Can someone help me please?

Mathematics

1 answer:

mafiozo [28]3 years ago

4 0

Answer:

1. The mathemagician is speaking

2. It says his name then puts a ":" than he talks

3. By the way he is speaking to Azaz, you can tell he disagrees with Azaz on something

4. I don't think I would because it seems he tries to force his opinions on people in a negative way. But that is a personal choice.

Step-by-step explanation:

Hope this Helped you :)

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$\left\{\begin{array}{ccc}3y+10x-54=0&\text{add 54 to both sides}\\5y-2x-34=0&\text{add 34 to both sides}\end{array}\right\\\left\{\begin{array}{ccc}3y+10x=54\\5y-2x=34&\text{multiply both sides by 5}\end{array}\right\\\underline{+\left\{\begin{array}{ccc}3y+10x=54\\25y-10x=170\end{array}\right}\ \ \ \ |\text{add both sides of the equations}\\.\ \ \ \ \ 28y=224\qquad\text{divide both sides by 28}\\.\ \ \ \ \ \ \ \dfrac{28y}{28}=\dfrac{224}{28}\\.\ \ \ \ \ \ \ \boxed{y=8}$

$\text{put the value of y to the first equation}\\\\3(8)+10x=54\\24+10x=54\qquad\text{subtract 24 from both sides}\\24-24+10x=54-24\\10x=30\qquad\text{divide both sides by 10}\\\dfrac{10x}{10}=\dfrac{30}{10}\\\boxed{x=3}$

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