Mister Exam
Other calculators
- Inequality Systems step by step
- Complex numbers step by step
- Differential equations Step by Step
- The system of differential equations in steps
-
How to use it?
- 3x+y=4; 5x−2y=14
-
Х+1=2у; 5ху+у^2-16=0
- 5x+3y-z=-7; 3x-4y+2z=10; x+y-3z=-5
- 2x+4y=30; X-5y=8
- The double integral of:
- 3x+y
- 3x+y
-
Identical expressions
- 3x+y= four ; 5x−2y= fourteen
- 3x plus y equally 4; 5x−2y equally 14
- 3x plus y equally four ; 5x−2y equally fourteen
-
Similar expressions
- 3x-y=4; 5x−2y=14
3x+y=4; 5x−2y=14
The solution
Detail solution
Given the system of equations
$$3 x + y = 4$$
$$5 x - 2 y = 14$$
Let's express from equation 1 x
$$3 x + y = 4$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$3 x = 4 - y$$
$$3 x = 4 - y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{3 x}{3} = \frac{4 - y}{3}$$
$$x = \frac{4}{3} - \frac{y}{3}$$
Let's try the obtained element x to 2-th equation
$$5 x - 2 y = 14$$
We get:
$$- 2 y + 5 \left(\frac{4}{3} - \frac{y}{3}\right) = 14$$
$$\frac{20}{3} - \frac{11 y}{3} = 14$$
We move the free summand 20/3 from the left part to the right part performing the sign change
$$- \frac{11 y}{3} = - \frac{20}{3} + 14$$
$$- \frac{11 y}{3} = \frac{22}{3}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{11}{3} y}{- \frac{11}{3}} = \frac{22}{\left(- \frac{11}{3}\right) 3}$$
$$y = -2$$
Because
$$x = \frac{4}{3} - \frac{y}{3}$$
then
$$x = \frac{4}{3} - - \frac{2}{3}$$
$$x = 2$$
The answer:
$$x = 2$$
$$y = -2$$
$$3 x + y = 4$$
$$5 x - 2 y = 14$$
Let's express from equation 1 x
$$3 x + y = 4$$
Let's move the summand with the variable y from the left part to the right part performing the sign change
$$3 x = 4 - y$$
$$3 x = 4 - y$$
Let's divide both parts of the equation by the multiplier of x
$$\frac{3 x}{3} = \frac{4 - y}{3}$$
$$x = \frac{4}{3} - \frac{y}{3}$$
Let's try the obtained element x to 2-th equation
$$5 x - 2 y = 14$$
We get:
$$- 2 y + 5 \left(\frac{4}{3} - \frac{y}{3}\right) = 14$$
$$\frac{20}{3} - \frac{11 y}{3} = 14$$
We move the free summand 20/3 from the left part to the right part performing the sign change
$$- \frac{11 y}{3} = - \frac{20}{3} + 14$$
$$- \frac{11 y}{3} = \frac{22}{3}$$
Let's divide both parts of the equation by the multiplier of y
$$\frac{\left(-1\right) \frac{11}{3} y}{- \frac{11}{3}} = \frac{22}{\left(- \frac{11}{3}\right) 3}$$
$$y = -2$$
Because
$$x = \frac{4}{3} - \frac{y}{3}$$
then
$$x = \frac{4}{3} - - \frac{2}{3}$$
$$x = 2$$
The answer:
$$x = 2$$
$$y = -2$$
Rapid solution
$$x_{1} = 2$$
=
$$2$$
=
$$y_{1} = -2$$
=
$$-2$$
=
=
$$2$$
=
2
$$y_{1} = -2$$
=
$$-2$$
=
-2
Cramer's rule
$$3 x + y = 4$$
$$5 x - 2 y = 14$$
We give the system of equations to the canonical form
$$3 x + y = 4$$
$$5 x - 2 y = 14$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 x_{1} + x_{2}\\5 x_{1} - 2 x_{2}\end{matrix}\right] = \left[\begin{matrix}4\\14\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}3 & 1\\5 & -2\end{matrix}\right] \right)} = -11$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}4 & 1\\14 & -2\end{matrix}\right] \right)}}{11} = 2$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & 4\\5 & 14\end{matrix}\right] \right)}}{11} = -2$$
$$5 x - 2 y = 14$$
We give the system of equations to the canonical form
$$3 x + y = 4$$
$$5 x - 2 y = 14$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 x_{1} + x_{2}\\5 x_{1} - 2 x_{2}\end{matrix}\right] = \left[\begin{matrix}4\\14\end{matrix}\right]$$
