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3 years ago

How many solutions are there to the system of equations graphed below if the lines are parallel

Mathematics

2 answers:

melamori03 [73]3 years ago

8 0

Answer:

Zero

C is the correct option.

Step-by-step explanation:

In the given graph, the given lines are parallel to each other and hence never intersect.

Now, we know that the intersection point (s) of two curves gives us the solution.

Here the lines are parallel to each other and hence do not intersect each other.

Thus, there would not be any intersection point.

Hence, the system of equations has no solution.

C is the correct option.

mash [69]3 years ago

3 0

If a consistent system has an infinite number of solutions, it is dependent . When you graph the equations, both equations represent the same line. If a system has no solution, it is said to be inconsistent . The graphs of the lines do not intersect, so the graphs are parallel and there is no solution.

Sorry, this is a general answer. I could not view the graph below...

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Verdich [7]

Answer:

$x=-1,\:x=-7,\:x=i,\:x=-i$

Step-by-step explanation:

Considering the equation

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Solving

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$\mathrm{Factor\:}x^4+8x^3+8x^2+8x+7:\quad \left(x+1\right)\left(x+7\right)\left(x^2+1\right)$

$\mathrm{Use\:the\:rational\:root\:theorem}$

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$\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:7,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1$

$\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:7}{1}$

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

$=\left(x+1\right)\frac{x^4+8x^3+8x^2+8x+7}{x+1}...[A]$

Solving

$\frac{x^4+8x^3+8x^2+8x+7}{x+1}$

$=x^3+7x^2+x+7$

Putting $\frac{x^4+8x^3+8x^2+8x+7}{x+1}$ = $x^3+7x^2+x+7$ in equation [A]

So,

$\left(x+1\right)\frac{x^4+8x^3+8x^2+8x+7}{x+1}...[A]$

$=\left(x+1\right)x^3+7x^2+x+7$

$x^3+7x^2+x+7=\left(x+7\right)\left(x^2+1\right)$

So,

Equation [A] becomes

$=\left(x+1\right)\left(x+7\right)\left(x^2+1\right)$

So, the polynomial equation becomes

$\left(x+1\right)\left(x+7\right)\left(x^2+1\right)=0$

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