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Alex787 [66]
1 year ago
5

If 2x+ y= 10 and y=3x find the value of both x and y

Mathematics
1 answer:
Marat540 [252]1 year ago
6 0

Given the set of equations:

2x + y = 10

y = 3x

Let's solve for x and y.

Here, let's use substitution method.

Substitute 3x for y in equation 1:

2x + y = 10

2x + 3x = 10

5x = 10

Divide both sides by 5:

\begin{gathered} \frac{5x}{5}=\frac{10}{5} \\  \\ x=2 \end{gathered}

In equation 2, substitute 2 for x to find y.

y = 3x

y = 3(2)

y = 6

Therefore, we have:

x = 2 and y = 6

ANSWER:

x = 2, y = 6

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If a+b+c=1,

a

2

+

b

2

+

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2

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b

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+

c

3

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3

then find the value of

a

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b

4

+

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?

we know

2

(

a

b

+

b

c

+

c

a

)

=

(

a

+

b

+

c

)

2

−

(

a

2

+

b

2

+

c

2

)

⇒

2

(

a

b

+

b

c

+

c

a

)

=

1

2

−

2

=

−

1

⇒

a

b

+

b

c

+

c

a

=

−

1

2

given

a

3

+

b

3

+

c

3

=

3

⇒

a

3

+

b

3

+

c

3

−

3

a

b

c

+

3

a

b

c

=

3

⇒

(

a

+

b

+

c

)

(

a

2

+

b

2

+

c

2

−

a

b

−

b

c

−

c

a

)

+

3

a

b

c

=

3

⇒

(

a

+

b

+

c

)

(

a

2

+

b

2

+

c

2

−

(

a

b

+

b

c

+

c

a

)

+

3

a

b

c

=

3

⇒

(

1

×

(

2

−

(

−

1

2

)

+

3

a

b

c

)

)

=

3

⇒

(

2

+

1

2

)

+

3

a

b

c

=

3

⇒

3

a

b

c

=

3

−

5

2

=

1

2

⇒

a

b

c

=

1

6

Now

(

a

2

b

2

+

b

2

c

2

+

c

2

a

2

)

=

(

a

b

+

b

c

+

c

a

)

2

−

2

a

b

2

c

−

2

b

c

2

a

−

2

c

a

2

b

=

(

a

b

+

b

c

+

c

a

)

2

−

2

a

b

c

(

b

+

c

+

a

)

=

(

−

1

2

)

2

−

2

×

1

6

×

1

=

1

4

−

1

3

=

−

1

12

Now

a

4

+

b

4

+

c

4

=

(

a

2

+

b

2

+

c

2

)

2

−

2

(

a

2

b

2

+

b

2

c

2

+

c

2

a

2

)

=

2

2

−

2

×

(

−

1

12

)

=

4

+

1

6

=

4

1

6

Extension

a

5

+

b

5

+

c

5

=

(

a

3

+

b

3

+

c

3

)

(

a

2

+

b

2

+

c

2

)

−

[

a

3

(

b

2

+

c

2

)

+

b

3

(

c

2

+

a

2

)

+

c

3

(

a

2

+

c

2

)

]

=

3

⋅

2

−

[

a

3

(

b

2

+

c

2

)

+

b

3

(

c

2

+

a

2

)

+

c

3

(

a

2

+

b

2

)

]

Now

a

3

(

b

2

+

c

2

)

+

b

3

(

c

2

+

a

2

)

+

c

3

(

a

2

+

b

2

)

=

a

2

b

2

(

a

+

b

)

+

b

2

c

2

(

b

+

c

)

+

c

2

a

2

(

a

+

c

)

=

a

2

b

2

(

1

−

c

)

+

b

2

c

2

(

1

−

a

)

+

c

2

a

2

(

1

−

b

)

=

a

2

b

2

+

b

2

c

2

+

c

2

a

2

−

(

a

2

b

2

c

+

b

2

c

2

a

+

c

2

a

2

b

)

=

−

1

12

−

a

b

c

(

a

b

+

b

c

+

c

a

)

=

−

1

12

−

1

6

⋅

(

−

1

2

)

=

0

So

a

5

+

b

5

+

c

5

=

6

−

0

=

6

Step-by-step explanation:

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