Question

for the problem 1/5g - 1/10 -g+ 1/3/10g -1/10 tyson created an equivalent expression using the folling steps is his final expression equivalent to the initial expression​

Answer 1

No, Tyson's final expression is not equivalent to the initial expression.

Fraction addition;

According to Tyson's evaluation after collecting like terms; we have;

• 1/5g +(-1g) + 1(3/10g) + (-1/10) + (-1/10)

open parenthesis

• 1/5g - 1g + 3/10g - 1/10 - 1/10

Upon addition of like terms; we have;

• (-1/2)g -1/5.

Hence, the correct expression is; (-1/2)g -1/5

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