Mister Exam
9х+4y=10; 7х+2y=0
The solution
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9*x + 4*y = 10
$$9 x + 4 y = 10$$
7*x + 2*y = 0
$$7 x + 2 y = 0$$
7*x + 2*y = 0
Detail solution
Given the system of equations
$$9 x + 4 y = 10$$
$$7 x + 2 y = 0$$
Let's express from equation 1 x
$$9 x + 4 y = 10$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$9 x = 10 - 4 y$$
$$9 x = 10 - 4 y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{9 x}{9} = \frac{10 - 4 y}{9}$$
$$x = \frac{10}{9} - \frac{4 y}{9}$$
Let's try the obtained element x to 2-th equation
$$7 x + 2 y = 0$$
We get:
$$2 y + 7 \left(\frac{10}{9} - \frac{4 y}{9}\right) = 0$$
$$\frac{70}{9} - \frac{10 y}{9} = 0$$
We move the free summand 70/9 from the left part to the right part performing the sign change
$$- \frac{10 y}{9} = - \frac{70}{9}$$
$$- \frac{10 y}{9} = - \frac{70}{9}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{10}{9} y}{- \frac{10}{9}} = - \frac{70}{\left(- \frac{10}{9}\right) 9}$$
$$y = 7$$
Because
$$x = \frac{10}{9} - \frac{4 y}{9}$$
then
$$x = \frac{10}{9} - \frac{28}{9}$$
$$x = -2$$
The answer:
$$x = -2$$
$$y = 7$$
$$9 x + 4 y = 10$$
$$7 x + 2 y = 0$$
Let's express from equation 1 x
$$9 x + 4 y = 10$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$9 x = 10 - 4 y$$
$$9 x = 10 - 4 y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{9 x}{9} = \frac{10 - 4 y}{9}$$
$$x = \frac{10}{9} - \frac{4 y}{9}$$
Let's try the obtained element x to 2-th equation
$$7 x + 2 y = 0$$
We get:
$$2 y + 7 \left(\frac{10}{9} - \frac{4 y}{9}\right) = 0$$
$$\frac{70}{9} - \frac{10 y}{9} = 0$$
We move the free summand 70/9 from the left part to the right part performing the sign change
$$- \frac{10 y}{9} = - \frac{70}{9}$$
$$- \frac{10 y}{9} = - \frac{70}{9}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{10}{9} y}{- \frac{10}{9}} = - \frac{70}{\left(- \frac{10}{9}\right) 9}$$
$$y = 7$$
Because
$$x = \frac{10}{9} - \frac{4 y}{9}$$
then
$$x = \frac{10}{9} - \frac{28}{9}$$
$$x = -2$$
The answer:
$$x = -2$$
$$y = 7$$
Rapid solution
$$x_{1} = -2$$
=
$$-2$$
=
$$y_{1} = 7$$
=
$$7$$
=
=
$$-2$$
=
-2
$$y_{1} = 7$$
=
$$7$$
=
7
Cramer's rule
$$9 x + 4 y = 10$$
$$7 x + 2 y = 0$$
We give the system of equations to the canonical form
$$9 x + 4 y = 10$$
$$7 x + 2 y = 0$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}9 x_{1} + 4 x_{2}\\7 x_{1} + 2 x_{2}\end{matrix}\right] = \left[\begin{matrix}10\\0\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}9 & 4\\7 & 2\end{matrix}\right] \right)} = -10$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}10 & 4\\0 & 2\end{matrix}\right] \right)}}{10} = -2$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}9 & 10\\7 & 0\end{matrix}\right] \right)}}{10} = 7$$
$$7 x + 2 y = 0$$
We give the system of equations to the canonical form
$$9 x + 4 y = 10$$
$$7 x + 2 y = 0$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}9 x_{1} + 4 x_{2}\\7 x_{1} + 2 x_{2}\end{matrix}\right] = \left[\begin{matrix}10\\0\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}9 & 4\\7 & 2\end{matrix}\right] \right)} = -10$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}10 & 4\\0 & 2\end{matrix}\right] \right)}}{10} = -2$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}9 & 10\\7 & 0\end{matrix}\right] \right)}}{10} = 7$$
Gaussian elimination
Given the system of equations
$$9 x + 4 y = 10$$
$$7 x + 2 y = 0$$
We give the system of equations to the canonical form
$$9 x + 4 y = 10$$
$$7 x + 2 y = 0$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}9 & 4 & 10\\7 & 2 & 0\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}9\\7\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}9 & 4 & 10\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}7 - \frac{7 \cdot 9}{9} & 2 - \frac{4 \cdot 7}{9} & - \frac{7 \cdot 10}{9}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{10}{9} & - \frac{70}{9}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}9 & 4 & 10\\0 & - \frac{10}{9} & - \frac{70}{9}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}4\\- \frac{10}{9}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{10}{9} & - \frac{70}{9}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}9 - \frac{\left(-18\right) 0}{5} & 4 - - -4 & 10 - - -28\end{matrix}\right] = \left[\begin{matrix}9 & 0 & -18\end{matrix}\right]$$
you get
$$\left[\begin{matrix}9 & 0 & -18\\0 & - \frac{10}{9} & - \frac{70}{9}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$9 x_{1} + 18 = 0$$
$$\frac{70}{9} - \frac{10 x_{2}}{9} = 0$$
We get the answer:
$$x_{1} = -2$$
$$x_{2} = 7$$
$$9 x + 4 y = 10$$
$$7 x + 2 y = 0$$
We give the system of equations to the canonical form
$$9 x + 4 y = 10$$
$$7 x + 2 y = 0$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}9 & 4 & 10\\7 & 2 & 0\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}9\\7\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}9 & 4 & 10\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}7 - \frac{7 \cdot 9}{9} & 2 - \frac{4 \cdot 7}{9} & - \frac{7 \cdot 10}{9}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{10}{9} & - \frac{70}{9}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}9 & 4 & 10\\0 & - \frac{10}{9} & - \frac{70}{9}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}4\\- \frac{10}{9}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{10}{9} & - \frac{70}{9}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}9 - \frac{\left(-18\right) 0}{5} & 4 - - -4 & 10 - - -28\end{matrix}\right] = \left[\begin{matrix}9 & 0 & -18\end{matrix}\right]$$
you get
$$\left[\begin{matrix}9 & 0 & -18\\0 & - \frac{10}{9} & - \frac{70}{9}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$9 x_{1} + 18 = 0$$
$$\frac{70}{9} - \frac{10 x_{2}}{9} = 0$$
We get the answer:
$$x_{1} = -2$$
$$x_{2} = 7$$