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alex41 [277]
3 years ago
6

The bus fare in a city is $2.00. People who use the bus have the option of purchasing a monthly coupon book for $28.00. With the

coupon book, the fare is reduced to $1.00. Determine the number of times in a month the bus can be used so that the total monthly cost without the coupon book is the same as the total monthly cost with the coupon book
Mathematics
1 answer:
LekaFEV [45]3 years ago
4 0

Answer:

56 = 56

Step-by-step explanation:

Given:

Bus fare = $2.00

coupon book = 28.00

bus fare w/ coupon book = $1.00

let x be the number of bus rides.

2.00x = 1.00x + 28.00

2.00x - 1.00x = 28.00

1x = 28.00

x = 28.00

24 bus rides for both to have the same cost.

2.00x = 1.00x + 28

2.00(28) = 1.00(28) + 28

56 = 28 + 28

56 = 56

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Step-by-step explanation:

The correct questions is as follows:

Carina begins to solve the equation -4-2/3x=-6 by adding 4 to both sides. Which statements regarding the rest of the solving process could be true? Check all that apply.

A.) After adding 4 to both sides, the equation is -2/3x=-2.

B.) After adding 4 to both sides, the equation is -2/3x=-10 .

C.) The equation can be solved for x using exactly one more step by multiplying both sides by -3/2.

D.) The equation can be solved for x using exactly one more step by dividing both sides by -2/3.

E.) The equation can be solved for x using exactly one more step by multiplying both sides by -2/3.

Given equation:

-4-\frac{2}{3}x=-6

To show the steps we will carry out in order to solve for x

Solution:

Solving for x

Step 1:

Adding both sides by 4

4-4-\frac{2}{3}x=-6+4

Thus, we get:

-\frac{2}{3}x=-2

Thus statement A is correct.

Step 2:

Multiplying both sides by -\frac{3}{2}

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Thus, we get:

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This proves that statement C is correct.

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Dividing both sides by -\frac{2}{3}

\frac{-\frac{2}{3}x}{-\frac{2}{3}}=\frac{-2}{-\frac{2}{3}}

Thus, we get:

x=-2\times -\frac{3}{2} [On dividing with a fractional divisor, we take reciprocal and multiply it with the dividend.]

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This prove that statement D is correct.

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