Solve each of the following systems of equations by using the method of cross multiplication:

$\frac{x}{6} + \frac{y}{15} = 4, \frac{x}{3} - \frac{y}{12} = \frac{19}{4} .$

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Solution

We have,

Therefore, we have
$a subscript 1 equals 1 over 6 space space semicolon space space b subscript 1 equals 1 over 15 space space semicolon space space c subscript 1 equals negative 4 space space semicolon a subscript 2 equals 1 third space space semicolon space space b subscript 2 equals fraction numerator negative 1 over denominator 12 end fraction space space semicolon space space c subscript 2 equals negative 19 over 4 space space semicolon$

We have,
$x equals fraction numerator 1 over 15 open parentheses fraction numerator negative 19 over denominator 4 end fraction close parentheses minus open parentheses fraction numerator negative 1 over denominator 12 end fraction close parentheses left parenthesis negative 4 right parenthesis over denominator begin display style 1 over 6 end style open parentheses fraction numerator negative 1 over denominator 12 end fraction close parentheses minus begin display style 1 third end style x open parentheses 1 over 15 close parentheses end fraction equals 18 space space semicolon y equals fraction numerator left parenthesis negative 4 right parenthesis asterisk times begin display style 1 third end style minus open parentheses fraction numerator negative 19 over denominator 4 end fraction close parentheses asterisk times begin display style 1 over 6 end style over denominator 1 over 6 open parentheses fraction numerator negative 1 over denominator 12 end fraction close parentheses minus 1 third x open parentheses 1 over 15 close parentheses end fraction equals 15 space space semicolon$

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