Question

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 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

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 2

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 3

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 4

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 5

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 6

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 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**.