Elliot has a total of 26 books. He has 12 more fiction books than nonfiction books. Let x represent the number of fiction books and y represent the number of nonfiction books. Which system of linear equations represents the situation?

Question

x + y = 26 and x + y = 12
x + y = 26 and x – y = 12
x + y = 26 and –x + y = 12
x – y = 26 and x + y = 12

Answer 1

To represent the situation with a system of linear equations, we can set up the equations based on the information given:

1. Elliot has a total of 26 books, which means: \[ x + y = 26 \] where \( x \) represents the number of fiction books and \( y \) represents the number of nonfiction books.

2. Elliot has 12 more fiction books than nonfiction books, which means: \[ x = y + 12 \] This can be rearranged to: \[ x - y = 12 \]

Now we can write the system of equations based on the two conditions:

1. \( x + y = 26 \) (total number of books)
2. \( x - y = 12 \) (12 more fiction books than nonfiction)

Thus, the correct system of linear equations that represents the situation is: \[ \begin{cases} x + y = 26 \ x - y = 12 \end{cases} \]

From the given options, this corresponds to: x + y = 26 and x – y = 12.

Answer 2

Elliot has a total of 26 books. He has 12 more fiction books than nonfiction books. Let x represent the number of fiction books and y represent the number of nonfiction books.

This system of equations models the number of books.

x + y = 26

x – y = 12

Elliot added the two equations and the result was 2x = 38.

Solve the equation. How many fiction books does Elliot have?

fiction books

How many nonfiction books does Elliot have?

nonfiction books

Answer 3

To solve the system of equations given:

1. \( x + y = 26 \) (total number of books)
2. \( x - y = 12 \) (12 more fiction books than nonfiction)

Let's add the two equations as mentioned:

\[ (x + y) + (x - y) = 26 + 12 \]

This simplifies to:

\[ 2x = 38 \]

Now, divide both sides by 2 to solve for \( x \):

\[ x = \frac{38}{2} = 19 \]

So, Elliot has 19 fiction books.

Next, we need to find the number of nonfiction books \( y \). We can substitute \( x \) into the first equation:

\[ x + y = 26 \] \[ 19 + y = 26 \]

Now, subtract 19 from both sides:

\[ y = 26 - 19 = 7 \]

So, Elliot has 7 nonfiction books.

In summary:

• Fiction books: 19
• Nonfiction books: 7

Answer 4

Use this system of equations to answer the questions that follow.

4x – 9y = 7

–2x + 3y = 4
What number would you multiply the second equation by in order to eliminate the x-terms when adding to the first equation?

What number would you multiply the second equation by in order to eliminate the y-terms when adding to the first equation?

Answer 5

To eliminate the \(x\) and \(y\) terms when adding the two equations, we need to make the coefficients of the respective variables equal.

The two equations are:

1. \( 4x - 9y = 7 \) (Equation 1)
2. \( -2x + 3y = 4 \) (Equation 2)

Eliminating the \(x\)-terms

To eliminate the \(x\)-terms, we want the coefficients of \(x\) in both equations to be equal (in magnitude). The coefficient of \(x\) in Equation 1 is \(4\), and in Equation 2 it is \(-2\).

We can multiply Equation 2 by \(2\) to make the coefficient of \(x\) equal to \(4\):

\[ 2 \cdot (-2x + 3y) = 2 \cdot 4 \]

This results in:

\[ -4x + 6y = 8 \]

Now, adding this to Equation 1, we have:

\[ (4x - 9y) + (-4x + 6y) = 7 + 8 \]

Final Result for \(x\)-term Elimination

Thus, you would multiply the second equation by \(2\) to eliminate the \(x\)-terms.

Eliminating the \(y\)-terms

Now, to eliminate the \(y\)-terms, we focus on the coefficients of \(y\). The coefficient of \(y\) in Equation 1 is \(-9\), and in Equation 2 it is \(3\).

To make these coefficients equal, we can multiply Equation 2 by \(3\) so that the \(y\) coefficient becomes \(9\):

\[ 3 \cdot (-2x + 3y) = 3 \cdot 4 \]

This results in:

\[ -6x + 9y = 12 \]

Now, adding this to Equation 1, we have:

\[ (4x - 9y) + (-6x + 9y) = 7 + 12 \]

Final Result for \(y\)-term Elimination

Thus, you would multiply the second equation by \(3\) to eliminate the \(y\)-terms.

