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nataly862011 [7]
3 years ago
5

3x + 2y = 9 x - 4y = 4

Mathematics
1 answer:
kherson [118]3 years ago
7 0

Answer:

Enter a problem...

Algebra Examples

Popular Problems Algebra Solve by Substitution 3x-4y=9 , -3x+2y=9

3

x

−

4

y

=

9

,

−

3

x

+

2

y

=

9

Solve for

x

in the first equation.

Tap for more steps...

x

=

3

+

4

y

3

−

3

x

+

2

y

=

9

Replace all occurrences of

x

in

−

3

x

+

2

y

=

9

with

3

+

4

y

3

.

x

=

3

+

4

y

3

−

3

(

3

+

4

y

3

)

+

2

y

=

9

Simplify

−

3

(

3

+

4

y

3

)

+

2

y

.

Tap for more steps...

x

=

3

+

4

y

3

−

9

−

2

y

=

9

Solve for

y

in the second equation.

Tap for fewer steps...

Move all terms not containing

y

to the right side of the equation.

Tap for more steps...

x

=

3

+

4

y

3

−

2

y

=

18

Divide each term by

−

2

and simplify.

Tap for more steps...

x

=

3

+

4

y

3

y

=

−

9

Replace all occurrences of

y

in

x

=

3

+

4

y

3

with

−

9

.

x

=

3

+

4

(

−

9

)

3

y

=

−

9

Simplify

3

+

4

(

−

9

)

3

.

Tap for more steps...

x

=

−

9

y

=

−

9

The solution to the system of equations can be represented as a point.

(

−

9

,

−

9

)

The result can be shown in multiple forms.

Point Form:

(

−

9

,

−

9

)

Equation Form:

x

=

−

9

,

y

=

−

9

image of graph

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7 0
2 years ago
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What is the minimum number of times you must throw three fair six-sided dice to ensure that the same sum is rolled twice?
Vladimir [108]
For three fair six-sided dice, the possible sum of the faces rolled can be any digit from 3 to 18.
For instance the minimum sum occurs when all three dices shows 1 (i.e. 1 + 1 + 1 = 3) and the maximum sum occurs when all three dces shows 6 (i.e. 6 + 6 + 6 = 18).

Thus, there are 16 possible sums when three six-sided dice are rolled.

Therefore, from the pigeonhole principle, <span>the minimum number of times you must throw three fair six-sided dice to ensure that the same sum is rolled twice is 16 + 1 = 17 times.

The pigeonhole principle states that </span><span>if n items are put into m containers, with n > m > 0, then at least one container must contain more than one item.

That is for our case, given that there are 16 possible sums when three six-sided dice is rolled, for there to be two same sums, the number of sums will be greater than 16 and the minimum number greater than 16 is 17.
</span>
4 0
3 years ago
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