Solve the system: 2x+3y=6 and x=-2y+3. I need help

Mathematics

1 answer:

Alex Ar [27]2 years ago

3 0

Answer:

$\large\boxed{x=3,\ y=0\to(3,\ 0)}$

Step-by-step explanation:

$\left\{\begin{array}{ccc}2x+3y=6&(1)\\x=-2y+3&(2)\end{array}\right\\\\\text{Substitute (2) to (1):}\\\\2(-2y+3)+3y=6\qquad\text{use distributive property}\\\\(2)(-2y)+(2)(3)+3y=6\\\\-4y+6+3y=6\qquad\text{subtract 6 from both sides}\\\\-y=0\to\boxed{y=0}\\\\\text{Put the value of y to (2):}\\\\x=-2(0)+3\\\\x=0+3\\\\\boxed{x=3}$

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Bingel [31]

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Step-by-step explanation:

First, it is important to know the definition of "Chord" and "Arcs" in circles.

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1) If two chords of a circle are congruent (which means that they have equal measure), then their intercepted Arcs are also congruent.

2) If the Arcs are congruent, then they have congruent chords.

In this case, you know that the Arc GH and the Arc JK are congruent. Therefore, you can conclude that the Chord GH and the Chord JK are also congruent.

Then, knowing that:

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