The first three steps in determining the solution set of the system of equations algebraically are shown.

2 answers:

8 0

Since both are equivalent to y, the equations must be equivalent.
x^2-x-3= -3x+5
x^2+2x-8=0

(x+4)(x-2)=0
x=-4, x=2

Plug the values of x in to either equation
y=-3(-4)+5
y= 12+5
y=17

y= -3(2)+5
y=-6+5
y=-1

Final answer: (-4,17) and (2,-1)

Viktor [21]3 years ago

4 0

Answer:

ayo the answer is C for ED

Step-by-step explanation:

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Find all the zeros of the equation x^4-6x^2-7x-6=0

rusak2 [61]

Answer:

The zeros are

$x=-2,\:x=3,\:x=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$

Step-by-step explanation:

We have been given the equation x^4-6x^2-7x-6=0

Use rational root theorem, we have

$a_0=6,\:\quad a_n=1$

$\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:2,\:3,\:6,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1$

$\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:2,\:3,\:6}{1}$

$-\frac{2}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+2$

$=\left(x+2\right)\frac{x^4-6x^2-7x-6}{x+2}\\$

$=x^3-2x^2-2x-3$

Again factor using the rational root test, we get

$=\left(x+2\right)\left(x-3\right)\left(x^2+x+1\right)$

Using the zero product rule, we have

$x+2=0:\quad x=-2\\x-3=0:\quad x=3\\x^2+x+1=0\\\\x_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\cdot \:1\cdot \:1}}{2\cdot \:1}\\\\=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$