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3 years ago

What are the solutions to the following system of equations? Select the correct answer. y=x^2-4x+8 4x+y=12

2 answers:

8 0

Reduce the second equation to y= -4x + 12, then subtract the two equations. (Quadratic on top), to get y=x^2 -4. Then, plug in an x,y coordinate if you have one.

lorasvet [3.4K]3 years ago

7 0

Y=x^2-4x+8 and 4x+y=12 so y=12-4x, insert y in the quadratic to get
12-4x=x^2-4x+8
12=x^2+8
0=x^2-4=(x-2)(x+2)
So the solutions are x=2 and x=-2.

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Answer:

$\frac{4x+2}{x^2 - 9x + 8} = \frac{4x}{(x-8)(x-1)} + \frac{2}{(x-8)(x-1)}$

Step-by-step explanation:

Given

$\frac{4x+2}{x^{2}-9+8 } = \frac{A}{()(x-1)} + \frac{B}{()(x-8)}$

Required

Fill in the gaps

Going by the given parameters, we have that

$\frac{4x+2}{x^{2}-9+8 } = \frac{A}{()(x-1)} + \frac{B}{()(x-8)}$

$x^2 - 9x + 8$ , when factorized is $(x-1)(x-8)$

Hence; the expression becomes

$\frac{4x+2}{(x-1)(x-8)} = \frac{A}{(x-8)(x-1)} + \frac{B}{(x-1)(x-8)}$

Combine Fractions

$\frac{4x+2}{(x-1)(x-8)} = \frac{A + B}{(x-8)(x-1)}$

Simplify the denominators

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By direct comparison

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$B = 2$

Hence, the complete expression is

$\frac{4x+2}{x^2 - 9x + 8} = \frac{4x}{(x-8)(x-1)} + \frac{2}{(x-8)(x-1)}$

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