Solve the system of two linear equations. −3x−4y=−23 3x−5y=5 (1 point) Responses (5,−2) left parenthesis 5 comma negative 2 right parenthesis (5, 2) left parenthesis 5 comma 2 right parenthesis (−5, 2) left parenthesis negative 5 comma 2 right parenthesis (−5,−2)

Answer 1

To solve the system of equations:

1. \(-3x - 4y = -23\)
2. \(3x - 5y = 5\)

We can use the method of substitution or elimination. Here, we will use elimination.

First, we can multiply the first equation by \(-1\) to make the coefficients of \(x\) in both equations match:

\[ 3x + 4y = 23 \quad \text{(1')} \]

Now we have:

1. \(3x + 4y = 23\)
2. \(3x - 5y = 5\)

Next, we'll subtract the second equation from the first:

\[ (3x + 4y) - (3x - 5y) = 23 - 5 \] \[ 4y + 5y = 18 \] \[ 9y = 18 \] \[ y = 2 \]

Now that we have \(y\), we can substitute it back into one of the original equations to solve for \(x\). We'll use the second equation:

\[ 3x - 5(2) = 5 \] \[ 3x - 10 = 5 \] \[ 3x = 15 \] \[ x = 5 \]

So, the solution to the system of equations is \( (x, y) = (5, 2) \).

Thus, the answer is \((5, 2)\).

Answer 2

How many solutions does the system of equations have? 8x−6y=48 8x+6y=48 (1 point) Responses one one two two none none infinite

Answer 3

To find out how many solutions the system of equations has, let's write down the equations clearly:

1. \(8x - 6y = 48\) (Equation 1)
2. \(8x + 6y = 48\) (Equation 2)

Now, let's analyze the two equations.

First, we can subtract Equation 1 from Equation 2:

\[ (8x + 6y) - (8x - 6y) = 48 - 48 \]

This simplifies to:

\[ 6y + 6y = 0 \] \[ 12y = 0 \] \[ y = 0 \]

Now that we have \(y = 0\), we can substitute this value back into either of the original equations to solve for \(x\). We'll use Equation 1:

\[ 8x - 6(0) = 48 \] \[ 8x = 48 \] \[ x = 6 \]

So, we found a single solution: \((x, y) = (6, 0)\).

Next, let's examine the nature of the system.

Both equations can be rearranged as follows:

1. \(y = \frac{4}{3}x - 8\) (from Equation 1)
2. \(y = -\frac{4}{3}x + 8\) (from Equation 2)

The first equation has a positive slope, while the second has a negative slope. Thus, they are lines that intersect at one point.

Therefore, the system of equations has exactly one solution.

The answer is: one.

Answer 4

Solve the system of two equations using elimination with multiplication. 4x−6y=−48 −2x+4y=30 (1 point) Responses (3, 9) left parenthesis 3 comma 9 right parenthesis (−21,−6) left parenthesis negative 21 comma negative 6 right parenthesis (−33,−9) left parenthesis negative 33 comma negative 9 right parenthesis (−3, 6)

Answer 5

To solve the system of equations using the elimination method, we start with the following two equations:

1. \(4x - 6y = -48\) (Equation 1)
2. \(-2x + 4y = 30\) (Equation 2)

Step 1: Make the coefficients of \(x\) or \(y\) the same.

To eliminate \(x\), we can multiply Equation 2 by \(2\) so that the coefficients of \(x\) will match:

\[ 2(-2x + 4y) = 2(30) \] This gives us: \[ -4x + 8y = 60 \quad \text{(Equation 3)} \]

Now the system of equations is:

1. \(4x - 6y = -48\) (Equation 1)
2. \(-4x + 8y = 60\) (Equation 3)

Step 2: Add Equations 1 and 3.

Now we can add Equation 1 and Equation 3:

\[ (4x - 6y) + (-4x + 8y) = -48 + 60 \] This simplifies to:

\[ 0x + 2y = 12 \] \[ 2y = 12 \] \[ y = 6 \]

Step 3: Substitute \(y\) back into one of the original equations to find \(x\).

We'll use Equation 2 for substitution:

\[ -2x + 4(6) = 30 \] \[ -2x + 24 = 30 \] \[ -2x = 30 - 24 \] \[ -2x = 6 \] \[ x = -3 \]

Step 4: Write the solution as an ordered pair.

The solution to the system of equations is \((x, y) = (-3, 6)\).

Thus, the answer is: \((-3, 6)\).

