Mister Exam
(3x)/4-(y-3x)/2=-6; (y-x)/3-(1)/6=y/2
The solution
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3*x y - 3*x --- - ------- = -6 4 2
$$\frac{3 x}{4} - \frac{- 3 x + y}{2} = -6$$
y - x y ----- - 0.166666666666667 = - 3 2
$$\frac{- x + y}{3} - 0.166666666666667 = \frac{y}{2}$$
(-x + y)/3 - 0.166666666666667 = y/2
Detail solution
Given the system of equations
$$\frac{3 x}{4} - \frac{- 3 x + y}{2} = -6$$
$$\frac{- x + y}{3} - 0.166666666666667 = \frac{y}{2}$$
Let's express from equation 1 x
$$\frac{3 x}{4} - \frac{- 3 x + y}{2} = -6$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$\frac{3 x}{2} + \frac{3 x}{4} = \left(\frac{3 x}{2} + \frac{- 3 x + y}{2}\right) - 6$$
$$\frac{9 x}{4} = \frac{y}{2} - 6$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{\frac{9}{4} x}{\frac{9}{4}} = \frac{\frac{y}{2} - 6}{\frac{9}{4}}$$
$$x = \frac{2 y}{9} - \frac{8}{3}$$
Let's try the obtained element x to 2-th equation
$$\frac{- x + y}{3} - 0.166666666666667 = \frac{y}{2}$$
We get:
$$\frac{y - \left(\frac{2 y}{9} - \frac{8}{3}\right)}{3} - 0.166666666666667 = \frac{y}{2}$$
$$\frac{7 y}{27} + 0.722222222222222 = \frac{y}{2}$$
Let's move the summand with the variable y from the right part to the left part performing the sign change
$$- \frac{y}{2} + \left(\frac{7 y}{27} + 0.722222222222222\right) = 0$$
$$0.722222222222222 - \frac{13 y}{54} = 0$$
We move the free summand 0.722222222222222 from the left part to the right part performing the sign change
$$- \frac{13 y}{54} = -0.722222222222222$$
$$- \frac{13 y}{54} = -0.722222222222222$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{13}{54} y}{- \frac{13}{54}} = - \frac{0.722222222222222}{- \frac{13}{54}}$$
$$y = 3$$
Because
$$x = \frac{2 y}{9} - \frac{8}{3}$$
then
$$x = - \frac{8}{3} + \frac{2 \cdot 3}{9}$$
$$x = -2$$
The answer:
$$x = -2$$
$$y = 3$$
$$\frac{3 x}{4} - \frac{- 3 x + y}{2} = -6$$
$$\frac{- x + y}{3} - 0.166666666666667 = \frac{y}{2}$$
Let's express from equation 1 x
$$\frac{3 x}{4} - \frac{- 3 x + y}{2} = -6$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$\frac{3 x}{2} + \frac{3 x}{4} = \left(\frac{3 x}{2} + \frac{- 3 x + y}{2}\right) - 6$$
$$\frac{9 x}{4} = \frac{y}{2} - 6$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{\frac{9}{4} x}{\frac{9}{4}} = \frac{\frac{y}{2} - 6}{\frac{9}{4}}$$
$$x = \frac{2 y}{9} - \frac{8}{3}$$
Let's try the obtained element x to 2-th equation
$$\frac{- x + y}{3} - 0.166666666666667 = \frac{y}{2}$$
We get:
$$\frac{y - \left(\frac{2 y}{9} - \frac{8}{3}\right)}{3} - 0.166666666666667 = \frac{y}{2}$$
$$\frac{7 y}{27} + 0.722222222222222 = \frac{y}{2}$$
Let's move the summand with the variable y from the right part to the left part performing the sign change
$$- \frac{y}{2} + \left(\frac{7 y}{27} + 0.722222222222222\right) = 0$$
$$0.722222222222222 - \frac{13 y}{54} = 0$$
We move the free summand 0.722222222222222 from the left part to the right part performing the sign change
$$- \frac{13 y}{54} = -0.722222222222222$$
$$- \frac{13 y}{54} = -0.722222222222222$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{13}{54} y}{- \frac{13}{54}} = - \frac{0.722222222222222}{- \frac{13}{54}}$$
$$y = 3$$
Because
$$x = \frac{2 y}{9} - \frac{8}{3}$$
then
$$x = - \frac{8}{3} + \frac{2 \cdot 3}{9}$$
$$x = -2$$
The answer:
$$x = -2$$
$$y = 3$$
Rapid solution
$$x_{1} = -2$$
=
$$-2$$
=
$$y_{1} = 3$$
=
$$3$$
=
=
$$-2$$
=
-2
$$y_{1} = 3$$
=
$$3$$
=
3
Gaussian elimination
Given the system of equations
$$\frac{3 x}{4} - \frac{- 3 x + y}{2} = -6$$
$$\frac{- x + y}{3} - 0.166666666666667 = \frac{y}{2}$$
We give the system of equations to the canonical form
$$\frac{9 x}{4} - \frac{y}{2} = -6$$
$$- \frac{x}{3} - \frac{y}{6} = 0.166666666666667$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}\frac{9}{4} & - \frac{1}{2} & -6\\- \frac{1}{3} & - \frac{1}{6} & \frac{1}{6}\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}\frac{9}{4}\\- \frac{1}{3}\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}\frac{9}{4} & - \frac{1}{2} & -6\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{1}{3} - \frac{\left(-4\right) 9}{4 \cdot 27} & - \frac{1}{6} - - \frac{-2}{27} & \frac{1}{6} - - \frac{-8}{9}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{13}{54} & - \frac{13}{18}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}\frac{9}{4} & - \frac{1}{2} & -6\\0 & - \frac{13}{54} & - \frac{13}{18}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}- \frac{1}{2}\\- \frac{13}{54}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{13}{54} & - \frac{13}{18}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\frac{9}{4} - \frac{0 \cdot 27}{13} & - \frac{1}{2} - - \frac{1}{2} & -6 - - \frac{3}{2}\end{matrix}\right] = \left[\begin{matrix}\frac{9}{4} & 0 & - \frac{9}{2}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}\frac{9}{4} & 0 & - \frac{9}{2}\\0 & - \frac{13}{54} & - \frac{13}{18}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$\frac{9 x_{1}}{4} + \frac{9}{2} = 0$$
