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3 years ago

Solve for x and y. y = 2/3x - 2 y = -x + 3

Mathematics

1 answer:

sesenic [268]3 years ago

5 0

Answer:

1. x= 3y+6/2

y=2/3x-2

2.x=−y+3

y=−x+3

Step-by-step explanation:

1. Let's solve for x.

y=2/3x-2

Step 1: Flip the equation.

2/3x-2=y

Step 2: Add 2 to both sides.

2/3x-2+2=y+2

2/3x=y+2

Step 3: Divide both sides by 2/3.

2/3x/2/3=y+2/2/3

x= 3y+6/2

Let's solve for y.

y=2/3x-2

2. Let's solve for x.

y=−x+3

Step 1: Flip the equation.

−x+3=y

Step 2: Add -3 to both sides.

−x+3+−3=y+−3

−x=y−3

Step 3: Divide both sides by -1.

-x/-1=y-3/-1

x=−y+3

Let's solve for y.

y=−x+3

here are two graphs if you need a better explanation

HOPE THIS HELPS

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3 years ago

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<h2>Answer:</h2>

Recursive equation for the pattern followed is given by,

$a_{n}=a_{n-1}+(n-1)^{2}$

<h2>Step-by-step explanation:</h2>

In the question,

The number of interaction for 1 child = 0

Number of interactions for 2 children = 1

Number of interactions for 3 children = 5

Number of interaction for 4 children = 14

So,

We need to find out the pattern for the recursive equation for the given conditions.

So,

We see that,

$a_{1}=0\\a_{2}=1\\a_{3}=5\\a_{4}=14\\$

Therefore, on checking, we observe that,

$a_{n}=a_{n-1}+(n-1)^{2}$

On checking the equation at the given values of 'n' of, 1, 2, 3 and 4.

At,

n = 1

$a_{n}=a_{n-1}+(n-1)^{2}\\a_{1}=a_{1-1}+(1-1)^{2}\\a_{1}=0+0=0\\a_{1}=0$

which is true.

At,

n = 2

$a_{n}=a_{n-1}+(n-1)^{2}\\a_{2}=a_{2-1}+(2-1)^{2}\\a_{2}=a_{1}+1\\a_{2}=1$

Which is also true.

At,

n = 3

$a_{n}=a_{n-1}+(n-1)^{2}\\a_{3}=a_{3-1}+(3-1)^{2}\\a_{3}=a_{2}+4\\a_{3}=5$

Which is true.

At,

n = 4

$a_{n}=a_{n-1}+(n-1)^{2}\\a_{4}=a_{4-1}+(4-1)^{2}\\a_{4}=a_{3}+9\\a_{4}=14$

This is also true at the given value of 'n'.

Therefore, the recursive equation for the pattern followed is given by,

$a_{n}=a_{n-1}+(n-1)^{2}$

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