Question

11(2x+6)+3(2x-3)-3(x-9)=50 . Necesito porfavor resolver esta ecuacion
y comprobarla me ayudan porfavor

Answer 1

Tarea:

resolver la ecuación de primer grado y verificarla:

$11 (2x + 6) + 3 (2x - 3) - 3( x - 9) = 50$

solución:

$11 (2x + 6) + 3 (2x - 3) - 3( x - 9) = 50$

→  desarrollo los paréntesis:

$25x+84 = 50$

→ resto 84 ambos lados:

$25x+84 -84= 50-84$

→ simplificar:

$25x = -34$              

→ dividir ambos lados entre 25:

$\dfrac{25x}{25} = \dfrac{-34}{25}$  

→ simplificar:

$\boxed{x=-\dfrac{34}{25} }$

Y listo la respuesta de la ecuación es -34/25.

comprobamos:

$11 (2x + 6) + 3 (2x - 3) - 3( x - 9) = 50$

$11\left(2\left(-\dfrac{34}{25}\right)+6\right)+3\left(2\left(-\dfrac{34}{25}\right)-3\right)-3\left(\left(-\dfrac{34}{25}\right)-9\right)=50$

$11\left(-2\cdot\dfrac{34}{25}+6\right)+3\left(-2\cdot\dfrac{34}{25}-3\right)-3\left(-\dfrac{34}{25}-9\right)=50$

$11\left(-\dfrac{34\cdot2}{25}+6\right)+3\left(-\dfrac{34\cdot2}{25}-3\right)-3\left(-\dfrac{34}{25}-9\right)=50$

$11\left(-\dfrac{68}{25}+\dfrac{6\cdot25}{25} \right)+3\left(-\dfrac{68}{25}-\dfrac{3\cdot25}{25} \right)-3\left(-\dfrac{34}{25}-\dfrac{9\cdot25}{25} \right)=50$

$11\left(\dfrac{-68+6\cdot25}{25}\right)+3\left(\dfrac{-68-3\cdot25}{25} \right)-3\left(\dfrac{-34-9\cdot25}{25} \right)=50$

$11\left(\dfrac{82}{25}\right)+3\left(-\dfrac{143}{25} \right)-3\left(-\dfrac{259}{25} \right)=50$

$\dfrac{11 \cdot82}{25}+\left(\dfrac{3\cdot(-143)}{25}\right) -\left(\dfrac{3\cdot(-259)}{25}\right) =50$

$\dfrac{902}{25}+\left(-\dfrac{429}{25}\right)-\left(-\dfrac{777}{25}\right) =50 \to\to\to \to \boldsymbol{\ \text{aplicamos la Ley de signos.}}$

$\dfrac{902}{25}-\dfrac{429}{25}+\dfrac{777}{25}=50 \to\to\to\boldsymbol{\ \text{ tenemos Fracciones homogeneas.}}$

$\dfrac{902-429+777}{25}}=50$  

$\dfrac{1250}{25}}=50$

$\boxed{50=50}$