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2 years ago

Find the solution to the system of equations x + y = 9 and x - y = -1.

Mathematics

2 answers:

Galina-37 [17]2 years ago

6 0

Answer:

(4, 5 )

Step-by-step explanation:

x + y = 9 → (1)

x - y = - 1 → (2)

adding the 2 equations term by term will eliminate y

2x + 0 = 8

2x = 8 ( divide both sides by 2 )

x = 4

substitute x = 4 into either of the 2 equations and solve for y

substituting into (1)

4 + y = 9 ( subtract 4 from both sides )

y = 5

solution is (4, 5 )

DerKrebs [107]2 years ago

3 0

Answer: x = 4, y = 5

Step-by-Step Explanation:

=> x + y = 9 (Eq. 1)
=> x - y = -1 (Eq. 2)

Lets Solve by Elimination Method.

Adding Eq. 1 and Eq. 2, we get :-

2x = 9 - 1
2x = 8
x = 8/2
=> x = 4

Therefore, x = 4

Substitute value of ‘x’ in Eq. 1 :-

x + y = 9
4 + y = 9
y = 9 - 4
=> y = 5

Therefore, y = 5

Hence, x = 4 and y = 5 (4, 5)

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Step-by-step explanation:

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3 years ago

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3 years ago

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3 years ago

How to do the inverse of a 3x3 matrix gaussian elimination.

nata0808 [166]

As an example, let's invert the matrix

$\begin{bmatrix}-3&2&1\\2&1&1\\1&1&1\end{bmatrix}$

We construct the augmented matrix,

$\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 2 & 1 & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array} \right]$

On this augmented matrix, we perform row operations in such a way as to transform the matrix on the left side into the identity matrix, and the matrix on the right will be the inverse that we want to find.

Now we can carry out Gaussian elimination.

• Eliminate the column 1 entry in row 2.

Combine 2 times row 1 with 3 times row 2 :

2 (-3, 2, 1, 1, 0, 0) + 3 (2, 1, 1, 0, 1, 0)

= (-6, 4, 2, 2, 0, 0) + (6, 3, 3, 0, 3, 0)

= (0, 7, 5, 2, 3, 0)

which changes the augmented matrix to

$\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array} \right]$

• Eliminate the column 1 entry in row 3.

Using the new aug. matrix, combine row 1 and 3 times row 3 :

(-3, 2, 1, 1, 0, 0) + 3 (1, 1, 1, 0, 0, 1)

= (-3, 2, 1, 1, 0, 0) + (3, 3, 3, 0, 0, 3)

= (0, 5, 4, 1, 0, 3)

$\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 5 & 4 & 1 & 0 & 3 \end{array} \right]$

• Eliminate the column 2 entry in row 3.

Combine -5 times row 2 and 7 times row 3 :

-5 (0, 7, 5, 2, 3, 0) + 7 (0, 5, 4, 1, 0, 3)

= (0, -35, -25, -10, -15, 0) + (0, 35, 28, 7, 0, 21)

= (0, 0, 3, -3, -15, 21)

$\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 0 & 3 & -3 & -15 & 21 \end{array} \right]$

• Multiply row 3 by 1/3 :

$\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]$

• Eliminate the column 3 entry in row 2.

Combine row 2 and -5 times row 3 :

(0, 7, 5, 2, 3, 0) - 5 (0, 0, 1, -1, -5, 7)

= (0, 7, 5, 2, 3, 0) + (0, 0, -5, 5, 25, -35)

= (0, 7, 0, 7, 28, -35)

$\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 0 & 7 & 28 & -35 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]$

• Multiply row 2 by 1/7 :

$\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]$

• Eliminate the column 2 and 3 entries in row 1.

Combine row 1, -2 times row 2, and -1 times row 3 :

(-3, 2, 1, 1, 0, 0) - 2 (0, 1, 0, 1, 4, -5) - (0, 0, 1, -1, -5, 7)

= (-3, 2, 1, 1, 0, 0) + (0, -2, 0, -2, -8, 10) + (0, 0, -1, 1, 5, -7)

= (-3, 0, 0, 0, -3, 3)

$\left[ \begin{array}{ccc|ccc} -3 & 0 & 0 & 0 & -3 & 3 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]$

• Multiply row 1 by -1/3 :

$\left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & 1 & -1 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]$

So, the inverse of our matrix is

$\begin{bmatrix}-3&2&1\\2&1&1\\1&1&1\end{bmatrix}^{-1} = \begin{bmatrix}0&1&-1\\1&4&-5\\-1&-5&7\end{bmatrix}$

6 0

2 years ago

In a sample of n = 6 scores, 5 of the scores are each above the mean by one point. Where is the 6th score located relative to th

den301095 [7]

The mean of a dataset is the sum of all data elements divided by the count of the elements.

The location of the 6th score relative to the mean is 5 points below the mean

Let:

$\bar x \to$ Mean

$a \to$ 5 scores

$b \to$ 6th scores

Given that:

$n = 6$

The 5 scores that are 1 above the mean implies that:

$a = \bar x + 1$

The mean of a dataset is calculated using:

$\bar x = \frac{\sum x}{n}$

So, we have:

$\bar x =\frac{5a + b}{6}$

$\bar x =\frac{5(\bar x + 1) + b}{6}$

Open brackets

$\bar x =\frac{5\bar x + 5 + b}{6}$

Multiply both sides by 6

$6\bar x =5\bar x + 5 + b$

Make b the subject

$b = 6\bar x -5\bar x - 5$

$b = \bar x - 5$

This means that the 6th score is 5 points below the mean