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Thepotemich [5.8K]
3 years ago
10

How many solutions does the equation 9x +3=9x + 5 have?

Mathematics
2 answers:
Sphinxa [80]3 years ago
8 0

Answer:

none

Step-by-step explanation:

Because X has the same coefficient on both sides but a different value is being added to it, there cannot be an solution.

9x+3=9x+5

-9x -5

-2=0, which doesnt work so theres no solution

kkurt [141]3 years ago
3 0

Answer:

This equation doesn't have any solution

Step-by-step explanation:

Given that,

9x +3=9x + 5

Subtract 3 from both sides

9x+3-3=9x+5-3

9x=9x+2

Since 9x cannot be equals to 9x+2, no matter the value of x.

Or further still divide through by 9

9x/9 = 9x/9 + 2/9

x=x + 2/9

it is not possible for x to be equal to x+2/9.

So it doesn't have a solution

x≠x+2/9

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5 0
3 years ago
A theater group made appearances into cities the hotel charge before tax and the second city was 1500 higher than the first the
Svetlanka [38]

Answer:

The hotel charge in each city before tax was <u>$5125</u> of the first city and <u>$3625</u> of the second city.

Step-by-step explanation:

Given:

A theater group made appearances into cities the hotel charge before tax and the second city was 1500 higher than the first.

The tax of the first city was 6% and the tax of the second city was 10%.

Total hotel tax paid for two cities with $670.

Now, to find the hotel charge in each city before tax.

Let the hotel charge in first city before tax be x.

And the hotel charge in second city before tax be y.

<em>So, as the hotel charge of the second city was 1500 higher than the first.</em>

<em>Thus</em>,

y=x-1500   ........(1)

<em>And as given, the tax of the first city was 6% and the tax of the second city was 10%, total hotel tax paid for two cities with $670.</em>

6% of x + 10% of y = $670.

\frac{6x}{100} +\frac{10y}{100} =670

0.06x+0.10y=670

Substituting the value of y from equation (1) we get:

0.06x+0.10(x-1500)=670

0.06x+0.10x-150=670

0.16x-150=670

<em>Adding both sides by 150 we get:</em>

0.16x=820

<em>Dividing both sides by 0.16 we get:</em>

x=5125.

<em>The hotel charge in first city before tax = $5125.</em>

Now, substituting the value of x in equation (1) we get:

y=x-1500

y=5125-1500

y=3625.

<em>The hotel charge in second city before tax = $3625.</em>

Therefore, the hotel charge in each city before tax was $5125 of the first city and $3625 of the second city.

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