Mister Exam
-3x+5y=9; 11x-3y=-13
The solution
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-3*x + 5*y = 9
$$- 3 x + 5 y = 9$$
11*x - 3*y = -13
$$11 x - 3 y = -13$$
11*x - 3*y = -13
Detail solution
Given the system of equations
$$- 3 x + 5 y = 9$$
$$11 x - 3 y = -13$$
Let's express from equation 1 x
$$- 3 x + 5 y = 9$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$- 3 x = 9 - 5 y$$
$$- 3 x = 9 - 5 y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{\left(-1\right) 3 x}{-3} = \frac{9 - 5 y}{-3}$$
$$x = \frac{5 y}{3} - 3$$
Let's try the obtained element x to 2-th equation
$$11 x - 3 y = -13$$
We get:
$$- 3 y + 11 \left(\frac{5 y}{3} - 3\right) = -13$$
$$\frac{46 y}{3} - 33 = -13$$
We move the free summand -33 from the left part to the right part performing the sign change
$$\frac{46 y}{3} = -13 + 33$$
$$\frac{46 y}{3} = 20$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{46}{3} y}{\frac{46}{3}} = \frac{20}{\frac{46}{3}}$$
$$y = \frac{30}{23}$$
Because
$$x = \frac{5 y}{3} - 3$$
then
$$x = -3 + \frac{5 \cdot 30}{3 \cdot 23}$$
$$x = - \frac{19}{23}$$
The answer:
$$x = - \frac{19}{23}$$
$$y = \frac{30}{23}$$
$$- 3 x + 5 y = 9$$
$$11 x - 3 y = -13$$
Let's express from equation 1 x
$$- 3 x + 5 y = 9$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$- 3 x = 9 - 5 y$$
$$- 3 x = 9 - 5 y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{\left(-1\right) 3 x}{-3} = \frac{9 - 5 y}{-3}$$
$$x = \frac{5 y}{3} - 3$$
Let's try the obtained element x to 2-th equation
$$11 x - 3 y = -13$$
We get:
$$- 3 y + 11 \left(\frac{5 y}{3} - 3\right) = -13$$
$$\frac{46 y}{3} - 33 = -13$$
We move the free summand -33 from the left part to the right part performing the sign change
$$\frac{46 y}{3} = -13 + 33$$
$$\frac{46 y}{3} = 20$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\frac{46}{3} y}{\frac{46}{3}} = \frac{20}{\frac{46}{3}}$$
$$y = \frac{30}{23}$$
Because
$$x = \frac{5 y}{3} - 3$$
then
$$x = -3 + \frac{5 \cdot 30}{3 \cdot 23}$$
$$x = - \frac{19}{23}$$
The answer:
$$x = - \frac{19}{23}$$
$$y = \frac{30}{23}$$
Rapid solution
$$x_{1} = - \frac{19}{23}$$
=
$$- \frac{19}{23}$$
=
$$y_{1} = \frac{30}{23}$$
=
$$\frac{30}{23}$$
=
=
$$- \frac{19}{23}$$
=
-0.826086956521739
$$y_{1} = \frac{30}{23}$$
=
$$\frac{30}{23}$$
=
1.30434782608696
Gaussian elimination
Given the system of equations
$$- 3 x + 5 y = 9$$
$$11 x - 3 y = -13$$
We give the system of equations to the canonical form
$$- 3 x + 5 y = 9$$
$$11 x - 3 y = -13$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}-3 & 5 & 9\\11 & -3 & -13\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}-3\\11\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}-3 & 5 & 9\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}11 - - -11 & -3 - \frac{\left(-11\right) 5}{3} & -13 - \frac{\left(-11\right) 9}{3}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{46}{3} & 20\end{matrix}\right]$$
you get
$$\left[\begin{matrix}-3 & 5 & 9\\0 & \frac{46}{3} & 20\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}5\\\frac{46}{3}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{46}{3} & 20\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-3 - \frac{0 \cdot 15}{46} & 5 - \frac{15 \cdot 46}{3 \cdot 46} & 9 - \frac{15 \cdot 20}{46}\end{matrix}\right] = \left[\begin{matrix}-3 & 0 & \frac{57}{23}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}-3 & 0 & \frac{57}{23}\\0 & \frac{46}{3} & 20\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$- 3 x_{1} - \frac{57}{23} = 0$$
$$\frac{46 x_{2}}{3} - 20 = 0$$
We get the answer:
$$x_{1} = - \frac{19}{23}$$
$$x_{2} = \frac{30}{23}$$
$$- 3 x + 5 y = 9$$
$$11 x - 3 y = -13$$
We give the system of equations to the canonical form
$$- 3 x + 5 y = 9$$
$$11 x - 3 y = -13$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}-3 & 5 & 9\\11 & -3 & -13\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}-3\\11\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}-3 & 5 & 9\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}11 - - -11 & -3 - \frac{\left(-11\right) 5}{3} & -13 - \frac{\left(-11\right) 9}{3}\end{matrix}\right] = \left[\begin{matrix}0 & \frac{46}{3} & 20\end{matrix}\right]$$
you get
$$\left[\begin{matrix}-3 & 5 & 9\\0 & \frac{46}{3} & 20\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}5\\\frac{46}{3}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & \frac{46}{3} & 20\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-3 - \frac{0 \cdot 15}{46} & 5 - \frac{15 \cdot 46}{3 \cdot 46} & 9 - \frac{15 \cdot 20}{46}\end{matrix}\right] = \left[\begin{matrix}-3 & 0 & \frac{57}{23}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}-3 & 0 & \frac{57}{23}\\0 & \frac{46}{3} & 20\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$- 3 x_{1} - \frac{57}{23} = 0$$
$$\frac{46 x_{2}}{3} - 20 = 0$$
We get the answer:
$$x_{1} = - \frac{19}{23}$$
$$x_{2} = \frac{30}{23}$$
Cramer's rule
$$- 3 x + 5 y = 9$$
$$11 x - 3 y = -13$$
We give the system of equations to the canonical form
$$- 3 x + 5 y = 9$$
$$11 x - 3 y = -13$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}- 3 x_{1} + 5 x_{2}\\11 x_{1} - 3 x_{2}\end{matrix}\right] = \left[\begin{matrix}9\\-13\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}-3 & 5\\11 & -3\end{matrix}\right] \right)} = -46$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}9 & 5\\-13 & -3\end{matrix}\right] \right)}}{46} = - \frac{19}{23}$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}-3 & 9\\11 & -13\end{matrix}\right] \right)}}{46} = \frac{30}{23}$$
$$11 x - 3 y = -13$$
We give the system of equations to the canonical form
$$- 3 x + 5 y = 9$$
$$11 x - 3 y = -13$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}- 3 x_{1} + 5 x_{2}\\11 x_{1} - 3 x_{2}\end{matrix}\right] = \left[\begin{matrix}9\\-13\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}-3 & 5\\11 & -3\end{matrix}\right] \right)} = -46$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}9 & 5\\-13 & -3\end{matrix}\right] \right)}}{46} = - \frac{19}{23}$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}-3 & 9\\11 & -13\end{matrix}\right] \right)}}{46} = \frac{30}{23}$$
Numerical answer
[src]
x1 = -0.8260869565217391 y1 = 1.304347826086957
x1 = -0.8260869565217391 y1 = 1.304347826086957