Mister Exam
(х-у)(х+у)-х(х+10)=у(5-у)+15; (х+1)²+(у-1)²=(х+4)²+(у+2)²-18
The solution
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(x - y)*(x + y) - x*(x + 10) = y*(5 - y) + 15
$$- x \left(x + 10\right) + \left(x - y\right) \left(x + y\right) = y \left(5 - y\right) + 15$$
2 2 2 2 (x + 1) + (y - 1) = (x + 4) + (y + 2) - 18
$$\left(x + 1\right)^{2} + \left(y - 1\right)^{2} = \left(\left(x + 4\right)^{2} + \left(y + 2\right)^{2}\right) - 18$$
(x + 1)^2 + (y - 1)^2 = (x + 4)^2 + (y + 2)^2 - 18
Detail solution
Given the system of equations
$$- x \left(x + 10\right) + \left(x - y\right) \left(x + y\right) = y \left(5 - y\right) + 15$$
$$\left(x + 1\right)^{2} + \left(y - 1\right)^{2} = \left(\left(x + 4\right)^{2} + \left(y + 2\right)^{2}\right) - 18$$
Let's express from equation 1 x
$$- x \left(x + 10\right) + \left(x - y\right) \left(x + y\right) = y \left(5 - y\right) + 15$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$x^{2} - x \left(x + 10\right) = \left(x^{2} - \left(x - y\right) \left(x + y\right)\right) + \left(y \left(5 - y\right) + 15\right)$$
$$x^{2} - x \left(x + 10\right) = x^{2} + y \left(5 - y\right) - \left(x - y\right) \left(x + y\right) + 15$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{x^{2} - x \left(x + 10\right)}{-10} = \frac{x^{2} + y \left(5 - y\right) - \left(x - y\right) \left(x + y\right) + 15}{-10}$$
$$- \frac{x^{2}}{10} + \frac{x \left(x + 10\right)}{10} = - \frac{x^{2}}{10} - \frac{y \left(5 - y\right)}{10} + \frac{\left(x - y\right) \left(x + y\right)}{10} - \frac{3}{2}$$
Let's try the obtained element x to 2-th equation
$$\left(x + 1\right)^{2} + \left(y - 1\right)^{2} = \left(\left(x + 4\right)^{2} + \left(y + 2\right)^{2}\right) - 18$$
We get:
$$\left(y - 1\right)^{2} + \left(\left(- \frac{x^{2}}{10} - \frac{y \left(5 - y\right)}{10} + \frac{\left(x - y\right) \left(x + y\right)}{10} - \frac{3}{2}\right) + 1\right)^{2} = \left(\left(y + 2\right)^{2} + \left(\left(- \frac{x^{2}}{10} - \frac{y \left(5 - y\right)}{10} + \frac{\left(x - y\right) \left(x + y\right)}{10} - \frac{3}{2}\right) + 4\right)^{2}\right) - 18$$
$$\frac{5 y^{2}}{4} - \frac{3 y}{2} + \frac{5}{4} = \frac{5 y^{2}}{4} + \frac{3 y}{2} - \frac{31}{4}$$
Let's move the summand with the variable y from the right part to the left part performing the sign change
$$\left(- \frac{5 y^{2}}{4} - \frac{3 y}{2}\right) + \left(\frac{5 y^{2}}{4} - \frac{3 y}{2} + \frac{5}{4}\right) = - \frac{31}{4}$$
$$\frac{5}{4} - 3 y = - \frac{31}{4}$$
We move the free summand 5/4 from the left part to the right part performing the sign change
$$- 3 y = - \frac{31}{4} - \frac{5}{4}$$
$$- 3 y = -9$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) 3 y}{-3} = - \frac{9}{-3}$$
$$y = 3$$
Because
$$- \frac{x^{2}}{10} + \frac{x \left(x + 10\right)}{10} = - \frac{x^{2}}{10} - \frac{y \left(5 - y\right)}{10} + \frac{\left(x - y\right) \left(x + y\right)}{10} - \frac{3}{2}$$
then
$$x = - \frac{x^{2}}{10} + \frac{\left(x + 3\right) \left(x - 3\right)}{10} - \frac{3}{2} - \frac{3 \left(5 - 3\right)}{10}$$
$$x = - \frac{x^{2}}{10} + \frac{\left(x - 3\right) \left(x + 3\right)}{10} - \frac{21}{10}$$
The answer:
$$x = - \frac{x^{2}}{10} + \frac{\left(x - 3\right) \left(x + 3\right)}{10} - \frac{21}{10}$$
$$y = 3$$
$$- x \left(x + 10\right) + \left(x - y\right) \left(x + y\right) = y \left(5 - y\right) + 15$$
