [3.6K]3 years ago
4 0

9514 1404 393

Answer:

  undo what is done to the variable

Step-by-step explanation:

As with any "solve for" problem, it is a good idea to take a look at what is being done to the variable you're solving for.

<u>Starting "as is"</u>

Here, the value 2.5 is added to it, and the sum is multiplied by -5 (which we notice is negative).

In general, you want to "undo" what is done to the variable — in reverse order. What was done:

  • add 2.5
  • multiply by -5

If we adopt that strategy, it means we first need to undo multiplication by -5. That is, we must either (a) multiply both sides by -1/5, or (b) divide both sides by -5. Either choice will require that the sense of the comparison be reversed, so you will have ...

  ( ) < (x + 2.5)

After that, you need to "undo" the addition of 2.5. That is accomplished by adding -2.5 to both sides of the equation.

__

<u>Simplifying first</u>

For many problems that have fractions and/or parentheses, it can work well to eliminate both before you start with your "undo" strategy. Here, a first step can be to simplify the inequality, eliminating parentheses. This means you will use the distributive property to multiply each term inside parentheses by the (negative) factor outside.

Having done that, you find the "what was done to the variable" list is slightly different:

  • multiply by -5
  • add -12.5

As a consequence the "undo" steps are slightly different. Addition is still reversed by adding the opposite. Multiplication is still reversed by multiplying by the reciprocal, or by division (same thing).

As before, (and as <em>always</em>), what you do to one side of the inequality, you must also do to the other side. If multiplication or division by a negative number is involved, the comparison symbol must be reversed.

_____

<u>About Negative Multiplication/Division</u>

Above, we have noted that multiplication or division by a negative number requires reversal of the comparison symbol. For example, ...

  starting with -1 > -2

multiplying both sides by -1 gives ...

  ending with 1 < 2

__

When the (only) variable term has a negative coefficient, there is another strategy you can use that does not require the comparison to be reversed. You can add the opposite of that term to both sides of the inequality. For example, ...

  starting with -5x < -38

we can add 5x to both sides to get ...

  0 < -38 +5x . . . . . now the variable has a <em>positive coefficient</em>

This lets us proceed with the solution without worrying about the sense of the inequality. (Note that the variable will be "greater than" some value.)

_____

<em>More about solving strategy</em>

The rules that make Algebra possible are <em>the rules of equality</em>. If you look them up, you may find separate versions for addition, multiplication, subtraction, division. If you realize that subtraction is a form of addition, and that division is a form of multiplication, there are really only two—and those can be simplified to ONE RULE: always do the same operation on both sides of the equal sign. <em>The same rule applies to inequalities</em>. You will stay out of trouble if you remember and follow this rule. (There are many "shortcut" versions that go by names like "cross-multiply" and others. They are basically this rule with some steps not shown.)

When it comes to <em>applying functions to inequalities</em>, you need to consider carefully whether the function preserves the ordering relationship (>, ≥, <, ≤). Some functions do not, or have different behavior in different domains, so paying attention is required. Multiplication by a negative number is just the tip of the iceberg here.

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a)

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Step-by-step explanation:

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# (5 , π/3) ⇒ 1st quadrant

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* Lets solve the problem

(a)

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∵ Ф = 4π/3 ⇒ 3rd quadrant

∴ The coordinates of the point in polar form is (-6 , 4π/3)

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