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raketka [301]
3 years ago
15

Solve 3(y-2)=6(y-1)-3y

Mathematics
1 answer:
ASHA 777 [7]3 years ago
3 0

Answer:

All real numbers are solutions.

Step-by-step explanation:

1. First you want to distribute what is inside the parenthesis. You would want to do 3 × y + 3 × -2 = 6 × y + 6 × -1 -3y.

2. Now that the equation has been distributed, you want to simplify it. 3 × y is equal to 3y, and 3 × -2 is equal to -6. So, 3y + -6 = 6 × y + 6 × -1 -3y.

3. Now you want to simplify the right side of the equation. So, 6 times y is equal to 6y, and 6 × -1 is equal to -6. Now the new equation is 3y + -6 =6y + -6-3y.

4. Also in an equation, if there is a + and - sign together like this +-, it would equal to -. So, 3y + -6 = 6y + -6-3y, is equal to 3y - 6 = 6y -6-3y.

5. Now you want to combine like terms on the right side of the equation. 3y - 6 = 3y -6 because 6y-3y is equal to 3y.

6. Now the rest of the equation has to be combined using like terms. Start by moving the -3y fromthe right side of the equation to the 3y on the left side of the equation. -3y plus 3y is equal to zero so it cancels out. The 3y on the left side plus another 3y is equal to 6y.

7. Now the equation is 6y - 6 = 6y - 6

8. You can add 6 to the -6 on both sides of the equation which would cancel out on both sides leaving you with an equation of 6y = 6y.

9. Divide both sides by 6 to get y = y.

10. This equation means that all real numbers are solutions to this equation. In other words, no matter what real number you use in this equation, it would come out to be true. For example, if y = 3, 3 = 3, so the equation is true.

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<h3><u>Question:</u></h3>

Cylinders A and B are similar solids. The base of cylinder A has a circumference of 4π units. The base of cylinder B has an area of 9π units.

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<h3><u>Solution:</u></h3>

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The base of cylinder A has a circumference of 4 \pi units

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Let "x" be the required factor

From given question,

Dimensions of cylinder A are multiplied by what factor to produce the corresponding dimensions of cylinder B

Therefore, we can say,

\text{Dimensions of cylinder A} \times x = \text{Dimensions of cylinder B }

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Given that  base of cylinder A has a circumference of 4 \pi units

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The area of base of cylinder (circle) is given as:

A = \pi r^2

Given that,  the base of cylinder B has an area of 9 \pi units

Therefore,

\pi r^2 = 9 \pi\\\\r^2 = 9\\\\r = 3

Thus the dimension of cylinder B is radius = 3 units

\text{Dimensions of cylinder A} \times x = \text{Dimensions of cylinder B }\\\\2 \times x = 3\\\\x = \frac{3}{2}

Thus dimensions of cylinder A are multiplied by \frac{3}{2}  to produce the corresponding dimensions of cylinder B

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