Question

How many solutions does the system of equations have?
8x−6y=48
8x+6y=48

options:

infinite
infinite

one
one

two
two

none

Answer 1

Solve the system of two equations using elimination with multiplication.
4x−6y=−48
−2x+4y=30

Answer 2

Solve the system of equations.
3x−4y=40
5x−5y=55

Answer 3

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

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

Let's rewrite both equations:

From Equation 1, we can isolate \( y \):
\[ 8x - 6y = 48 \]
\[ -6y = 48 - 8x \]
\[ y = \frac{8x - 48}{6} \]
\[ y = \frac{4x - 24}{3} \]

From Equation 2, isolating \( y \) gives:
\[ 8x + 6y = 48 \]
\[ 6y = 48 - 8x \]
\[ y = \frac{48 - 8x}{6} \]
\[ y = 8 - \frac{4x}{3} \]

Now we will examine the two expressions for \( y \):

1. \( y = \frac{4x - 24}{3} \)
2. \( y = 8 - \frac{4x}{3} \)

To find the number of solutions, we can set the two expressions for \( y \) equal to each other:

\[
\frac{4x - 24}{3} = 8 - \frac{4x}{3}
\]

Multiplying both sides by 3 to eliminate the fraction:

\[
4x - 24 = 24 - 4x
\]

Next, add \( 4x \) to both sides:

\[
4x + 4x - 24 = 24
\]
\[
8x - 24 = 24
\]
\[
8x = 48
\]
\[
x = 6
\]

Now we can substitute \( x = 6 \) back into either equation to find \( y \):

Using Equation 1:
\[
8(6) - 6y = 48
\]
\[
48 - 6y = 48
\]
\[
-6y = 0
\]
\[
y = 0
\]

Thus, we have one solution: \( (x, y) = (6, 0) \).

Next, let's look at another method – examining the slopes of the lines represented by these equations:
- The equations represent two lines.
- If they are parallel, they will not intersect, yielding no solutions.
- If they are the same line (which would be infinitely many solutions).

Rearranging the equations into slope-intercept form:

1. From \( 8x - 6y = 48 \), we can get:
\[
y = \frac{4}{3}x - 8
\]
2. From \( 8x + 6y = 48 \):
\[
y = -\frac{4}{3}x + 8
\]

The slopes are different (\( \frac{4}{3} \) and \( -\frac{4}{3} \)), indicating that the lines are not parallel and will intersect at exactly one point.

**Conclusion:** The system of equations has **one solution**. The correct answer is:

**one**.

Answer 4

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

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

We will manipulate these equations to eliminate one of the variables. Let's eliminate \( x \) by making the coefficients of \( x \) in both equations equal.

### Step 1: Make the coefficients of \( x \) the same

We can multiply the second equation by \( 2 \) to make the coefficient of \( x \) in Equation 2 equal to that in Equation 1:

Multiply Equation 2 by \( 2 \):

\[
2(-2x + 4y) = 2(30)
\]

This gives us:

\[
-4x + 8y = 60 \quad \text{(Equation 2')}
\]

Now we have:

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

### Step 2: Add the two equations

Now, we can add Equation 1 and Equation 2':

\[
(4x - 6y) + (-4x + 8y) = -48 + 60
\]

This simplifies to:

\[
0x + 2y = 12
\]

\[
2y = 12
\]

### Step 3: Solve for \( y \)

Dividing by \( 2 \):

\[
y = 6
\]

### Step 4: Substitute \( y \) back into one of the original equations

Now that we have \( y \), we can substitute \( y = 6 \) back into one of the original equations to solve for \( x \). We'll use Equation 1:

\[
4x - 6y = -48
\]

Substituting \( y = 6 \):

\[
4x - 6(6) = -48
\]

\[
4x - 36 = -48
\]

### Step 5: Solve for \( x \)

Add \( 36 \) to both sides:

\[
4x = -48 + 36
\]
\[
4x = -12
\]

Now, divide by \( 4 \):

\[
x = -3
\]

### Step 6: Write the solution

The solution to the system of equations is:

\[
(x, y) = (-3, 6)
\]

Thus, the final answer is:

**\( x = -3 \) and \( y = 6 \)**.

Answer 5

To solve the system of equations

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

we can use either substitution or elimination. In this case, let's use the elimination method.

### Step 1: Simplify Equation 2
First, let's simplify Equation 2:

\[
5x - 5y = 55
\]

We can divide every term by \( 5 \):

\[
x - y = 11 \quad \text{(Equation 2')}
\]

### Step 2: Rearrange Equation 2'
Now, we can isolate \( x \) from Equation 2':

\[
x = y + 11 \quad \text{(Equation 2'')}
\]

### Step 3: Substitute \( x \) in Equation 1
Now substitute Equation 2'' into Equation 1:

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

### Step 4: Distribute and solve for \( y \)

Distributing \( 3 \):

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

Combine like terms:

\[
-y + 33 = 40
\]

Now subtract \( 33 \) from both sides:

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

Multiply by \( -1 \):

\[
y = -7
\]

### Step 5: Substitute \( y \) back to find \( x \)

Now that we have \( y \), substitute \( y = -7 \) back into Equation 2'':

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

### Step 6: Present the solution

The solution to the system of equations is:

\[
(x, y) = (4, -7)
\]

Thus, the final answer is:

**\( x = 4 \) and \( y = -7 \)**.