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Burka [1]
3 years ago
9

Jada solved the equation Negative StartFraction 4 over 9 EndFraction = StartFraction x over 108 EndFraction for x using the step

s below. What was Jada’s error?
Negative StartFraction 4 over 9 EndFraction = StartFraction x over 108 EndFraction
Negative StartFraction 4 over 9 EndFraction (Negative StartFraction 9 over 4 EndFraction) = StartFraction x over 108 EndFraction (Negative StartFraction 9 over 4 EndFraction)
x = negative StartFraction 1 over 48 EndFraction

Jada should have multiplied both sides of the equation by 108.
Jada should have multiplied both sides of the equation by Negative StartFraction 4 over 9 EndFraction.
The product of Negative StartFraction 9 over 4 EndFraction and StartFraction 1 over 108 EndFraction is not equal to Negative StartFraction 1 over 48 EndFraction.
The product of Negative StartFraction 9 over 4 EndFraction and 108 should have been the value of x.
Mark this and return

Mathematics
2 answers:
zavuch27 [327]3 years ago
6 0

Answer:first option

Step-by-step explanation:

I taked test

Lilit [14]3 years ago
3 0

Answer: First option.

Step-by-step explanation:

The complete exercise is attached.

In order to solve this exercise, it is necessary to remember the following property:

The Multiplication property of Equality states that:

If\ a=b\ then\ a*c=b*c

In this case, the equation that Jada had is the folllowing:

-\frac{4}{9}=\frac{x}{108}

Jada needed to solve for the variable "x" in order to find its value.

The correct procedure to solve for for "x" is to multiply both sides of the equation by 108. Then, you get:

(108)(-\frac{4}{9})=(\frac{x}{108})(108)\\\\-48=x

As you can notice in the picture, Jada did not multiply both sides of the equation by 108, but multiplied the left side by -\frac{4}{9}<em>  </em>and the right side by -\frac{9}{4}.

Therefore,you can conclude that Jada should have multiplied both sides of the equation by 108.

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PLEASE DONT COPY THE ANSWER FROM SOMEWHERE ELSE PLEASE
klio [65]
2x+x=36
Kim's Sister age

2x+x=36 Add "x and 2x"
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÷3 ÷3
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Soo....
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