Summary:

• To eliminate \(x\)-terms: Multiply the second equation by \(2\)
• To eliminate \(y\)-terms: Multiply the second equation by \(3\)

Answer 6

An equivalent equation has been written by multiplying the second equation by 2.

4x – 9y = 7 → 4x – 9y = 7

2(–2x + 3y = 4) → –4x + 6y = 8

What is the solution to the system?

((negative StartFraction 19 over 2 EndFraction, negative 5), –5)
(–5, (negative 5, negative StartFraction 19 over 2 EndFraction))
((negative StartFraction 15 over 2 EndFraction, negative StartFraction 11 over 3 EndFraction), )
((negative StartFraction 19 over 10 EndFraction, StartFraction 1 over 15 EndFraction), )

Answer 7

To find the solution to the given system of equations:

1. \( 4x - 9y = 7 \) (Equation 1)
2. \( -2x + 3y = 4 \) (Equation 2)

We will solve the system using the equivalent equation obtained by multiplying Equation 2 by \(2\):

\[ -4x + 6y = 8 \quad \text{(from multiplying Equation 2 by 2)} \]

So now we have:

\[ 4x - 9y = 7 \quad \text{(Equation 1)} \] \[ -4x + 6y = 8 \quad \text{(modified Equation 2)} \]

Adding the Two Equations

Now, we can add both equations to eliminate \(x\):

\[ (4x - 9y) + (-4x + 6y) = 7 + 8 \]

This simplifies to:

\[ 0x - 3y = 15 \]

or

\[ -3y = 15 \]

Dividing both sides by \(-3\):

\[ y = -5 \]

Substituting \(y\) back into One of the Original Equations

Now we substitute \(y = -5\) back into one of the equations to find \(x\). We can use Equation 1:

\[ 4x - 9(-5) = 7 \] \[ 4x + 45 = 7 \] \[ 4x = 7 - 45 \] \[ 4x = -38 \] \[ x = -\frac{38}{4} = -\frac{19}{2} \]

Solution

Therefore, the solution to the system is:

\[ \left( -\frac{19}{2}, -5 \right) \]

Final Answer

Among the options provided, the correct solution is:

\(\left( -\frac{19}{2}, -5 \right)\).

Answer 8

The drama club is selling short-sleeved shirts for $5 each, and long-sleeved shirts for $10 each. They hope to sell all of the shirts they ordered, to earn a total of $1,750. After the first week of the fundraiser, they sold StartFraction one-third EndFraction of the short-sleeved shirts and StartFraction one-half EndFraction of the long-sleeved shirts, for a total of 100 shirts.

Let x represent the number of short-sleeved shirts ordered and let y represent the number of long-sleeved shirts ordered. Which system of linear equations represents the situation?

5x + 10y = 100 and 5 x plus 10 y equals 100 and StartFraction one-third EndFraction x plus StartFraction one-half EndFraction y equals 1,750.x + StartFraction one-half EndFractiony = 1,750
5x + 10y = 1,750 and 5 x plus 10 y equals 1,750 and StartFraction one-third EndFraction x plus StartFraction one-half EndFraction y equals 100.x + y = 100
x + y = 100 and x plus y equals 100 and StartFraction one-third EndFraction x plus StartFraction one-half EndFraction y equals 1,750.x + y = 1,750
x + y = 1,750 and x plus y equals 1,750 and StartFraction one-third EndFraction x plus StartFraction one-half EndFraction y equals 100.x + y = 100

Answer 9

To set up the system of equations based on the information given in the problem, we need to analyze the situation carefully.

1. Understanding the Problem:

   • Short-sleeved shirts sell for $5 each, and long-sleeved shirts sell for $10 each. Their goal is to earn a total of $1,750.
   • After the first week, they sold \( \frac{1}{3} \) of the short-sleeved shirts and \( \frac{1}{2} \) of the long-sleeved shirts, for a total of 100 shirts.

2. Defining Variables:

   • Let \( x \) represent the number of short-sleeved shirts ordered.
   • Let \( y \) represent the number of long-sleeved shirts ordered.

