Solve for all values of x. (9)/(7x)=(9x)/(7) Answer: x=

Set up equation: First, we need to set up the equation given by the problem: (9)/(7x) = (9x)/(7).Cross-multiply to eliminate fractions: To solve for x, we can cross-multiply to get rid of the fractions. This means we multiply both sides of the equation by 7x and by 7, which gives us: 9 * 7 = 9x * 7x.Simplify equation: Simplify both sides of the equation: 63 = 63x^2.Isolate x: To find the values of x, we need to isolate x. We can do this by dividing both sides of the equation by 63: 63/63 = 63x^2/63.Take square root: Simplify the equation: 1 = x^2.Consider positive and negative roots: To find the values of x, we take the square root of both sides of the equation: √1 = √x^2.The square root of 1 is 1, and the square root of x^2 is x. However, we must consider both the positive and negative square roots, so x can be 1 or -1: x = ±1.

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Algebra 2

Simplify variable expressions using properties

Full solution

Q. Solve for all values of

x

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\frac{9}{7 x}=\frac{9 x}{7}

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Answer:

x=

Set up equation: First, we need to set up the equation given by the problem: $\frac{9}{7x} = \frac{9x}{7}$ .
Cross-multiply to eliminate fractions: To solve for $x$ , we can cross-multiply to get rid of the fractions. This means we multiply both sides of the equation by $7x$ and by $7$ , which gives us: $9 \times 7 = 9x \times 7x$ .
Simplify equation: Simplify both sides of the equation: $63 = 63x^2$ .
Isolate x: To find the values of x, we need to isolate x. We can do this by dividing both sides of the equation by $63$ : $\frac{63}{63} = \frac{63x^2}{63}$ .
Take square root: Simplify the equation: $1 = x^2$ .
Consider positive and negative roots: To find the values of $x$ , we take the square root of both sides of the equation: $\sqrt{1} = \sqrt{x^2}$ .
Consider positive and negative roots: To find the values of $x$ , we take the square root of both sides of the equation: $\sqrt{1} = \sqrt{x^2}$ .The square root of $1$ is $1$ , and the square root of $x^2$ is $x$ . However, we must consider both the positive and negative square roots, so $x$ can be $1$ or $-1$ : $x = \pm1$ .