$$\frac{13}{18} - \frac{13 x_{2}}{54} = 0$$
We get the answer:
$$x_{1} = -2$$
$$x_{2} = 3$$
$$\frac{3 x}{4} - \frac{- 3 x + y}{2} = -6$$
$$\frac{- x + y}{3} - 0.166666666666667 = \frac{y}{2}$$
We give the system of equations to the canonical form
$$\frac{9 x}{4} - \frac{y}{2} = -6$$
$$- \frac{x}{3} - \frac{y}{6} = 0.166666666666667$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}\frac{9}{4} & - \frac{1}{2} & -6\\- \frac{1}{3} & - \frac{1}{6} & \frac{1}{6}\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}\frac{9}{4}\\- \frac{1}{3}\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}\frac{9}{4} & - \frac{1}{2} & -6\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}- \frac{1}{3} - \frac{\left(-4\right) 9}{4 \cdot 27} & - \frac{1}{6} - - \frac{-2}{27} & \frac{1}{6} - - \frac{-8}{9}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{13}{54} & - \frac{13}{18}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}\frac{9}{4} & - \frac{1}{2} & -6\\0 & - \frac{13}{54} & - \frac{13}{18}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}- \frac{1}{2}\\- \frac{13}{54}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{13}{54} & - \frac{13}{18}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}\frac{9}{4} - \frac{0 \cdot 27}{13} & - \frac{1}{2} - - \frac{1}{2} & -6 - - \frac{3}{2}\end{matrix}\right] = \left[\begin{matrix}\frac{9}{4} & 0 & - \frac{9}{2}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}\frac{9}{4} & 0 & - \frac{9}{2}\\0 & - \frac{13}{54} & - \frac{13}{18}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$\frac{9 x_{1}}{4} + \frac{9}{2} = 0$$
$$\frac{13}{18} - \frac{13 x_{2}}{54} = 0$$
We get the answer:
$$x_{1} = -2$$
$$x_{2} = 3$$
Cramer's rule
$$\frac{3 x}{4} - \frac{- 3 x + y}{2} = -6$$
$$\frac{- x + y}{3} - 0.166666666666667 = \frac{y}{2}$$
We give the system of equations to the canonical form
$$\frac{9 x}{4} - \frac{y}{2} = -6$$
$$- \frac{x}{3} - \frac{y}{6} = 0.166666666666667$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}\frac{9 x_{1}}{4} - \frac{x_{2}}{2}\\- 0.333333333333333 x_{1} - 0.166666666666667 x_{2}\end{matrix}\right] = \left[\begin{matrix}-6\\0.166666666666667\end{matrix}\right]$$
- this is the system of equations that has the form
A*x = B
Let´s find a solution of this matrix equations using Cramer´s rule:
Since the determinant of the matrix:
$$A = \operatorname{det}{\left(\left[\begin{matrix}\frac{9}{4} & - \frac{1}{2}\\-0.333333333333333 & -0.166666666666667\end{matrix}\right] \right)} = -0.541666666666667$$
, then
The root xi is obtained by dividing the determinant of the matrix Ai. by the determinant of the matrix A.
( Ai we get it by replacement in the matrix A i-th column with column B )
$$x_{1} = - 1.84615384615385 \operatorname{det}{\left(\left[\begin{matrix}-6 & - \frac{1}{2}\\0.166666666666667 & -0.166666666666667\end{matrix}\right] \right)} = -2$$
$$x_{2} = - 1.84615384615385 \operatorname{det}{\left(\left[\begin{matrix}\frac{9}{4} & -6\\-0.333333333333333 & 0.166666666666667\end{matrix}\right] \right)} = 3$$
$$\frac{- x + y}{3} - 0.166666666666667 = \frac{y}{2}$$
We give the system of equations to the canonical form
$$\frac{9 x}{4} - \frac{y}{2} = -6$$
$$- \frac{x}{3} - \frac{y}{6} = 0.166666666666667$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}\frac{9 x_{1}}{4} - \frac{x_{2}}{2}\\- 0.333333333333333 x_{1} - 0.166666666666667 x_{2}\end{matrix}\right] = \left[\begin{matrix}-6\\0.166666666666667\end{matrix}\right]$$
- this is the system of equations that has the form
A*x = B
Let´s find a solution of this matrix equations using Cramer´s rule:
Since the determinant of the matrix:
$$A = \operatorname{det}{\left(\left[\begin{matrix}\frac{9}{4} & - \frac{1}{2}\\-0.333333333333333 & -0.166666666666667\end{matrix}\right] \right)} = -0.541666666666667$$
, then
The root xi is obtained by dividing the determinant of the matrix Ai. by the determinant of the matrix A.
( Ai we get it by replacement in the matrix A i-th column with column B )
$$x_{1} = - 1.84615384615385 \operatorname{det}{\left(\left[\begin{matrix}-6 & - \frac{1}{2}\\0.166666666666667 & -0.166666666666667\end{matrix}\right] \right)} = -2$$
$$x_{2} = - 1.84615384615385 \operatorname{det}{\left(\left[\begin{matrix}\frac{9}{4} & -6\\-0.333333333333333 & 0.166666666666667\end{matrix}\right] \right)} = 3$$