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

Question
Forms of linear equations
asked 2021-02-21
Cramer’s Rule to solve (if possible) the system of linear equations.
\(\displaystyle{\frac{{{5}}}{{{6}}}}{x}_{{1}}-{x}_{{2}}=-{20}\)
\(\displaystyle{\frac{{{3}}}{{{4}}}}{x}_{{1}}-{\frac{{{7}}}{{{2}}}}{x}_{{2}}=-{51}\)

Answers (1)

2021-02-22
Cramers rule
The solution system \(\displaystyle{b}{e}{g}\in{\left\lbrace{c}{a}{s}{e}{s}\right\rbrace}{a}{x}+{b}{y}={s}\backslash{c}{x}+{\left.{d}{y}\right.}={t}{e}{n}{d}{\left\lbrace{c}{a}{s}{e}{s}\right\rbrace}\) is given by
\(\displaystyle{x}={\frac{{{D}_{{x}}}}{{{D}}}},{y}={\frac{{{D}_{{y}}}}{{{D}}}}{w}{h}{e}{r}{e}{D}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{a}&{b}\backslash{c}&{d}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}\)
\(\displaystyle{D}_{{x}}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{s}&{b}\backslash{t}&{d}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}\)
\(\displaystyle{A}{n}{d}{D}_{{y}}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{a}&{s}\backslash{c}&{t}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{p}{r}{o}{v}{i}{d}{e}{d}{D}\ne{q}{0}\)
Consider the system of equations \(\displaystyle{b}{e}{g}\in{\left\lbrace{c}{a}{s}{e}{s}\right\rbrace}{\frac{{{5}}}{{{6}}}}{x}_{{1}}-{x}_{{2}}=-{20}---{\left({1}\right)}\backslash{\frac{{{4}}}{{{3}}}}{x}_{{1}}-{\frac{{{7}}}{{{2}}}}{x}_{{2}}=-{51}---{\left({2}\right)}{e}{n}{d}{\left\lbrace{c}{a}{s}{e}{s}\right\rbrace}\)
\(\displaystyle{D}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{\frac{{{5}}}{{{6}}}}&-{1}\backslash{\frac{{{4}}}{{{3}}}}&-{\frac{{{7}}}{{{2}}}}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}=-{\frac{{{19}}}{{{12}}}}\)
similarly
\(\displaystyle{D}_{{x}}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}-{20}&-{1}\backslash-{51}&-{\frac{{{7}}}{{{2}}}}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}\)
\(\displaystyle{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{\frac{{{5}}}{{{6}}}}&-{20}\backslash{\frac{{{4}}}{{{3}}}}&-{51}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}=-{\frac{{{95}}}{{{6}}}}\)
Note that \(\displaystyle{D}\ne{q}{0}\)
By the above formula
\(\displaystyle{x}{\frac{{{19}}}{{-\frac{{19}}{{12}}}}}\)
\(\displaystyle=-{12}\)
SIMILARLY
\(\displaystyle{y}={\frac{{{D}_{{y}}}}{{{D}}}}\)
\(\displaystyle{\frac{{-\frac{{95}}{{6}}}}{{-\frac{{19}}{{12}}}}}\)
=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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