Answer 6

Solve the system of equations. 3x−4y=40 5x−5y=55 (1 point) Responses (8,−3) left parenthesis 8 comma negative 3 right parenthesis (14, 3) left parenthesis 14 comma 3 right parenthesis (4,−7) left parenthesis 4 comma negative 7 right parenthesis (18, 7)

Answer 7

To solve the system of equations:

1. \(3x - 4y = 40\) (Equation 1)
2. \(5x - 5y = 55\) (Equation 2)

Let's start by simplifying Equation 2:

Divide the entire Equation 2 by 5:

\[ x - y = 11 \quad \text{(Equation 3)} \]

Now we can express \(x\) in terms of \(y\) from Equation 3:

\[ x = y + 11 \]

Step 1: Substitute \(x\) in Equation 1.

Now we can substitute \(x\) back into Equation 1:

\[ 3(y + 11) - 4y = 40 \]

Expanding this gives:

\[ 3y + 33 - 4y = 40 \] \[ -y + 33 = 40 \]

Step 2: Solve for \(y\):

\[ -y = 40 - 33 \] \[ -y = 7 \] \[ y = -7 \]

Step 3: Substitute \(y\) back to find \(x\):

Now substitute \(y = -7\) back into Equation 3:

\[ x = -7 + 11 \] \[ x = 4 \]

Step 4: Write the solution as an ordered pair:

The solution to the system of equations is \((x, y) = (4, -7)\).

Thus, the answer is: \((4, -7)\).

Answer 8

How many solutions does the system have? 12x−15y=18 4x−5y=6 (1 point) Responses infinitely many solutions infinitely many solutions one solution one solution two solutions two solutions no solution

Answer 9

To determine how many solutions the given system of equations has, we can analyze the equations:

1. \(12x - 15y = 18\) (Equation 1)
2. \(4x - 5y = 6\) (Equation 2)

Step 1: Simplify the equations.

Let's see if we can relate these two equations. Notice that we can multiply Equation 2 by \(3\) to see if it helps us compare directly with Equation 1:

\[ 3(4x - 5y) = 3(6) \]

This gives us:

\[ 12x - 15y = 18 \quad \text{(This is exactly Equation 1)} \]

Step 2: Analyze the relationship.

Since multiplying Equation 2 by \(3\) gives us exactly Equation 1, both equations represent the same line. Therefore, every point on the line \(12x - 15y = 18\) is also a solution to \(4x - 5y = 6\).

Conclusion:

Since the two equations represent the same line, the system has infinitely many solutions.

The answer is: infinitely many solutions.

Answer 10

The senior class at Hartford High School is planning their prom. They’ve narrowed the location down to two choices. The Regal Hotel charges, C , $500 for the ballroom rental fee and $75 per student, s , for the meal. The Greenhill Country Club charges, C , $1,000 for the ballroom rental fee and $65 per student, s , for the meal. Create a pair of linear equations that represent this real-world problem.(1 point) Responses Regal Hotel: C=75s+1,000; Greenhill Country Club: C=65s+500 Regal Hotel: upper C equals 75 s plus 1,000 ; Greenhill Country Club: upper C equals 65 s plus 500 Regal Hotel: C=1,000s+75; Greenhill Country Club: C=500s+65 Regal Hotel: upper C equals 1,000 s plus 75 ; Greenhill Country Club: upper C equals 500 s plus 65 Regal Hotel: C=75s+500; Greenhill Country Club: C=65s+1,000 Regal Hotel: upper C equals 75 s plus 500 ; Greenhill Country Club: upper C equals 65 s plus 1,000 Regal Hotel: C=500s+75; Greenhill Country Club: C=1,000s+65

Answer 11

To create the pair of linear equations representing the cost for each location based on the number of students, we can analyze the information provided for both venues.

Regal Hotel:

• Fixed ballroom rental fee: \(500\)
• Cost per student for the meal: \(75\)

The equation for the total cost \(C\) at the Regal Hotel as a function of the number of students \(s\) is:

\[ C = 75s + 500 \]

Greenhill Country Club:

• Fixed ballroom rental fee: \(1,000\)
• Cost per student for the meal: \(65\)

The equation for the total cost \(C\) at the Greenhill Country Club as a function of the number of students \(s\) is:

\[ C = 65s + 1,000 \]

Summary of the Equations:

• Regal Hotel: \(C = 75s + 500\)
• Greenhill Country Club: \(C = 65s + 1,000\)

So, the correct response representing these two equations is:

Regal Hotel: \(C = 75s + 500\); Greenhill Country Club: \(C = 65s + 1,000\).