$$\left(x + 1\right)^{2} + \left(y - 1\right)^{2} = \left(\left(x + 4\right)^{2} + \left(y + 2\right)^{2}\right) - 18$$
Let's express from equation 1 x
$$- x \left(x + 10\right) + \left(x - y\right) \left(x + y\right) = y \left(5 - y\right) + 15$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$x^{2} - x \left(x + 10\right) = \left(x^{2} - \left(x - y\right) \left(x + y\right)\right) + \left(y \left(5 - y\right) + 15\right)$$
$$x^{2} - x \left(x + 10\right) = x^{2} + y \left(5 - y\right) - \left(x - y\right) \left(x + y\right) + 15$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{x^{2} - x \left(x + 10\right)}{-10} = \frac{x^{2} + y \left(5 - y\right) - \left(x - y\right) \left(x + y\right) + 15}{-10}$$
$$- \frac{x^{2}}{10} + \frac{x \left(x + 10\right)}{10} = - \frac{x^{2}}{10} - \frac{y \left(5 - y\right)}{10} + \frac{\left(x - y\right) \left(x + y\right)}{10} - \frac{3}{2}$$
Let's try the obtained element x to 2-th equation
$$\left(x + 1\right)^{2} + \left(y - 1\right)^{2} = \left(\left(x + 4\right)^{2} + \left(y + 2\right)^{2}\right) - 18$$
We get:
$$\left(y - 1\right)^{2} + \left(\left(- \frac{x^{2}}{10} - \frac{y \left(5 - y\right)}{10} + \frac{\left(x - y\right) \left(x + y\right)}{10} - \frac{3}{2}\right) + 1\right)^{2} = \left(\left(y + 2\right)^{2} + \left(\left(- \frac{x^{2}}{10} - \frac{y \left(5 - y\right)}{10} + \frac{\left(x - y\right) \left(x + y\right)}{10} - \frac{3}{2}\right) + 4\right)^{2}\right) - 18$$
$$\frac{5 y^{2}}{4} - \frac{3 y}{2} + \frac{5}{4} = \frac{5 y^{2}}{4} + \frac{3 y}{2} - \frac{31}{4}$$
Let's move the summand with the variable y from the right part to the left part performing the sign change
$$\left(- \frac{5 y^{2}}{4} - \frac{3 y}{2}\right) + \left(\frac{5 y^{2}}{4} - \frac{3 y}{2} + \frac{5}{4}\right) = - \frac{31}{4}$$
$$\frac{5}{4} - 3 y = - \frac{31}{4}$$
We move the free summand 5/4 from the left part to the right part performing the sign change
$$- 3 y = - \frac{31}{4} - \frac{5}{4}$$
$$- 3 y = -9$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) 3 y}{-3} = - \frac{9}{-3}$$
$$y = 3$$
Because
$$- \frac{x^{2}}{10} + \frac{x \left(x + 10\right)}{10} = - \frac{x^{2}}{10} - \frac{y \left(5 - y\right)}{10} + \frac{\left(x - y\right) \left(x + y\right)}{10} - \frac{3}{2}$$
then
$$x = - \frac{x^{2}}{10} + \frac{\left(x + 3\right) \left(x - 3\right)}{10} - \frac{3}{2} - \frac{3 \left(5 - 3\right)}{10}$$
$$x = - \frac{x^{2}}{10} + \frac{\left(x - 3\right) \left(x + 3\right)}{10} - \frac{21}{10}$$
The answer:
$$x = - \frac{x^{2}}{10} + \frac{\left(x - 3\right) \left(x + 3\right)}{10} - \frac{21}{10}$$
$$y = 3$$
Rapid solution
$$x_{1} = -3$$
=
$$-3$$
=
$$y_{1} = 3$$
=
$$3$$
=
=
$$-3$$
=
-3
$$y_{1} = 3$$
=
$$3$$
=
3
Gaussian elimination
Given the system of equations
$$- x \left(x + 10\right) + \left(x - y\right) \left(x + y\right) = y \left(5 - y\right) + 15$$
$$\left(x + 1\right)^{2} + \left(y - 1\right)^{2} = \left(\left(x + 4\right)^{2} + \left(y + 2\right)^{2}\right) - 18$$
We give the system of equations to the canonical form
$$- 10 x - 5 y = 15$$
$$- 6 x - 6 y = 0$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}-10 & -5 & 15\\-6 & -6 & 0\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}-10\\-6\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}-10 & -5 & 15\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-6 - - 6 & -6 - - 3 & - \frac{3 \cdot 15}{5}\end{matrix}\right] = \left[\begin{matrix}0 & -3 & -9\end{matrix}\right]$$