3. Setting Up Equations:

   • From the total number of shirts sold: \[ \frac{1}{3}x + \frac{1}{2}y = 100 \]
   • From the total revenue desired: \[ 5x + 10y = 1,750 \]

4. Final System of Equations:

   • Thus, the system of linear equations representing the situation is:
     1. \( \frac{1}{3}x + \frac{1}{2}y = 100 \) (Equation for shirts sold)
     2. \( 5x + 10y = 1,750 \) (Equation for revenue)

Let's analyze the provided options:

• The correct response from the choices would be: \( \frac{1}{3} x + \frac{1}{2} y = 100 \) and \( 5x + 10y = 1,750 \).

However, since \( \frac{1}{3} x + \frac{1}{2} y = 100 \) can be multiplied by a common denominator to clear the fractions if need be (that is \(6\)), it can be transformed to: \[ 2x + 3y = 600 \]

Nonetheless, based on the original forms of the equations noted above, the exact set of equations preferred in simplified forms should reflect the stated amounts in the revenue and units sold.

Therefore, the correct option that most matches with the above statements is:

• \( 5x + 10y = 1,750 \) and \( \frac{1}{3} x + \frac{1}{2} y = 100 \).

Answer 10

The drama club is selling short-sleeved shirts for $5 each, and long-sleeved shirts for $10 each. They hope to sell all of the shirts they ordered, to earn a total of $1,750. After the first week of the fundraiser, they sold StartFraction one-third EndFraction of the short-sleeved shirts and StartFraction one-half EndFraction of the long-sleeved shirts, for a total of 100 shirts.

This system of equations models the situation.

5x + 10y = 1,750

x + A system of equations. 5 x plus 10 y equals 1,750. StartFraction one-third EndFraction x plus StartFraction one-half EndFraction y equals 100.y = 100

Let x represent the number of short-sleeved shirts ordered and let y represent the number of long-sleeved shirts ordered.

How many short-sleeved shirts were ordered?

How many long-sleeved shirts were ordered?

Answer 11

Let’s start with the system of equations representing the situation.

1. Equations:

   • The first equation, which represents the total revenue from the shirts, is: \[ 5x + 10y = 1750 \]
   • The second equation, which accounts for the shirts sold after the first week, is: \[ \frac{1}{3}x + \frac{1}{2}y = 100 \]

2. Clearing Fractions: To eliminate the fractions in the second equation, we can multiply the entire equation by 6 (the least common multiple of 3 and 2): \[ 6\left(\frac{1}{3}x\right) + 6\left(\frac{1}{2}y\right) = 6(100) \] This simplifies to: \[ 2x + 3y = 600 \]

So, we now have a system of equations:

1. \( 5x + 10y = 1750 \)

2. \( 2x + 3y = 600 \)

3. Solving the System: We can solve this system using either substitution or elimination. Let's use the elimination method.

First, we can multiply the second equation by 5 to align the coefficients of \(x\): \[ 5(2x + 3y = 600) \implies 10x + 15y = 3000 \]

Now we have:

1. \( 5x + 10y = 1750 \)
2. \( 10x + 15y = 3000 \)

Next, we can multiply the first equation by 2 so that the coefficient of \(x\) aligns with the second equation: \[ 2(5x + 10y = 1750) \implies 10x + 20y = 3500 \]

Now we have:

1. \( 10x + 20y = 3500 \)
2. \( 10x + 15y = 3000 \)

Now, we can subtract the second equation from the first equation: \[ (10x + 20y) - (10x + 15y) = 3500 - 3000 \] This simplifies to: \[ 5y = 500 \] Now, divide by 5: \[ y = 100 \]

4. Substituting to Find \(x\): Now that we know \(y = 100\) (the number of long-sleeved shirts ordered), we can substitute this value back into one of the original equations to find \(x\). Let’s use the simpler second equation: \[ 2x + 3(100) = 600 \] This simplifies to: \[ 2x + 300 = 600 \] Subtracting 300 from both sides gives: \[ 2x = 300 \] Dividing by 2 yields: \[ x = 150 \]

5. Final Answers:

• The number of short-sleeved shirts ordered: 150
• The number of long-sleeved shirts ordered: 100