For the graph shown, select the statement that best represents the given system of equations. 3y – x = 2 9y – 3x = 6

1 answer:

5 0

You can find that out by first putting two equations in slope-intercept form y=mx+b by solving for y.
3y - x = 2
3y = x+2
y = (1/3)x + 2/3

9y = 3x + 6
y = (3/9)x + 6/9
y = (1/3)x + 2/3

They are the same line as you can see in the final step.
Since the two lines are the same line, (they "coincide") they are coincident.
Coincident means there are an infinite number of solutions, and since they are the same line, there are an infinite number of solutions.

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I know that real numbers consist of the natural or counting numbers, whole numbers, integers, rational numbers and irrational nu

ra1l [238]

The imaginary unit $i$ belongs to the set of complex numbers, denoted by $\mathbb C$ . These numbers take the form $a+bi$ , where $a,b$ are any real numbers.

The set of real numbers, $\mathbb R$ , is a subset of $\mathbb C$ , where each number in $\mathbb R$ can be obtained by taking $b=0$ and letting $a$ be any real number.

But any number in $\mathbb C$ with non-zero imaginary part is not a real number. This includes $i$ .

"is it possible that i can use an imaginary number for a real number"

I'm not sure what you mean by this part of your question. It is possible to represent any real number as a complex number, but not a purely imaginary one. All real numbers are complex, but not all complex numbers are real. For example, 2 is real and complex because $2=2+0i$ .

There are some operations that you can carry out on purely imaginary numbers to get a purely real number. A famous example is raising $i$ to the $i$ -th power. Since $i=e^{i\pi/2}$ , we have