Question

Por favor necesito ayuda con cualquier método 7x+2y=42
3x-2y= 1


7x+3y=100
3x-7y=20. Agradecido de antemano

Answer 1

SOLUCIÓN

♛ HØlα!! ✌

A)     7x + 2y = 42

        3x - 2y = 1

                         $Despejaremos\:"x"\:de\:la\:primera\:ecuaci\'on\\\\7x+2y=42\\\\\\7x=42-2y\\\\\\x=\dfrac{42-2y}{7}\\\\\\Reemplazamos\:x\:en\:la\:segunda\:ecuaci\'on\\\\\\3x-2y=1\\\\\\(3)\left(\dfrac{42-2y}{7}\right)-2y=1\\\\\\\dfrac{(3)(42-2y)+[(7)(-2y)]}{7}=1\\\\\\(126-6y)+(-14y)=(1)(7)\\\\\\126-20y=7\\\\\\-20y=-119\\\\\\y=\dfrac{-119}{-20}\\\\\\\boxed{\boldsymbol{y=5.95}}$

                        $Reemplazamos\:"y"\:en\:"x"\\\\\\x=\dfrac{42-2y}{7}\\\\\\x=\dfrac{42-2(5.95)}{7}\\\\\\x=\dfrac{42-11.9}{7}\\\\\\x=\dfrac{30.1}{7}\\\\\\\boxed{\boldsymbol{x=4.3}}$

B)         7x + 3y = 100

            3x - 7y = 20

                      $Despejaremos\:"x"\:de\:la\:primera\:ecuaci\'on\\\\7x+3y=100\\\\\\7x=100-3y\\\\\\x=\dfrac{100-3y}{7}\\\\\\Reemplazamos\:x\:en\:la\:segunda\:ecuaci\'on\\\\\\3x-7y=20\\\\\\(3)\left(\dfrac{100-3y}{7}\right)-7y=20\\\\\\\dfrac{(3)(100-3y)+[(7)(-7y)]}{7}=20\\\\\\(300-9y)+(-49y)=(20)(7)\\\\\\300-58y=140\\\\\\-58y=-160\\\\\\y=\dfrac{-160}{-58}\\\\\\\boxed{\boldsymbol{y=2.75862}}$

                     $Reemplazamos\:"y"\:en\:"x"\\\\\\x=\dfrac{100-3y}{7}\\\\\\x=\dfrac{100-3(2.75862)}{7}\\\\\\x=\dfrac{100-8.27586}{7}\\\\\\x=\dfrac{91.7241}{7}\\\\\\\boxed{\boldsymbol{x=13.1034}}$