- 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:
$$A = \operatorname{det}{\left(\left[\begin{matrix}3 & 1\\5 & -2\end{matrix}\right] \right)} = -11$$
, 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 )
$$x_{1} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}4 & 1\\14 & -2\end{matrix}\right] \right)}}{11} = 2$$
$$x_{2} = - \frac{\operatorname{det}{\left(\left[\begin{matrix}3 & 4\\5 & 14\end{matrix}\right] \right)}}{11} = -2$$
Gaussian elimination
Given the system of equations
$$3 x + y = 4$$
$$5 x - 2 y = 14$$
We give the system of equations to the canonical form
$$3 x + y = 4$$
$$5 x - 2 y = 14$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & 1 & 4\\5 & -2 & 14\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\5\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & 1 & 4\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{3 \cdot 5}{3} & -2 + \frac{\left(-1\right) 5}{3} & 14 - \frac{4 \cdot 5}{3}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{11}{3} & \frac{22}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 1 & 4\\0 & - \frac{11}{3} & \frac{22}{3}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}1\\- \frac{11}{3}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{11}{3} & \frac{22}{3}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{\left(-3\right) 0}{11} & 1 - - -1 & 4 - \frac{\left(-3\right) 22}{3 \cdot 11}\end{matrix}\right] = \left[\begin{matrix}3 & 0 & 6\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & 6\\0 & - \frac{11}{3} & \frac{22}{3}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} - 6 = 0$$
$$- \frac{11 x_{2}}{3} - \frac{22}{3} = 0$$
We get the answer:
$$x_{1} = 2$$
$$x_{2} = -2$$
$$3 x + y = 4$$
$$5 x - 2 y = 14$$
We give the system of equations to the canonical form
$$3 x + y = 4$$
$$5 x - 2 y = 14$$
Rewrite the system of linear equations as the matrix form
$$\left[\begin{matrix}3 & 1 & 4\\5 & -2 & 14\end{matrix}\right]$$
In 1 -th column
$$\left[\begin{matrix}3\\5\end{matrix}\right]$$
let’s convert all the elements, except
1 -th element into zero.
- To do this, let’s take 1 -th line
$$\left[\begin{matrix}3 & 1 & 4\end{matrix}\right]$$
,
and subtract it from other lines:
From 2 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}5 - \frac{3 \cdot 5}{3} & -2 + \frac{\left(-1\right) 5}{3} & 14 - \frac{4 \cdot 5}{3}\end{matrix}\right] = \left[\begin{matrix}0 & - \frac{11}{3} & \frac{22}{3}\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 1 & 4\\0 & - \frac{11}{3} & \frac{22}{3}\end{matrix}\right]$$
In 2 -th column
$$\left[\begin{matrix}1\\- \frac{11}{3}\end{matrix}\right]$$
let’s convert all the elements, except
2 -th element into zero.
- To do this, let’s take 2 -th line
$$\left[\begin{matrix}0 & - \frac{11}{3} & \frac{22}{3}\end{matrix}\right]$$
,
and subtract it from other lines:
From 1 -th line. Let’s subtract it from this line:
$$\left[\begin{matrix}3 - \frac{\left(-3\right) 0}{11} & 1 - - -1 & 4 - \frac{\left(-3\right) 22}{3 \cdot 11}\end{matrix}\right] = \left[\begin{matrix}3 & 0 & 6\end{matrix}\right]$$
you get
$$\left[\begin{matrix}3 & 0 & 6\\0 & - \frac{11}{3} & \frac{22}{3}\end{matrix}\right]$$
It is almost ready, all we have to do is to find variables, solving the elementary equations:
$$3 x_{1} - 6 = 0$$
$$- \frac{11 x_{2}}{3} - \frac{22}{3} = 0$$
We get the answer:
$$x_{1} = 2$$
$$x_{2} = -2$$