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ad-work [718]
3 years ago
13

How to solve to system y=x^2-2 y=2x+1

Mathematics
1 answer:
vovikov84 [41]3 years ago
7 0

Answer:

the solutions are (3, 7) and (-1, -1)

Step-by-step explanation:

Insert the " = " symbol between the two equations, obtaining:

y = x^2 - 2 = y = 2x + 1

Then x^2 - 2 = 2x + 1, or

x^2 - 2x - 3 = 0, and this can be factored into (x - 3)(x + 1) = 0.

Thus, the x values that satisfy this system are {-1, 3}.

Use y = 2x + 1 to find the corresponding y values:

y = 2(-1) + 1 = -1

and

y = 2(3) + 1 = 7

Then the solutions are (3, 7) and (-1, -1)

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Below is the graph of y=b^x. Find b.
omeli [17]
Y = <span>b^x

when x = 1
y = b^1
y  = b

Therefore, the value of b is the same as the value of y when x =1
From the graph,
When x = 1, y = 0.5
Therefore, b = 0.5


To confirm this

From the graph,

When x = -1, y = 2
Since </span>y = b^x<span>
2 = </span>b^-1
2 = 1/b
2b = 1
b = 0.5

When x = -2, y = 4
Since y = b^x
4 = b^-2
4 = 1/(b^2)
b^2 = 1/4
b = √(1/4)
b = 1/2
b = 0.5

Therefore, it is conformed that b = 0.5

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What are the solutions to the system?
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