Find the solution(s) to the system of equations below:
3x - 5y = 3 and y = 3x2 - 4x + 1

The solutions I found to this system of equations are (1, 0) and (⁸/₁₅, -7).  

I keep second guessing and recalculating the second point there, so I'd suggest double-checking it by going through the work yourself.  The first one checks out though.

Step-by-step explanation:

Let's solve this by substitution, where one equation is solved for one variable and substituted in to the other.  As the latter equation is already solved for y, we can simply substitute it into the other:

given:

   y = 3x² - 4x + 1

   3x - 5y = 3

then:

  3x - 5(3x² - 4x + 1) = 3

Which we can then simplify and solve:

3x - 15x² + 20x - 5 = 3

-3x + 15x² - 20x + 5 = -3

15x² - 23x = -8

Now we can find find the solution by completing the square:

15(15x² - 23x) = 15(-8)

225x² - 345x = -120

225x² - 345x  + (23/2)² = -120 + (23/2)²

225x² - 345x  + 529/4 = -120 + 529/4

(15x - 11.5)² = -120 + 132.25

(15x - 11.5)² = 12.25

15x - 11.5 = ± √12.25

15x - 11.5 = ± 3.5

15x = 11.5 ± 3.5

x = (11.5 ± 3.5) / 15

x = 15 / 15, and 8 / 15

x = 1, and 0.5333...

Now that we have x, we can solve for y.

given: 3x - 5y = 3

for x = 1:

3 - 5y = 3

y = 0

for x = ⁸/₁₅:

3(⁸/₁₅) - 5y = 3

5y = 3(⁸/₁₅) - 3

5y = ²⁴/₁₅ - ⁴⁵/₁₅

5y = -²¹/₁₅

y = -²¹/₃

y = -7

So the solutions to this system of equations are (1, 0) and (⁸/₁₅, -7).

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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step-by-step explanation: