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Andrews [41]
2 years ago
11

Suppose x, y and z are integers. Prove that, if 3x−y + 5z is even, then at least one of x, y or z is even.

Mathematics
1 answer:
Elden [556K]2 years ago
4 0

Given:

x, y and z are integers.

To prove:

If 3x-y+5z is even, then at least one of x, y or z is even.

Solution:

We know that,

Product of two odd integers is always odd.          ...(i)

Difference of two odd integers is always even.          ...(ii)

Sum of an even integer and an odd integer is odd.      ...(iii)

Let as assume x, y and z all are odd, then 3x-y+5z is even.

3x is always odd.          [Using (i)]

5z is always odd.          [Using (i)]

3x-y is always even.       [Using (ii)]

(3x-y)+5z is always odd.       [Using (iii)]

3x-y+5z is always odd.

So, out assumption is incorrect.

Thus, at least one of x, y or z is even.

Hence proved.

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12x + 4 divided by 4 plus 20x + 5 divided by 5
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12x + 4 ➗ 4 + 20x + 5 ➗ 5

1. PEMDAS (order of operations) solve the division problems

(4 divided by 4 & 5 divided by 5)
12x + 1 + 20x + 1

2. Swap around the order of operations as commutative property states that you can swap the order of an addition problem and get the same answer.

12x + 20x + 1 + 1

3. Add like terms

32x + 2

This is the most simplified answer you could get as you can not add 32x and 2 as they are not like terms.
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2 years ago
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viva [34]

Answer:

We can order 35 different types of pizza based on the toppings.

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3 years ago
How do you solve an equation like<br> 2+6+7+20x
Leokris [45]
The answer would be 15 +20x. You have to combine like terms.
 
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                              8+7+20x
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Which fraction is less 2/9 1/3
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Answer:

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Step-by-step explanation:

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Please help this is really important
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D is 4/4, or just simply 1

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Read 2 more answers
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