4x-8y=12

-x+2y=-3

Maiclubk
2021-02-04
Answered

Solve the system of equations using Gaussian elimination

4x-8y=12

-x+2y=-3

4x-8y=12

-x+2y=-3

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Obiajulu

Answered 2021-02-05
Author has **98** answers

Step 1

The system of equations is given by,

4x-8y=12

-x+2y=-3

Step 2

Find the solution of system of equations, it is need to write an equivalent matrix equation AX=B .

$\left[\begin{array}{cc}4& -8\\ -1& 2\end{array}\right]\cdot \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}12\\ -3\end{array}\right]$

Find the inverse matrix by using Gauss-Jordan elimination method.

That is, apply the row equivalent operations to the matrix.

Consider the matrix$A=\left[\begin{array}{cc}4& -8\\ -1& 2\end{array}\right]$ .

Form of an augment matrix, in order to find the inverse matrix.

That is, form an matrix contains of A on the left side and the$2\times 2$ identify matrix on the right side.

$\left[\begin{array}{ccc}4& -8& 12\\ -1& 2& -3\end{array}\right]$

Step 3

Divide the first row by 4 as follows.

$\left[\begin{array}{ccc}1& -2& 3\\ -1& 2& -3\end{array}\right]$ New row $1=\frac{1}{4}$ row 1

Add the first and second rows as follows.

$\left[\begin{array}{ccc}1& -2& 3\\ 0& 0& 0\end{array}\right]$ New row 2 = row 1 + row 2

Thus, the system of equations has a solution set is$\{x-2y=3.$ .

Step 4

Answer:

The system of equations has a solution set$\{x-2y=3.$ .

The system of equations is given by,

4x-8y=12

-x+2y=-3

Step 2

Find the solution of system of equations, it is need to write an equivalent matrix equation AX=B .

Find the inverse matrix by using Gauss-Jordan elimination method.

That is, apply the row equivalent operations to the matrix.

Consider the matrix

Form of an augment matrix, in order to find the inverse matrix.

That is, form an matrix contains of A on the left side and the

Step 3

Divide the first row by 4 as follows.

Add the first and second rows as follows.

Thus, the system of equations has a solution set is

Step 4

Answer:

The system of equations has a solution set

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I am trying to solve this equation but I am stuck little bit right now. This is how I did it:

${7}^{2x}\cdot {4}^{x-2}={11}^{x}\phantom{\rule{0ex}{0ex}}\text{log both sides}\phantom{\rule{0ex}{0ex}}\mathrm{log}({7}^{2x}\cdot {4}^{x-2})=\mathrm{log}({11}^{x})\phantom{\rule{0ex}{0ex}}\mathrm{log}({7}^{2x})+\mathrm{log}({4}^{x-2})=\mathrm{log}({11}^{x})\phantom{\rule{0ex}{0ex}}2x\mathrm{log}(7)+(x-2)\mathrm{log}(4)=x\mathrm{log}(11)\phantom{\rule{0ex}{0ex}}\text{I got this far}\phantom{\rule{0ex}{0ex}}$

Can somebody give me idea, am I on right track, and if I am, how can I solve it further. Thanks.

I am trying to solve this equation but I am stuck little bit right now. This is how I did it:

${7}^{2x}\cdot {4}^{x-2}={11}^{x}\phantom{\rule{0ex}{0ex}}\text{log both sides}\phantom{\rule{0ex}{0ex}}\mathrm{log}({7}^{2x}\cdot {4}^{x-2})=\mathrm{log}({11}^{x})\phantom{\rule{0ex}{0ex}}\mathrm{log}({7}^{2x})+\mathrm{log}({4}^{x-2})=\mathrm{log}({11}^{x})\phantom{\rule{0ex}{0ex}}2x\mathrm{log}(7)+(x-2)\mathrm{log}(4)=x\mathrm{log}(11)\phantom{\rule{0ex}{0ex}}\text{I got this far}\phantom{\rule{0ex}{0ex}}$

Can somebody give me idea, am I on right track, and if I am, how can I solve it further. Thanks.

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So I've got an inequality: $\mathrm{ln}(2x-5)>\mathrm{ln}(7-2x)$ and I attempt to solve by doing the following:

$\frac{\mathrm{ln}(2x)}{\mathrm{ln}(5)}>\frac{\mathrm{ln}(7)}{\mathrm{ln}(2x)}$

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I thought I could multiply by ${e}^{2}$ to get rid of ${\mathrm{ln}}^{2}$ but I guess not...I thought they were inverses??? So anyways, my solution seems WAY off as the solution to the problem is: $3<x<\frac{7}{2}$

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