# Solving a system of equations of first degree with two unknowns

This calculator solves a system of equations of first degree with two unknowns.

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To solve the equations such as this
$\left\{{ax+by=c\atop a_1x+b_1y=c_1$
there are common formulas
$x=\frac{b_1c-bc_1}{ab_1-a_1b}$,
$y=\frac{ac_1-a_1c}{ab_1-a_1b}$
These formulas are easy to remember, if you introduce the concept of determinant of the second order as
$\left|\begin{matrix} p & q \\ r & s \end{matrix} \right| = ps-rq$
Then the solution of the equations can be written as
$x=\frac{\left|\begin{matrix} c & b \\ c_1 & b_1 \end{matrix} \right|}{\left|\begin{matrix} a & b \\ a_1 & b_1 \end{matrix} \right|}\\y=\frac{\left|\begin{matrix} a & c \\ a_1 & c_1 \end{matrix} \right|}{\left|\begin{matrix} a & b \\ a_1 & b_1 \end{matrix} \right|}$
ie each of the unknowns is equal to the fraction, the denominator of which is the determinant consisting of the coefficients of the unknowns and the numerator is obtained from this determinant by replacement of coefficients of the corresponding unknown to the absolute term.

There are three different solitions possible:

1. Coefficients at unknowns in equations are disproportionat
$\frac{a}{a_1}<>\frac{b}{b_1}$
in this case the system of equations has a single solution, presented by formula

2. Coefficients at unknowns are proportional, but disproportionate to free terms
$\frac{a}{a_1}=\frac{b}{b_1}<>\frac{c}{c_1}$
in this case the system of equations has no solutions, because we have here contradictory equations.

3. All coefficients of equations are proportional
$\frac{a}{a_1}=\frac{b}{b_1}=\frac{c}{c_1}$
The system of equations has an infinite set of solutions, because we have actually one equation instead of two.

Calculator:

#### Solving a system of equations of first degree with two uknknowns

Digits after the decimal point: 2
x

y

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