Ax+by+c=0 solve for y

Mathematics

2 answers:

vodomira [7]2 years ago

8 0

Answer:

bY=-C-aX

Y=-(C+aX)/b

MAXImum [283]2 years ago

3 0

Answer: $y=\frac{-ax-c}{b}$

Step-by-step explanation:

$ax+by+c=0\\$

isolate (by) by substracting (ax) and (c) on both sides.

$(-ax-c)+ax+by+c=0+(-ax-c)$

$by=-ax-c$

divide by (b)

$\frac{by}{b}=\frac{-ax-c}{b}$

$y=\frac{-ax-c}{b}$

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Andre45 [30]

Answer:

the one real zero is in the interval (-1, 0)

Step-by-step explanation:

Descartes' rule of signs tells you there are 0 or 2 positive real zeros. Changing the signs of the odd-degree terms and applying that rule again tells you there is one negative real zero. At the same time, those coefficients (-3, -5, -5, +7) have a negative sum, so you know ...

f(-1) = -6

f(0) = +7

so there is a zero in the interval (-1, 0).

You can try a few values between x=0 and x=10 to see what the function does in that part of the graph. You find ...

f(1) = 10

f(2) = 21

f(3) = 58

So, it is safe to conclude that there are no real zeros for x > 0.

The only real zero of f(x) is in the interval (-1, 0).

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I like to use a graphing calculator for problems like this.