A bag contains 18 coins consisting of quarters and dimes. The total value of the coins is $2.85. Which system of equations can be used to determine the number of quarters, q, and the number of dimes, d, in the bag?

Question

Question 3 options:

0.25q + 0.10d = 18
q + d = 2.85

0.10q + 0.25d = 2.85
q + d = 18

0.25q + 0.10d = 2.85
q + d = 18

0.10q + 0.25d = 18
q + d = 2.85

Answer 1

To set up the correct system of equations, we need to consider two pieces of information: the total number of coins and the total value of the coins.

1. The total number of coins (quarters and dimes) is 18, which gives us the equation: \[ q + d = 18 \]

2. The total value of the coins is $2.85. Since quarters are worth $0.25 and dimes are worth $0.10, the value equation can be expressed as: \[ 0.25q + 0.10d = 2.85 \]

Therefore, the correct system of equations that can be used to determine the number of quarters \( q \) and the number of dimes \( d \) in the bag is: \[ \begin{align*} q + d &= 18 \ 0.25q + 0.10d &= 2.85 \end{align*} \]

From the given options, this corresponds to: \[ \text{0.25q + 0.10d = 2.85} \ \text{q + d = 18} \]

Thus, the correct answer is option 3: 0.25q + 0.10d = 2.85 and q + d = 18.

Answer 2

Two customers went to a post office to buy postcards and large envelopes. Each postcard costs the same amount, and each large envelope costs the same amount.

The first customer paid $12 for 14 postcards and 5 large envelopes.
The second customer paid $24.80 for 10 postcards and 15 large envelopes.
What was the cost in dollars of each large envelope?

Question 4 options:

$1.15

$0.35

$0.63

$1.42

Answer 3

Let's set up a system of equations to find the cost of each postcard and each large envelope.

Let:

• \( p \) = cost of each postcard (in dollars)
• \( e \) = cost of each large envelope (in dollars)

From the information given about the first customer, we have the equation: \[ 14p + 5e = 12 \]

For the second customer, we have: \[ 10p + 15e = 24.80 \]

Now we can solve this system of equations.

First, let's simplify the first equation: \[ 14p + 5e = 12 \tag{1} \] We can multiply this entire equation by 2 to make it easier to eliminate \( p \): \[ 28p + 10e = 24 \tag{2} \]

Now, let's look at the second equation and also multiply it by 2 to get: \[ 20p + 30e = 49.60 \tag{3} \]

Now we have: \[ \begin{align*} 28p + 10e &= 24 \ 20p + 30e &= 49.60 \end{align*} \]

Next, we can solve either equation for \( p \). Let's start by isolating \( e \) from Equation (2): \[ 10e = 24 - 28p \] \[ e = \frac{24 - 28p}{10} \tag{4} \]

Now substitute (4) into Equation (3): \[ 20p + 30\left(\frac{24 - 28p}{10}\right) = 49.60 \] Simplifying: \[ 20p + 3(24 - 28p) = 49.60 \] \[ 20p + 72 - 84p = 49.60 \] \[ -64p + 72 = 49.60 \] \[ -64p = 49.60 - 72 \] \[ -64p = -22.40 \] \[ p = \frac{22.40}{64} \approx 0.35 \]

Next, plug the value of \( p \) back into Equation (4): \[ e = \frac{24 - 28(0.35)}{10} \] \[ e = \frac{24 - 9.8}{10} \approx \frac{14.2}{10} = 1.42 \]

Thus, the cost of each large envelope is: \[ \boxed{1.42} \]

So the correct option is $1.42.

Answer 4

What is the solution to this system of equations?

2x + y = 40
x - 2y = -20

Question 5 options:

There are an infinite number of solutions.

(15, 17.5)

There is no solution.

(12, 16)

Answer 5

To solve the system of equations

1. \( 2x + y = 40 \)
2. \( x - 2y = -20 \)

we can use either substitution or elimination. Here, I'll use the substitution method.

From the first equation (1), we can express \( y \) in terms of \( x \):

\[ y = 40 - 2x \tag{1} \]

Next, we can substitute \( y \) from equation (1) into the second equation (2):

\[ x - 2(40 - 2x) = -20 \]

Now simplify this expression:

\[ x - 80 + 4x = -20 \]

Combine like terms:

\[ 5x - 80 = -20 \]

Add 80 to both sides:

\[ 5x = 60 \]

Now divide by 5:

\[ x = 12 \]

Now that we have \( x \), substitute it back into equation (1) to find \( y \):

\[ y = 40 - 2(12) = 40 - 24 = 16 \]

Thus, the solution to the system of equations is:

\[ (x, y) = (12, 16) \]

Therefore, the correct option is (12, 16).

Answer 6

What is the value of x in the solution to this system of equations?

6y + x = -59

x = -2y + 9

Question 6 options:

8.5

-12.5

-17

43

Answer 7

To find the value of \( x \) in the given system of equations:

1. \( 6y + x = -59 \)
2. \( x = -2y + 9 \)

We can substitute the expression for \( x \) from the second equation into the first equation.

Substituting \( x \) from equation (2) into equation (1): \[ 6y + (-2y + 9) = -59 \]

Now simplify: \[ 6y - 2y + 9 = -59 \] \[ 4y + 9 = -59 \]

Next, subtract 9 from both sides: \[ 4y = -59 - 9 \] \[ 4y = -68 \]

Now divide by 4: \[ y = -\frac{68}{4} = -17 \]

Now that we have the value of \( y \), we can substitute it back into either equation to find \( x \). We'll use equation (2): \[ x = -2y + 9 \] Substituting \( y = -17 \): \[ x = -2(-17) + 9 \] \[ x = 34 + 9 \] \[ x = 43 \]

Thus, the value of \( x \) in the solution to this system of equations is: \[ \boxed{43} \]

So the correct answer is 43.