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Murrr4er [49]
2 years ago
15

25 points!!!

Mathematics
1 answer:
professor190 [17]2 years ago
3 0

Answer:

x+y=12

Step-by-step explanation:

First, substitute 3y for the spots with x:

2(3y) + 2y = 24

6y+ 2y = 24

8y= 24

y= 3

Then sub in 3 in the y place:

x= 3(3)

x= 9

Then plug in the x and y values in the equation:

9 + 3 =12

Good Luck!!!

plz mark me Brainliest!

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Step 1. Add 6 to both sides

a ≥ 1 + 6

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a ≥ 7

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What is the equation of the line written in general form?
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Answer:

option 2 or y + x -2 = 0

Step-by-step explanation:

Given a line on a graph, it is easiest to start with point-slope form.

y = mx + b

where m is the slope (rise/run) and b is the y-axis intercept.

From the plot m = -2/2 = -1 and b = 2 SO

y = mx + b

y = -x  + 2

Then the general form is

y + x -2 = 0

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Solve the following systems of equations using the matrix method: a. 3x1 + 2x2 + 4x3 = 5 2x1 + 5x2 + 3x3 = 17 7x1 + 2x2 + 2x3 =
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Answer:

a. The solutions are

\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]=\begin{pmatrix}\frac{11}{13}\\ \frac{50}{13}\\ -\frac{17}{13}\end{pmatrix}

b. The solutions are

\left[\begin{array}{c}x&y&z\\\end{array}\right]=\begin{pmatrix}\frac{54}{235}\\ \frac{6}{47}\\ \frac{24}{235}\end{pmatrix}

c. The solutions are

\left[\begin{array}{c}x_1&x_2&x_3&x_4\\\end{array}\right]=\begin{pmatrix}\frac{22}{9}\\ \frac{164}{9}\\ \frac{139}{9}\\ -\frac{37}{3}\end{pmatrix}

Step-by-step explanation:

Solving a system of linear equations using matrix method, we may define a system of equations with the same number of equations as variables as:

A\cdot X=B

where X is the matrix representing the variables of the system,  B is the matrix representing the constants, and A is the coefficient matrix.

Then the solution is this:

X=A^{-1}B

a. Given the system:

3x_1 + 2x_2 + 4x_3 = 5 \\2x_1 + 5x_2 + 3x_3 = 17 \\7x_1 + 2x_2 + 2x_3 = 11

The coefficient matrix is:

A=\left[\begin{array}{ccc}3&2&4\\2&5&3\\7&2&2\end{array}\right]

The variable matrix is:

X=\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]

The constant matrix is:

B=\left[\begin{array}{c}5&17&11\\\end{array}\right]

First, we need to find the inverse of the A matrix. To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

So, augment the matrix with identity matrix:

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

This matrix can be transformed by a sequence of elementary row operations to the matrix

\left[ \begin{array}{ccc|ccc}1&0&0&- \frac{2}{39}&- \frac{2}{39}&\frac{7}{39} \\\\ 0&1&0&- \frac{17}{78}&\frac{11}{39}&\frac{1}{78} \\\\ 0&0&1&\frac{31}{78}&- \frac{4}{39}&- \frac{11}{78}\end{array}\right]

And the inverse of the A matrix is

A^{-1}=\left[ \begin{array}{ccc} - \frac{2}{39} & - \frac{2}{39} & \frac{7}{39} \\\\ - \frac{17}{78} & \frac{11}{39} & \frac{1}{78} \\\\ \frac{31}{78} & - \frac{4}{39} & - \frac{11}{78} \end{array} \right]

Next, multiply A^ {-1} by B

X=A^{-1}\cdot B

\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]=\left[ \begin{array}{ccc} - \frac{2}{39} & - \frac{2}{39} & \frac{7}{39} \\\\ - \frac{17}{78} & \frac{11}{39} & \frac{1}{78} \\\\ \frac{31}{78} & - \frac{4}{39} & - \frac{11}{78} \end{array} \right] \cdot \left[\begin{array}{c}5&17&11\end{array}\right]

\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]=\begin{pmatrix}-\frac{2}{39}&-\frac{2}{39}&\frac{7}{39}\\ -\frac{17}{78}&\frac{11}{39}&\frac{1}{78}\\ \frac{31}{78}&-\frac{4}{39}&-\frac{11}{78}\end{pmatrix}\begin{pmatrix}5\\ 17\\ 11\end{pmatrix}=\begin{pmatrix}\frac{11}{13}\\ \frac{50}{13}\\ -\frac{17}{13}\end{pmatrix}

The solutions are

\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]=\begin{pmatrix}\frac{11}{13}\\ \frac{50}{13}\\ -\frac{17}{13}\end{pmatrix}

b. To solve this system of equations

x -y - z = 0 \\30x + 40y = 12 \\30x + 50z = 12

The coefficient matrix is:

A=\left[\begin{array}{ccc}1&-1&-1\\30&40&0\\30&0&50\end{array}\right]

The variable matrix is:

X=\left[\begin{array}{c}x&y&z\\\end{array}\right]

The constant matrix is:

B=\left[\begin{array}{c}0&12&12\\\end{array}\right]

The inverse of the A matrix is

A^{-1}=\left[ \begin{array}{ccc} \frac{20}{47} & \frac{1}{94} & \frac{2}{235} \\\\ - \frac{15}{47} & \frac{4}{235} & - \frac{3}{470} \\\\ - \frac{12}{47} & - \frac{3}{470} & \frac{7}{470} \end{array} \right]

The solutions are

\left[\begin{array}{c}x&y&z\\\end{array}\right]=\begin{pmatrix}\frac{54}{235}\\ \frac{6}{47}\\ \frac{24}{235}\end{pmatrix}

c. To solve this system of equations

4x_1 + 2x_2 + x_3 + 5x_4 = 0 \\3x_1 + x_2 + 4x_3 + 7x_4 = 1\\ 2x_1 + 3x_2 + x_3 + 6x_4 = 1 \\3x_1 + x_2 + x_3 + 3x_4 = 4\\

The coefficient matrix is:

A=\left[\begin{array}{cccc}4&2&1&5\\3&1&4&7\\2&3&1&6\\3&1&1&3\end{array}\right]

The variable matrix is:

X=\left[\begin{array}{c}x_1&x_2&x_3&x_4\\\end{array}\right]

The constant matrix is:

B=\left[\begin{array}{c}0&1&1&4\\\end{array}\right]

The inverse of the A matrix is

A^{-1}=\left[ \begin{array}{cccc} - \frac{1}{9} & - \frac{1}{9} & - \frac{1}{9} & \frac{2}{3} \\\\ - \frac{32}{9} & - \frac{5}{9} & \frac{13}{9} & \frac{13}{3} \\\\ - \frac{28}{9} & - \frac{1}{9} & \frac{8}{9} & \frac{11}{3} \\\\ \frac{7}{3} & \frac{1}{3} & - \frac{2}{3} & -3 \end{array} \right]

The solutions are

\left[\begin{array}{c}x_1&x_2&x_3&x_4\\\end{array}\right]=\begin{pmatrix}\frac{22}{9}\\ \frac{164}{9}\\ \frac{139}{9}\\ -\frac{37}{3}\end{pmatrix}

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