$a^2x+(a-8)=(a+8)x\\\\(a+8)x=a^2x+(a-8)\ \ \ \ |subtract\ a^2x\ from\ both\ sides\\\\(a+8)x-a^2x=a-8\\\\(a+8-a^2)x=a-8\ \ \ \ |divide\ both\ sides\ by\ (a+8-a^2)\\\\\huge\boxed{x=\frac{a-8}{a+8-a^2}}\to\huge\boxed{x=\frac{8-x}{a^2-a-8}}$

6 0

3 years ago

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Answer quickly! It very urgent

ra1l [238]

Answer: Because the graphs of the two equations overlap each other

Step-by-step explanation:

Every point on the first line also satisfies the second equation.

This is the same as equivalent fractions

where 1/2 = 2/4 = 3/6 = 4/8

Graphing these on a number line would only show one dot.

6 0

3 years ago

using the numbers 3, 5, and 8, can you write nine proper fractions and nine improper fractions? You may use each number only onc

Bond [772]

Answer:

First, we can write a fraction as a/b

Where a is the numerator, and b is the denominator.

A proper fraction is a fraction where the numerator is smaller than the denominator.

Using only the given numbers (only once per fraction), some examples of proper fractions are:

3/5

3/8

5/8

3/85

3/58

5/83

5/38

8/35

8/53

You can see that in all of them the denominator is larger than the numerator.

The improper fractions are those where the numerator is equal or larger than the denominator.

The 9 examples using the given numbers are:

5/3

8/5

8/3

35/8

38/5

53/8

58/3

83/5

85/3

6 0

3 years ago