Solve the equation below for y.

fraction numerator y over denominator y space plus space 1 end fraction space equals space 2 over 3

y = 1

y = -1

y = 2

No solution

Solve the equation below for b.

b space plus space 2 space equals space 8 over b

b = 4, b = 2

b = -4, b = -2

b = 4, b = -2

b = -4, b = 2

Solve the equation for y.

fraction numerator 1 over denominator y space plus thin space 4 end fraction space equals space fraction numerator 2 over denominator y squared space plus space 3 y space minus space 4 end fraction space minus space fraction numerator 1 over denominator 1 space minus space y end fraction

y = -1/2

y = -1

y = 6

No solution

Solve the equation for n.

fraction numerator 1 over denominator n space minus space 2 end fraction space equals space fraction numerator 2 n space plus thin space space 1 over denominator n squared space plus space 2 n space minus space 8 end fraction space plus thin space fraction numerator 2 over denominator n space plus space 4 end fraction

n = 7/3

n = -7/3

n = 11/3

n = -11/3

Amanda wanted to determine the average of her 6 test scores. She added the scores correctly to get T, but divided by 7 instead of 6. The result was 12 less than her actual average. Which equation could be used to determine the value of T?

A. 6 T space plus space 12 space equals space 7 T

B. T over 7 space equals space fraction numerator T space minus space 12 over denominator 6 end fraction

C. T over 7 space plus thin space 12 space equals space T over 6

D. T over 6 space equals space fraction numerator T space minus space 12 over denominator 7 end fraction

Solve the inequality for b.

7 space minus space 2 over b space less than space 5 over b

b < 1 0 < b < 1 b 1

b > -1

Solve the inequality for y.

fraction numerator 2 over denominator 3 y end fraction space plus space fraction numerator 5 over denominator 6 y end fraction space greater than thin space 3 over 4

y 2

y > 2

0 < y 0

Solve the inequality for b.

fraction numerator 12 over denominator b space minus space 1 end fraction space less than space 6

b 3

1 < b < 3 b < 3 0 < b < 3 Solve the inequality for m. fraction numerator 52 over denominator m space plus thin space 4 end fraction space greater than space 13 m 0

m 0

-4 < m < 0 The ratio of 3 more than a number to the square of 1 more than the number is less than 1. Find the numbers which satisfy this statement. 1 < n < 3 -2 < n < 1 n 1

n 1

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