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mote1985 [20]
3 years ago
8

Question 6 of 10- POOS

Mathematics
2 answers:
Serga [27]3 years ago
8 0

Answer:

The correct option is c. Combining like terms in the first expression and dividing each term by 5.

Step-by-step explanation:

Joe concluded that the expression 3x+15+2x is equivalent to the expression 5(x+3).

Let's analyze each option :

a. Dividing each term in the first expression by 3.

If we perform this operation it will lead us to :

\frac{3x+15+2x}{3}=x+5+\frac{2}{3}x

And if we sum each term ⇒ x+5+\frac{2}{3}x=5+\frac{5}{3}x=5(1+\frac{1}{3}x)

Which isn't equivalent to the expression 5(x+3)

b. Combining like terms in the first expression and dividing each term by 3.

If we perform this operation it will lead us to :

3x+15+2x=15+5x ⇒

\frac{15+5x}{3}=5+\frac{5}{3}x=5(1+\frac{1}{3}x)

Which isn't equivalent to the expression 5(x+3)

c. Combining like terms in the first expression and dividing each term by 5.

If we perform this operation it will lead us to :

3x+15+2x=15+5x ⇒ Now if we divide each term by 5 and then we multiply the expression by 5 (in order to keep the equivalence) ⇒

\frac{15+5x}{5}=3+x

And then we multiply by 5 to keep the equivalence ⇒

5(3+x)

Finally, using commutative properties ⇒ 5(3+x)=5(x+3)

Which is the same expression that Joe found.

The correct option is c. Combining like terms in the first expression and dividing each term by 5.

Anyway, let's analyze option d. Subtracting 20 and applying the distributive property to the first expression :

3x+15+2x-20=-5+5x=5(-1+x)

Which isn't equivalent to 5(x+3)

Yuri [45]3 years ago
6 0

Answer: The correct option is c. Combining like terms in the first expression and dividing each term by 5.

Step-by-step explanation:

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Step-by-step explanation:

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What is x+y=10;y=x+10
Alenkasestr [34]

[ Answer ]

\boxed{\bold{X \ = \ 0, \ Y \ = \ 10}}

[ Explanation ]

  • System Of Equations \bold{\left \{ {X \ + \ Y \ = \ 10} \atop {Y \ + \ X \ = \ 10}} \right.}

-----------------------------------

  • [Substitute] Y = X + 10

\bold{\begin{bmatrix}x+x+10=10\end{bmatrix}}

  • Isolate x for x + x + 10 = 10: x = 0

For y = 10

Substitute x = 0

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X = 0

Y = 10

\boxed{\bold{[] \ Eclipsed \ []}}

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