How many solutions does an equation have when you isolate the variable and it equals a constant?

Mathematics

1 answer:

OLEGan [10]3 years ago

3 0

1 solution is available when variable equals a constant.

Answer: Option B.

<u>Explanation:</u>

You will be able to determine if an equation has one solution (which is when one variable equals one number), or if it has no solution (the two sides of the equation are not equal to each other) or infinite solutions (the two sides of the equation are identical).

The ordered pair that is the solution of both equations is the solution of the system. A system of two linear equations can have one solution, an infinite number of solutions, or no solution. If a consistent system has exactly one solution, it is independent.

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Please solve the problem with steps

Debora [2.8K]

Answer:

Infinite series equals 4/5

Step-by-step explanation:

Notice that the series can be written as a combination of two geometric series, that can be found independently:

$\frac{3^{n-1}-1}{6^{n-1}} =\frac{3^{n-1}}{6^{n-1}} -\frac{1}{6^{n-1}} =(\frac{1}{2})^{n-1} -\frac{1}{6^{n-1}}$

The first one: $(\frac{1}{2})^{n-1}$ is a geometric sequence of first term ( $a_1$ ) "1" and common ratio (r) " $\frac{1}{2}$ ", so since the common ratio is smaller than one, we can find an answer for the infinite addition of its terms, given by: $Infinite\,Sum=\frac{a_1}{1-r} = \frac{1}{1-\frac{1}{2} } =\frac{1}{\frac{1}{2} } =2$

The second one: $\frac{1}{6^{n-1}}$ is a geometric sequence of first term "1", and common ratio (r) " $\frac{1}{6}$ ". Again, since the common ratio is smaller than one, we can find its infinite sum: