Solve absolute value equation or indicate the equation has no solution.

4|1-3/4x|+7=10

4|1-3/4x|+7=10

kloseyq
2021-12-14
Answered

Solve absolute value equation or indicate the equation has no solution.

4|1-3/4x|+7=10

4|1-3/4x|+7=10

You can still ask an expert for help

Durst37

Answered 2021-12-15
Author has **37** answers

Step 1

Consider the following equation:

$4|1-\frac{3}{4x}|+7=10$

$4|1-\frac{3}{4x}|=3$

$|1-\frac{3}{4x}|=\frac{3}{4}$

Step 2

$1-\frac{3}{4x}=\frac{3}{4}\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}1-\frac{3}{4x}=-\frac{3}{4}$

$\frac{4x-3}{4x}=\frac{3}{4}\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}\frac{4x-3}{4x}=-\frac{3}{4}$

4x-3=3x or 4x-3=-3x

$x=3\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}x=\frac{3}{7}$

Step 3

Hence, the solution is

$x=3,\frac{3}{7}$

Consider the following equation:

Step 2

4x-3=3x or 4x-3=-3x

Step 3

Hence, the solution is

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Solve, please:

1) $\frac{8}{x+1}=\frac{4}{3}$

2) $\frac{x+3}{3x-6}>0$

3) $2x+3=\frac{x}{4}$

4) $\frac{1}{{x}^{2}-4}\le 0$

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10) $\frac{12{x}^{3}+16{x}^{2}-3x-4}{8{x}^{3}+12{x}^{2}+10x+15}<0$

1) $\frac{8}{x+1}=\frac{4}{3}$

2) $\frac{x+3}{3x-6}>0$

3) $2x+3=\frac{x}{4}$

4) $\frac{1}{{x}^{2}-4}\le 0$

5) $\frac{4}{x}+\frac{1}{3x}=9$

6) $\frac{x+32}{x+6}\le 6$

7) $\frac{4}{x}+\frac{1}{{x}^{2}}=\frac{1}{5{x}^{2}}$

8) $1+\frac{2}{x+1}<\frac{2}{x}$

9) $\frac{x-3}{2x+10}+2x-12=\frac{{x}^{2}+3x-18}{2x+10}$

10) $\frac{12{x}^{3}+16{x}^{2}-3x-4}{8{x}^{3}+12{x}^{2}+10x+15}<0$

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Solution of a system of differential equations

${x}^{\prime}=-y+x{y}^{2}\phantom{\rule{0ex}{0ex}}{y}^{\prime}=4x-4{x}^{2}y$

${x}^{\prime}=-y+x{y}^{2}\phantom{\rule{0ex}{0ex}}{y}^{\prime}=4x-4{x}^{2}y$proposed solution to this which uses the fact that

$\frac{dy}{dx}=\frac{{y}^{\prime}}{{x}^{\prime}}=\frac{4x(1-xy)}{y(xy-1)}=-\frac{4x}{y}$

and then solves this as a separable equation, which is straight forward enough. What I am struggling to understand is why one chooses to calculate $\frac{dy}{dx}$. It seems to me quite arbitrary to calculate this derivative. What is the connection between this derivative and the original system?

${x}^{\prime}=-y+x{y}^{2}\phantom{\rule{0ex}{0ex}}{y}^{\prime}=4x-4{x}^{2}y$

${x}^{\prime}=-y+x{y}^{2}\phantom{\rule{0ex}{0ex}}{y}^{\prime}=4x-4{x}^{2}y$proposed solution to this which uses the fact that

$\frac{dy}{dx}=\frac{{y}^{\prime}}{{x}^{\prime}}=\frac{4x(1-xy)}{y(xy-1)}=-\frac{4x}{y}$

and then solves this as a separable equation, which is straight forward enough. What I am struggling to understand is why one chooses to calculate $\frac{dy}{dx}$. It seems to me quite arbitrary to calculate this derivative. What is the connection between this derivative and the original system?