Solve this system of equations by graphing. First graph the equations, and then type the solution.

To solve the system of equations by graphing, we need to graph both equations:

1. Equation 1: \(x - 2y = 4\)

   • To rewrite it in slope-intercept form (\(y = mx + b\)), we can solve for \(y\): \[ -2y = -x + 4 \ y = \frac{1}{2}x - 2 \]
   • This means the slope is \(\frac{1}{2}\) and the y-intercept is \(-2\).

2. Equation 2: \(y = 0\)

   • This is the x-axis.

Graphing the equations:

• Plot the equation \(y = \frac{1}{2}x - 2\). Start at the y-intercept (0, -2) and use the slope (rise/run) to find another point. From (0, -2), you can go up 1 and right 2 to get to (2, -1), or go to (4, 0) directly.

• Plot the line through those points.

• Since the second equation is \(y = 0\), this is a horizontal line along the x-axis.

Finding the intersection:

• The solution to the system of equations is where the two lines intersect.
• From the graph, it looks like they intersect at the point (4, 0).

Final Solution: \[ \text{The solution is } (4, 0). \]

the graph reaches out to 10 on both the y and x axis, however number 1,3,5,7,and 9 aren't visible on the graph, but the space belonging to the number still is

Thank you for the clarification regarding the axes.

Let's work through the equations again, taking into consideration that the graph range is limited to even numbers in a given space.

Equations

1. First equation: \(x - 2y = 4\)
   • Rearranged to slope-intercept form: \[ y = \frac{1}{2}x - 2 \]
   • Points you can plot from this equation include:
     • When \(x = 0\), \(y = -2\) (not visible)
     • When \(x = 2\), \(y = -1\) (not visible)
     • When \(x = 4\), \(y = 0\) (on the x-axis)
     • When \(x = 6\), \(y = 1\) (not visible)
     • When \(x = 8\), \(y = 2\) (visible)

Using these points, we can plot part of the line. Since \(y = \frac{1}{2}x - 2\) increases slowly, connecting the dots (4, 0) and (8, 2) would give the visual representation of the line within the visible graph.

2. Second equation: \(y = 0\)
   • The line is the x-axis.

Finding the Intersection

Given that the first equation \(y = \frac{1}{2}x - 2\) passes through (4, 0), this point is also on the line \(y = 0\), thus it is the intersection.

Conclusion

Therefore, even with restrictions on the visible plotting points, the intersection point remains identifiable.

\[ \text{The solution to the system of equations is } (4, 0). \]

Write a system of equations to describe the situation below, solve using substitution, and fill in the blanks.

Riley is going to hire a makeup artist for a fashion show and is comparing prices. Charlotte charges $17 as a booking fee and an additional $93 per hour. Belle charges $81 per hour, plus a booking fee of $53. Depending on the length of the show, the cost could end up being the same for either artist. How long would the show be? What would the cost be? If the show lasted for ( ____) hours, the cost would always be $(_____)

To represent the situation with a system of equations, we define the variables:

• Let \( h \) be the number of hours for the show.
• Let \( C_C \) be the total cost when hiring Charlotte.
• Let \( C_B \) be the total cost when hiring Belle.

Costs of the Two Artists:

1. Charlotte's cost: \[ C_C = 17 + 93h \] (booking fee + hourly rate × hours)

2. Belle's cost: \[ C_B = 53 + 81h \] (booking fee + hourly rate × hours)

Set the two equations equal to each other since we want to find out when the costs are the same:

\[ 17 + 93h = 53 + 81h \]

Solve for \( h \):

1. Rearranging the equation: \[ 93h - 81h = 53 - 17 \] This simplifies to: \[ 12h = 36 \]

2. Divide both sides by 12: \[ h = 3 \]

Calculate the cost for either artist when \( h = 3 \):

Using Charlotte's cost equation:

1. Calculate Charlotte's cost: \[ C_C = 17 + 93(3) = 17 + 279 = 296 \]

Using Belle's cost equation: 2. Calculate Belle's cost: \[ C_B = 53 + 81(3) = 53 + 243 = 296 \]

Conclusion:

If the show lasted for ( 3 ) hours, the cost would always be $( 296 ).

