Mister Exam
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-
How to use it?
- 4x+у=3; 6x-5y=11
- 3x-7y=32; X=-5y-4
-
х^2-xy=-8; y^2-xy=24
- log7(36−y^2)=log7(36−a^2x^2); x^2+y^2=2x+6y
-
Identical expressions
- 4x+у= three ; 6x-5y= eleven
- 4x plus у equally 3; 6x minus 5y equally 11
- 4x plus у equally three ; 6x minus 5y equally eleven
-
Similar expressions
- 4x+у=3; 6x+5y=11
- 4x-у=3; 6x-5y=11
4x+у=3; 6x-5y=11
The solution
Detail solution
Given the system of equations
$$4 x + y = 3$$
$$6 x - 5 y = 11$$
Let's express from equation 1 x
$$4 x + y = 3$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$4 x = 3 - y$$
$$4 x = 3 - y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{4 x}{4} = \frac{3 - y}{4}$$
$$x = \frac{3}{4} - \frac{y}{4}$$
Let's try the obtained element x to 2-th equation
$$6 x - 5 y = 11$$
We get:
$$- 5 y + 6 \left(\frac{3}{4} - \frac{y}{4}\right) = 11$$
$$\frac{9}{2} - \frac{13 y}{2} = 11$$
We move the free summand 9/2 from the left part to the right part performing the sign change
$$- \frac{13 y}{2} = - \frac{9}{2} + 11$$
$$- \frac{13 y}{2} = \frac{13}{2}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{13}{2} y}{- \frac{13}{2}} = \frac{13}{\left(- \frac{13}{2}\right) 2}$$
$$y = -1$$
Because
$$x = \frac{3}{4} - \frac{y}{4}$$
then
$$x = \frac{3}{4} - - \frac{1}{4}$$
$$x = 1$$
The answer:
$$x = 1$$
$$y = -1$$
$$4 x + y = 3$$
$$6 x - 5 y = 11$$
Let's express from equation 1 x
$$4 x + y = 3$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$4 x = 3 - y$$
$$4 x = 3 - y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{4 x}{4} = \frac{3 - y}{4}$$
$$x = \frac{3}{4} - \frac{y}{4}$$
Let's try the obtained element x to 2-th equation
$$6 x - 5 y = 11$$
We get:
$$- 5 y + 6 \left(\frac{3}{4} - \frac{y}{4}\right) = 11$$
$$\frac{9}{2} - \frac{13 y}{2} = 11$$
We move the free summand 9/2 from the left part to the right part performing the sign change
$$- \frac{13 y}{2} = - \frac{9}{2} + 11$$
$$- \frac{13 y}{2} = \frac{13}{2}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{13}{2} y}{- \frac{13}{2}} = \frac{13}{\left(- \frac{13}{2}\right) 2}$$
$$y = -1$$
Because
$$x = \frac{3}{4} - \frac{y}{4}$$
then
$$x = \frac{3}{4} - - \frac{1}{4}$$
$$x = 1$$
The answer:
$$x = 1$$
$$y = -1$$
Rapid solution
$$x_{1} = 1$$
=
$$1$$
=
$$y_{1} = -1$$
=
$$-1$$
=
=
$$1$$
=
1
$$y_{1} = -1$$
=
$$-1$$
=
-1
Gaussian elimination
Given the system of equations
$$4 x + y = 3$$
$$6 x - 5 y = 11$$
We give the system of equations to the canonical form
$$4 x + y = 3$$
$$6 x - 5 y = 11$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}4 & 1 & 3\\6 & -5 & 11\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}4\\6\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}4 & 1 & 3\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}6 - \frac{3 \cdot 4}{2} & -5 + \frac{\left(-1\right) 3}{2} & 11 - \frac{3 \cdot 3}{2}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{13}{2} & \frac{13}{2}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & 1 & 3\\0 & - \frac{13}{2} & \frac{13}{2}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}1\\- \frac{13}{2}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{13}{2} & \frac{13}{2}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}4 - \frac{\left(-2\right) 0}{13} & 1 - - -1 & 3 - \frac{\left(-2\right) 13}{2 \cdot 13}\end{matrix}\right] = \left[\begin{matrix}4 & 0 & 4\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & 0 & 4\\0 & - \frac{13}{2} & \frac{13}{2}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$4 x_{1} - 4 = 0$$
$$- \frac{13 x_{2}}{2} - \frac{13}{2} = 0$$
We get the answer:
$$x_{1} = 1$$
$$x_{2} = -1$$
$$4 x + y = 3$$
$$6 x - 5 y = 11$$
We give the system of equations to the canonical form
$$4 x + y = 3$$
$$6 x - 5 y = 11$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}4 & 1 & 3\\6 & -5 & 11\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}4\\6\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}4 & 1 & 3\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}6 - \frac{3 \cdot 4}{2} & -5 + \frac{\left(-1\right) 3}{2} & 11 - \frac{3 \cdot 3}{2}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{13}{2} & \frac{13}{2}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & 1 & 3\\0 & - \frac{13}{2} & \frac{13}{2}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}1\\- \frac{13}{2}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{13}{2} & \frac{13}{2}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}4 - \frac{\left(-2\right) 0}{13} & 1 - - -1 & 3 - \frac{\left(-2\right) 13}{2 \cdot 13}\end{matrix}\right] = \left[\begin{matrix}4 & 0 & 4\end{matrix}\right]$$
you get
$$\left[\begin{matrix}4 & 0 & 4\\0 & - \frac{13}{2} & \frac{13}{2}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$4 x_{1} - 4 = 0$$
$$- \frac{13 x_{2}}{2} - \frac{13}{2} = 0$$
We get the answer:
$$x_{1} = 1$$
$$x_{2} = -1$$
Cramer's rule
$$4 x + y = 3$$
$$6 x - 5 y = 11$$
We give the system of equations to the canonical form
$$4 x + y = 3$$
$$6 x - 5 y = 11$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}4 x_{1} + x_{2}\\6 x_{1} - 5 x_{2}\end{matrix}\right] = \left[\begin{matrix}3\\11\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}4 & 1\\6 & -5\end{matrix}\right] \right)} = -26$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & 1\\11 & -5\end{matrix}\right] \right)}}{26} = 1$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}4 & 3\\6 & 11\end{matrix}\right] \right)}}{26} = -1$$
$$6 x - 5 y = 11$$
We give the system of equations to the canonical form
$$4 x + y = 3$$
$$6 x - 5 y = 11$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}4 x_{1} + x_{2}\\6 x_{1} - 5 x_{2}\end{matrix}\right] = \left[\begin{matrix}3\\11\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}4 & 1\\6 & -5\end{matrix}\right] \right)} = -26$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & 1\\11 & -5\end{matrix}\right] \right)}}{26} = 1$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}4 & 3\\6 & 11\end{matrix}\right] \right)}}{26} = -1$$