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3 years ago

Solve the equation: 7x = 42. A. x = – 7 B. x = – 6 C. x = 6 D. x = 7

Mathematics

2 answers:

sergeinik [125]3 years ago

5 0

7x=42
Divide both sides by 7
42÷7 =6
6 x 7 = 42
C. x = 6

jenyasd209 [6]3 years ago

3 0

7x = 42

7x/7 = 42/7
x = 42/7
x = 6

Answer: x = 6

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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 =

lara [203]

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

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

7 0

3 years ago

Read 2 more answers

Todd begins piano practice at 4:15pm.and ends at 5:50pm.How long dose he practice

xxTIMURxx [149]

Answer:

If He began at 4:15 and ended at 5:50 then that would be 95 minutes

7 0

3 years ago

The ratio of areas of two triangles with common base is 4:5. The height of smaller triangle is 6cm. Find height of greater trian

8_murik_8 [283]

Answer:

Height of the greater triangle is 7.5 cm (Approximately).

Step-by-step explanation:

Given that

The ratio of two triangles with common base = 4:5

Let they be 4x cm² and 5x cm²

The area of the bigger triangle = 5x cm²

The area of the smaller triangle = 4x cm²

also,

Given that

The base of two triangles are same

Let the base of the two triangles be b cm

Height of the smaller triangle = 6 cm

Now,

Area of the triangle = (1/2)×base×height sq.units

⇛Area of the smaller triangle

⇛(1/2)×b×6 = 4

⇛6b/2 = 4

⇛3×b = 4

⇛b= 4/3 cm

.°. b = 4/3 cm

Now,

The length of the base = 4/3 cm

Area of the greater triangle = 5 cm²

⇛(1/2)×(4/3)×h = 5

⇛(4/6)×h = 5

⇛(2/3)×h = 5

⇛h = 5/(2/3)

= 5×3/2

= 15/2

⇛h = 7.5 cm (Approximately)

<u>Answer</u><u>:</u> Therefore, Height of the greater triangle is 7.5 cm.

Please let me know if you have any other questions.

6 0

3 years ago