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