Mister Exam
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How to use it?
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Equation (x-5)^2=(x-8)^2
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Equation k^2-4*k+13=0
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Equation -x+2x-8=0
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- -4*x+18*y=-14
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Identical expressions
- |(two x- five)/ three |=|(3x+ four)/2|
- module of (2x minus 5) divide by 3| equally |(3x plus 4) divide by 2|
- module of (two x minus five) divide by three | equally |(3x plus four) divide by 2|
- |2x-5/3|=|3x+4/2|
- |(2x-5) divide by 3|=|(3x+4) divide by 2|
-
Similar expressions
- |(2x-5)/3|=|(3x-4)/2|
- |(2x+5)/3|=|(3x+4)/2|
|(2x-5)/3|=|(3x+4)/2| equation
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
|2*x - 5| |3*x + 4| |-------| = |-------| | 3 | | 2 |
$$\left|{\frac{2 x - 5}{3}}\right| = \left|{\frac{3 x + 4}{2}}\right|$$
Detail solution
For every modulo expressions in the equation
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.
1.
$$\frac{2 x}{3} - \frac{5}{3} \geq 0$$
$$\frac{3 x}{2} + 2 \geq 0$$
or
$$\frac{5}{2} \leq x \wedge x < \infty$$
we get the equation
$$\left(\frac{2 x}{3} - \frac{5}{3}\right) - \left(\frac{3 x}{2} + 2\right) = 0$$
after simplifying we get
$$- \frac{5 x}{6} - \frac{11}{3} = 0$$
the solution in this interval:
$$x_{1} = - \frac{22}{5}$$
but x1 not in the inequality interval
2.
$$\frac{2 x}{3} - \frac{5}{3} \geq 0$$
$$\frac{3 x}{2} + 2 < 0$$
The inequality system has no solutions, see the next condition
3.
$$\frac{2 x}{3} - \frac{5}{3} < 0$$
$$\frac{3 x}{2} + 2 \geq 0$$
or
$$- \frac{4}{3} \leq x \wedge x < \frac{5}{2}$$
we get the equation
$$\left(\frac{5}{3} - \frac{2 x}{3}\right) - \left(\frac{3 x}{2} + 2\right) = 0$$
after simplifying we get
$$- \frac{13 x}{6} - \frac{1}{3} = 0$$
the solution in this interval:
$$x_{2} = - \frac{2}{13}$$
4.
$$\frac{2 x}{3} - \frac{5}{3} < 0$$
$$\frac{3 x}{2} + 2 < 0$$
or
$$-\infty < x \wedge x < - \frac{4}{3}$$
we get the equation
$$\left(\frac{5}{3} - \frac{2 x}{3}\right) - \left(- \frac{3 x}{2} - 2\right) = 0$$
after simplifying we get
$$\frac{5 x}{6} + \frac{11}{3} = 0$$
the solution in this interval:
$$x_{3} = - \frac{22}{5}$$
The final answer:
$$x_{1} = - \frac{2}{13}$$
$$x_{2} = - \frac{22}{5}$$
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.
1.
$$\frac{2 x}{3} - \frac{5}{3} \geq 0$$
$$\frac{3 x}{2} + 2 \geq 0$$
or
$$\frac{5}{2} \leq x \wedge x < \infty$$
we get the equation
$$\left(\frac{2 x}{3} - \frac{5}{3}\right) - \left(\frac{3 x}{2} + 2\right) = 0$$
after simplifying we get
$$- \frac{5 x}{6} - \frac{11}{3} = 0$$
the solution in this interval:
$$x_{1} = - \frac{22}{5}$$
but x1 not in the inequality interval
2.
$$\frac{2 x}{3} - \frac{5}{3} \geq 0$$
$$\frac{3 x}{2} + 2 < 0$$
The inequality system has no solutions, see the next condition
3.
$$\frac{2 x}{3} - \frac{5}{3} < 0$$
$$\frac{3 x}{2} + 2 \geq 0$$
or
$$- \frac{4}{3} \leq x \wedge x < \frac{5}{2}$$
we get the equation
$$\left(\frac{5}{3} - \frac{2 x}{3}\right) - \left(\frac{3 x}{2} + 2\right) = 0$$
after simplifying we get
$$- \frac{13 x}{6} - \frac{1}{3} = 0$$
the solution in this interval:
$$x_{2} = - \frac{2}{13}$$
4.
$$\frac{2 x}{3} - \frac{5}{3} < 0$$
$$\frac{3 x}{2} + 2 < 0$$
or
$$-\infty < x \wedge x < - \frac{4}{3}$$
we get the equation
$$\left(\frac{5}{3} - \frac{2 x}{3}\right) - \left(- \frac{3 x}{2} - 2\right) = 0$$
after simplifying we get
$$\frac{5 x}{6} + \frac{11}{3} = 0$$
the solution in this interval:
$$x_{3} = - \frac{22}{5}$$
The final answer:
$$x_{1} = - \frac{2}{13}$$
$$x_{2} = - \frac{22}{5}$$
Rapid solution
[src]
x1 = -22/5
$$x_{1} = - \frac{22}{5}$$
x2 = -2/13
$$x_{2} = - \frac{2}{13}$$
x2 = -2/13
Sum and product of roots
[src]
sum
-22/5 - 2/13
$$- \frac{22}{5} - \frac{2}{13}$$
=
-296 ----- 65
$$- \frac{296}{65}$$
product
-22*(-2) -------- 5*13
$$- \frac{-44}{65}$$
=
44 -- 65
$$\frac{44}{65}$$
44/65