Mister Exam
3x-5y=7; 6x-10y=14
The solution
You have entered
[src]
3*x - 5*y = 7
$$3 x - 5 y = 7$$
6*x - 10*y = 14
$$6 x - 10 y = 14$$
6*x - 10*y = 14
Detail solution
Given the system of equations
$$3 x - 5 y = 7$$
$$6 x - 10 y = 14$$
Let's express from equation 1 x
$$3 x - 5 y = 7$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$3 x = 5 y + 7$$
$$3 x = 5 y + 7$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{3 x}{3} = \frac{5 y + 7}{3}$$
$$x = \frac{5 y}{3} + \frac{7}{3}$$
Let's try the obtained element x to 2-th equation
$$6 x - 10 y = 14$$
We get:
$$- 10 y + 6 \left(\frac{5 y}{3} + \frac{7}{3}\right) = 14$$
so
is always executed
$$3 x - 5 y = 7$$
$$6 x - 10 y = 14$$
Let's express from equation 1 x
$$3 x - 5 y = 7$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$3 x = 5 y + 7$$
$$3 x = 5 y + 7$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{3 x}{3} = \frac{5 y + 7}{3}$$
$$x = \frac{5 y}{3} + \frac{7}{3}$$
Let's try the obtained element x to 2-th equation
$$6 x - 10 y = 14$$
We get:
$$- 10 y + 6 \left(\frac{5 y}{3} + \frac{7}{3}\right) = 14$$
so
is always executed
Rapid solution
$$x_{1} = \frac{5 y}{3} + \frac{7}{3}$$
=
$$\frac{5 y}{3} + \frac{7}{3}$$
=
=
$$\frac{5 y}{3} + \frac{7}{3}$$
=
2.33333333333333 + 1.66666666666667*y
Gaussian elimination
Given the system of equations
$$3 x - 5 y = 7$$
$$6 x - 10 y = 14$$
We give the system of equations to the canonical form
$$3 x - 5 y = 7$$
$$6 x - 10 y = 14$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & -5 & 7\\6 & -10 & 14\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\6\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & -5 & 7\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}6 - 2 \cdot 3 & -10 - - 10 & 14 - 2 \cdot 7\end{matrix}\right] = \left[\begin{matrix}0 & 0 & 0\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & -5 & 7\\0 & 0 & 0\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\0\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & -5 & 7\end{matrix}\right]$$
,
and subtract it from other lines:
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} - 5 x_{2} - 7 = 0$$
$$0 - 0 = 0$$
We get the answer:
$$x_{1} = \frac{5 x_{2}}{3} + \frac{7}{3}$$
where x2 - the free variables
$$3 x - 5 y = 7$$
$$6 x - 10 y = 14$$
We give the system of equations to the canonical form
$$3 x - 5 y = 7$$
$$6 x - 10 y = 14$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & -5 & 7\\6 & -10 & 14\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\6\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & -5 & 7\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}6 - 2 \cdot 3 & -10 - - 10 & 14 - 2 \cdot 7\end{matrix}\right] = \left[\begin{matrix}0 & 0 & 0\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & -5 & 7\\0 & 0 & 0\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\0\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & -5 & 7\end{matrix}\right]$$
,
and subtract it from other lines:
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} - 5 x_{2} - 7 = 0$$
$$0 - 0 = 0$$
We get the answer:
$$x_{1} = \frac{5 x_{2}}{3} + \frac{7}{3}$$
where x2 - the free variables