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inna [77]
3 years ago
8

Which list of ordered pairs represents solutions to x + y = 2 ?

Mathematics
2 answers:
riadik2000 [5.3K]3 years ago
7 0

First step is to solve the given equation for y. So, subtract x from each sides to isolate y. Therefore,

x + y - x = 2 - x

y = 2 -x.

Next step is to plug in the values of x = -4, 0 and 4 to get the corresponding values of y's so that we can get the correct order pair.

Let's plug in x = -4 now.

So, y = 2 - ( - 4) = 2 + 4 = 6

So, the first order pair is (- 4, 6).

For x = 0, y = 2 - 0 = 2

Second order pair is (0, 2).

For x = 4, y = 2 - 4 = -2.

Third order pair is (4, -2)

So, third list of order pairs (-4, 6), (0, 2), (4, -2) represent the solutions.

Hope this helps you!

DiKsa [7]3 years ago
3 0
(-4,6); (0,2); (4,-2)
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Answer:

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Step-by-step explanation:

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Add '7y' to each side of the equation.

3x + -7y + 7y = 11 + 7y

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Answer:

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Step-by-step explanation:

42/63 = 2/3

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IgorLugansk [536]

Answer:

See Explanation

Step-by-step explanation:

(a) Proof: Product of two rational numbers

Using direct proofs.

Let the two rational numbers be A and B.

Such that:

A = \frac{1}{2}

B = \frac{2}{3}

The product:

A * B = \frac{1}{2} * \frac{2}{3}

A * B = \frac{1}{1} * \frac{1}{3}

A * B = 1 * \frac{1}{3}

A * B = \frac{1}{3}

Proved, because 1/3 is rational

(b) Proof: Quotient of a rational number and a non-zero rational number

Using direct proofs.

Let the two rational numbers be A and B.

Such that:

A = \frac{1}{2}

B = \frac{2}{3}

The quotient:

A / B = \frac{1}{2} / \frac{2}{3}

Express as product

A / B = \frac{1}{2} / \frac{3}{2}

A / B = \frac{1*3}{2*2}

A / B = \frac{3}{4}

Proved, because 3/4 is rational

(c) x + y is rational (missing from the question)

Using direct proofs.

Let x and y be

Such that:

x = \frac{1}{2}

y = \frac{2}{3}

The sum:

x + y = \frac{1}{2} + \frac{2}{3}

Take LCM

x + y = \frac{3+4}{6}

x + y = \frac{7}{6}

Proved, because 7/6 is rational

<em>The above proof works for all values of A, B, x and y; as long as they are rational values</em>

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