Cramer’s Rule to solve (if possible) the system of linear equations. 56x1−x2=−20 34x1−72x2=−51

Cramers rule The solution system {ax+by=scx+dy=t is given by x=DxD,y=DyDwhereD=[abcd] Dx[sbtd] And Dy=[asct]providedD≠0 Consider the system of equations {56x1−x2=−20−−−(1)43x1−72x2=−51−−−(2) D=[56−143−72]=−1912 similarly Dx=[−20−1−51−72] [56−2043−51]=−956 Note that D≠q0 By the above formula x19−1912 =−12 SIMILARLY y=DyD −956−1912 =10 thus the solution of the system is x=−12 and y=10.This solution can be written as ordered pair (x,y)=(−12,10)

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High School
Algebra
Algebra I
Forms of linear equations

Josalynn

Answered question

2021-02-21

Cramer’s Rule to solve (if possible) the system of linear equations.

\frac{5}{6} x_{1} - x_{2} = - 20

\frac{3}{4} x_{1} - \frac{7}{2} x_{2} = - 51

Answer & Explanation

aprovard

Skilled2021-02-22Added 94 answers

Cramers rule
The solution system ${\begin{cases} a x + b y = s \\ c x + d y = t \end{cases}$ is given by
$x = \frac{D_{x}}{D}, y = \frac{D_{y}}{D} w h e r e D = [\begin{matrix} a & b \\ c & d \end{matrix}]$
$D_{x} [\begin{matrix} s & b \\ t & d \end{matrix}]$
And $D_{y} = [\begin{matrix} a & s \\ c & t \end{matrix}] p r o v i d e d D \neq 0$
Consider the system of equations ${\begin{cases} \frac{5}{6} x_{1} - x_{2} = - 20 - - - (1) \\ \frac{4}{3} x_{1} - \frac{7}{2} x_{2} = - 51 - - - (2) \end{cases}$
$D = [\begin{matrix} \frac{5}{6} & - 1 \\ \frac{4}{3} & - \frac{7}{2} \end{matrix}] = - \frac{19}{12}$
similarly
$D_{x} = [\begin{matrix} - 20 & - 1 \\ - 51 & - \frac{7}{2} \end{matrix}]$
$[\begin{matrix} \frac{5}{6} & - 20 \\ \frac{4}{3} & - 51 \end{matrix}] = - \frac{95}{6}$
Note that $D \neq q 0$
By the above formula
$x \frac{19}{- \frac{19}{12}}$
$= - 12$
SIMILARLY
$y = \frac{D_{y}}{D}$
$\frac{- \frac{95}{6}}{- \frac{19}{12}}$
=10
thus the solution of the system is $x = - 12$ and $y = 10$ .This solution can be written as ordered pair $(x, y) = (- 12, 10)$

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