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kari74 [83]
2 years ago
10

23/7y state restrictions

Mathematics
1 answer:
qwelly [4]2 years ago
4 0

The restrictions on the variable of the given rational fraction is y ≠ 0.

<h3>The types of numbers.</h3>

In Mathematics, there are six (6) common types of numbers and these include the following:

  • <u>Natural (counting) numbers:</u> these include 1, 2, 3, 4, 5, 6, .....114, ....560.
  • <u>Whole numbers:</u> these comprises all natural numbers and 0.
  • <u>Integers:</u> these are whole numbers that may either be positive, negative, or zero such as ....-560, ...... -114, ..... -4, -3, -2, -1, 0, 1, 2, 3, 4, .....114, ....560.
  • <u>Irrational numbers:</u> these comprises non-terminating or non-repeating decimals.
  • <u>Real numbers:</u> these comprises both rational numbers and irrational numbers.
  • <u>Rational numbers:</u> these comprises fractions, integers, and terminating (repeating) decimals such as ....-560, ...... -114, ..... -4, -3, -2, -1, -1/2, 0, 1, 1/2, 2, 3, 4, .....114, ....560.

This ultimately implies that, a rational fraction simply comprises a real number and it can be defined as a quotient which consist of two integers x and y.

<h3>What are restrictions?</h3>

In Mathematics, restrictions can be defined as all the real numbers that are not part of the domain because they produces a value of 0 in the denominator of a rational fraction.

In order to determine the restrictions for this rational fraction, we would equate the denominator to 0 and then solve:

23/7y;

7y = 0

y = 0/7

y ≠ 0.

Read more on restrictions here: brainly.com/question/10957518

#SPJ1

Complete Question:

State any restrictions on the variables 23/7y

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The solution to the system of equations is:

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The graph is attached below.

Step-by-step explanation:

Given the system of equations

2y = 2x - 8

2x + y = 5

Let us solve the system of equations using the elimination method

\begin{bmatrix}2y=2x-8\\ 2x+y=5\end{bmatrix}

Arrange equation variables for elimination

\begin{bmatrix}2y-2x=-8\\ y+2x=5\end{bmatrix}

Multiply y + 2x = 5 by 2:  2y + 4x = 10

\begin{bmatrix}2y-2x=-8\\ 2y+4x=10\end{bmatrix}

subtracting the equations

2y+4x=10

-

\underline{2y-2x=-8}

6x=18

solve 6x = 18 for x

6x=18

Divide both sides by 6

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2y-6=-8

Add 6 to both sides

2y-6+6=-8+6

Simplify

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Divide both sides by 2

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Simplify

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Therefore, the solution to the system of equations is:

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The graph is attached below.

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