you get
$$\left[\begin{matrix}-10 & -5 & 15\\0 & -3 & -9\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-5\\-3\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & -3 & -9\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-10 - \frac{0 \cdot 5}{3} & -5 - - 5 & 15 - - 15\end{matrix}\right] = \left[\begin{matrix}-10 & 0 & 30\end{matrix}\right]$$
you get
$$\left[\begin{matrix}-10 & 0 & 30\\0 & -3 & -9\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$- 10 x_{1} - 30 = 0$$
$$9 - 3 x_{2} = 0$$
We get the answer:
$$x_{1} = -3$$
$$x_{2} = 3$$
$$- x \left(x + 10\right) + \left(x - y\right) \left(x + y\right) = y \left(5 - y\right) + 15$$
$$\left(x + 1\right)^{2} + \left(y - 1\right)^{2} = \left(\left(x + 4\right)^{2} + \left(y + 2\right)^{2}\right) - 18$$
We give the system of equations to the canonical form
$$- 10 x - 5 y = 15$$
$$- 6 x - 6 y = 0$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}-10 & -5 & 15\\-6 & -6 & 0\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}-10\\-6\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}-10 & -5 & 15\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-6 - - 6 & -6 - - 3 & - \frac{3 \cdot 15}{5}\end{matrix}\right] = \left[\begin{matrix}0 & -3 & -9\end{matrix}\right]$$
you get
$$\left[\begin{matrix}-10 & -5 & 15\\0 & -3 & -9\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}-5\\-3\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & -3 & -9\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}-10 - \frac{0 \cdot 5}{3} & -5 - - 5 & 15 - - 15\end{matrix}\right] = \left[\begin{matrix}-10 & 0 & 30\end{matrix}\right]$$
you get
$$\left[\begin{matrix}-10 & 0 & 30\\0 & -3 & -9\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$- 10 x_{1} - 30 = 0$$
$$9 - 3 x_{2} = 0$$
We get the answer:
$$x_{1} = -3$$
$$x_{2} = 3$$
Cramer's rule
$$- x \left(x + 10\right) + \left(x - y\right) \left(x + y\right) = y \left(5 - y\right) + 15$$
$$\left(x + 1\right)^{2} + \left(y - 1\right)^{2} = \left(\left(x + 4\right)^{2} + \left(y + 2\right)^{2}\right) - 18$$
We give the system of equations to the canonical form
$$- 10 x - 5 y = 15$$
$$- 6 x - 6 y = 0$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}- 10 x_{1} - 5 x_{2}\\- 6 x_{1} - 6 x_{2}\end{matrix}\right] = \left[\begin{matrix}15\\0\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}-10 & -5\\-6 & -6\end{matrix}\right] \right)} = 30$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}15 & -5\\0 & -6\end{matrix}\right] \right)}}{30} = -3$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}-10 & 15\\-6 & 0\end{matrix}\right] \right)}}{30} = 3$$
$$\left(x + 1\right)^{2} + \left(y - 1\right)^{2} = \left(\left(x + 4\right)^{2} + \left(y + 2\right)^{2}\right) - 18$$
We give the system of equations to the canonical form
$$- 10 x - 5 y = 15$$
$$- 6 x - 6 y = 0$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}- 10 x_{1} - 5 x_{2}\\- 6 x_{1} - 6 x_{2}\end{matrix}\right] = \left[\begin{matrix}15\\0\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}-10 & -5\\-6 & -6\end{matrix}\right] \right)} = 30$$
, 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 )
$$x_{1} = \frac{\operatorname{det}{\left(\left[\begin{matrix}15 & -5\\0 & -6\end{matrix}\right] \right)}}{30} = -3$$
$$x_{2} = \frac{\operatorname{det}{\left(\left[\begin{matrix}-10 & 15\\-6 & 0\end{matrix}\right] \right)}}{30} = 3$$