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tatyana61 [14]
2 years ago
10

For the problem 1/5g- 1/10- g + 1 3/10g -1/10, Tyson created an equivalent expression using the following steps. 1/5g+-1g+1 3/10

g+-1/10-4/5g+1 1/10. Is his final expression equivalent to the initial expression? Show how you know. If the tow expressions are not equivalent, find Tyson’s mistake and correct it
Mathematics
1 answer:
professor190 [17]2 years ago
3 0

The true statements are:

  • Tyson's expression is not equivalent to the original expression
  • The equivalent expression is:\frac{1}{2}g- \frac 1{5}

<h3>What are equivalent expressions?</h3>

Equivalent expressions are expressions that have equal values

The original expression is given as:

\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10}

Collect like terms

\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10} = \frac 15g - g + 1 \frac{3}{10}g- \frac 1{10}  -\frac{1}{10}

Evaluate the like terms

\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10} = \frac{1}{2}g- \frac 1{5}

Tyler's equivalent expression is given as:

\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10}

Collect like terms

\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10} = \frac15g-g+ 1 \frac3{10}g-\frac 45g-\frac 1{10}+1 \frac{1}{10}

Evaluate the like terms

\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10} = -\frac 3{10}g

The simplified expressions of the original expression, and Tyson's equivalent expressions are not equal.

Hence, Tyson's expression is not equivalent to the original expression

Read more about equivalent expressions at:

brainly.com/question/9603710

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