Thus, filling in the blanks:

If the show lasted for ( 3 ) hours, the cost would always be $( 296 ).

Solve using substitution.

y = –3x + 8
y = –x − 4
(____),(______)

To solve the system of equations using substitution, we'll substitute one equation into the other:

Given equations:

1. \( y = -3x + 8 \)
2. \( y = -x - 4 \)

Since both expressions equal \( y \), we can set them equal to each other:

\[ -3x + 8 = -x - 4 \]

Step 1: Solve for \( x \)

1. Add \( 3x \) to both sides: \[ 8 = 2x - 4 \]

2. Add \( 4 \) to both sides: \[ 12 = 2x \]

3. Divide by \( 2 \): \[ x = 6 \]

Step 2: Substitute \( x \) back to find \( y \)

Now that we have \( x = 6 \), we can substitute it back into either equation to find \( y \). We will use the first equation:

\[ y = -3(6) + 8 \]

1. Calculate: \[ y = -18 + 8 = -10 \]

Step 3: Write the solution

The solution to the system of equations is:

\[ (6, -10) \]

Thus, filling in the blanks: ( 6, -10 )

How many solutions does the system of equations below have?

–2x = –6y − 2
x = 12y − 15
no solution
one solution
infinite many solutions

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

1. First equation: \(-2x = -6y - 2\)
   Let's rearrange this equation into the slope-intercept form \(y = mx + b\).

   Starting from: \[ -2x = -6y - 2 \]

   Add \(6y\) to both sides: \[ 6y - 2x = -2 \]

   Now, rearranging gives: \[ 6y = 2x - 2 \]

   Divide by \(6\): \[ y = \frac{1}{3}x - \frac{1}{3} \]

2. Second equation: \(x = 12y - 15\)
   Rearranging this equation into the slope-intercept form also: \[ 12y = x + 15 \]

   Divide by \(12\): \[ y = \frac{1}{12}x + \frac{15}{12} \]

Analysis of the Equations

Now we have two equations:

1. \(y = \frac{1}{3}x - \frac{1}{3}\) (Equation 1)
2. \(y = \frac{1}{12}x + \frac{15}{12}\) (Equation 2)

Comparing slopes:

• The slope of the first equation is \(\frac{1}{3}\).
• The slope of the second equation is \(\frac{1}{12}\).

Since the slopes are different, the lines represented by these equations will intersect at exactly one point, indicating that the system of equations has one solution.

Conclusion

The system of equations has one solution.

Solve using elimination.

7x + 3y = –16
10x + 8y = –8

To solve the system of equations using the elimination method, we need to manipulate the equations so that adding or subtracting them will eliminate one of the variables.

Given equations:

1. \(7x + 3y = -16 \quad \text{(Equation 1)}\)
2. \(10x + 8y = -8 \quad \text{(Equation 2)}\)

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

To eliminate \(y\), we can manipulate Equation 1 and Equation 2. We will multiply Equation 1 by 8 and Equation 2 by 3 so that both equations will have the same coefficient for \(y\):

Multiply Equation 1 by 8: \[ 8(7x + 3y) = 8(-16) \ 56x + 24y = -128 \quad \text{(Equation 3)} \]

Multiply Equation 2 by 3: \[ 3(10x + 8y) = 3(-8) \ 30x + 24y = -24 \quad \text{(Equation 4)} \]

Step 2: Subtract the new equations

Now, we can subtract Equation 4 from Equation 3 to eliminate \(y\): \[ (56x + 24y) - (30x + 24y) = -128 - (-24) \ 56x - 30x = -128 + 24 \ 26x = -104 \]

Step 3: Solve for \(x\)

Now, divide both sides by 26: \[ x = \frac{-104}{26} \ x = -4 \]

Step 4: Substitute \(x\) back to find \(y\)

Now we can substitute \(x = -4\) back into either of the original equations. We’ll use Equation 1: \[ 7(-4) + 3y = -16 \ -28 + 3y = -16 \]

Add 28 to both sides: \[ 3y = -16 + 28 \ 3y = 12 \]

Now divide by 3: \[ y = \frac{12}{3} \ y = 4 \]

Final Solution

The solution to the system of equations is: \[ (x, y) = (-4, 4) \]

Thus, \[ \text{The final answer is } (-4, 4). \]