Question

drag the system of equation to the point that represents the solution to the system.
3x + 2y = 10
6x - y = 10

y = 3x - 7
y = 2x - 5

y = 6
x = -5

4x - 3y = 5
3x + 2y = -9

2x - y = -13
y = x + 9

x + y = 7
x - y = -1

Explanation:

To solve each system of equations, we can use methods like substitution or elimination. Let's go through each system:

System 1: \(

$$\begin{cases} 3x + 2y = 10 \\ 6x - y = 10 \end{cases}$$

\)

Step 1: Solve the second equation for \( y \)

From \( 6x - y = 10 \), we get \( y = 6x - 10 \).

Step 2: Substitute \( y = 6x - 10 \) into the first equation

Substitute into \( 3x + 2y = 10 \):
\( 3x + 2(6x - 10) = 10 \)
\( 3x + 12x - 20 = 10 \)
\( 15x = 30 \)
\( x = 2 \)

Step 3: Find \( y \) using \( x = 2 \)

Substitute \( x = 2 \) into \( y = 6x - 10 \):
\( y = 6(2) - 10 = 12 - 10 = 2 \)
Solution: \( (2, 2) \) (matches the point at \( x=2, y=2 \))

System 2: \(

$$\begin{cases} y = 3x - 7 \\ y = 2x - 5 \end{cases}$$

\)

Step 1: Set the two equations equal (since both equal \( y \))

\( 3x - 7 = 2x - 5 \)

Step 2: Solve for \( x \)

\( 3x - 2x = -5 + 7 \)
\( x = 2 \)

Step 3: Find \( y \) using \( x = 2 \)

Substitute \( x = 2 \) into \( y = 2x - 5 \):
\( y = 2(2) - 5 = 4 - 5 = -1 \)
Solution: \( (2, -1) \) (matches the point at \( x=2, y=-1 \))

System 3: \(

$$\begin{cases} y = 6 \\ x = -5 \end{cases}$$

\)
The solution is the point where \( x = -5 \) and \( y = 6 \), so \( (-5, 6) \) (matches the left - most purple point).

System 4: \(

$$\begin{cases} 4x - 3y = 5 \\ 3x + 2y = -9 \end{cases}$$

\)

Step 1: Eliminate a variable. Multiply the first equation by 2 and the second by 3

First equation * 2: \( 8x - 6y = 10 \)
Second equation * 3: \( 9x + 6y = -27 \)

Step 2: Add the two new equations

\( (8x - 6y)+(9x + 6y)=10+( - 27) \)
\( 17x=-17 \)
\( x = - 1 \)

Step 3: Find \( y \) using \( x = -1 \)

Substitute \( x = -1 \) into \( 4x - 3y = 5 \):
\( 4(-1)-3y = 5 \)
\( -4 - 3y = 5 \)
\( -3y = 9 \)
\( y=-3 \)
Solution: \( (-1, -3) \) (matches the point at \( x = - 1,y=-3 \))

System 5: \(

$$\begin{cases} 2x - y = -13 \\ y = x + 9 \end{cases}$$

\)

Step 1: Substitute \( y = x + 9 \) into \( 2x - y = -13 \)

\( 2x-(x + 9)=-13 \)
\( 2x - x-9=-13 \)
\( x-9=-13 \)
\( x=-13 + 9=-4 \)

Step 2: Find \( y \) using \( x = -4 \)

Substitute \( x = -4 \) into \( y = x + 9 \):
\( y=-4 + 9 = 5 \)
Solution: \( (-4, 5) \) (matches the point at \( x=-4,y = 5 \))

System 6: \(

$$\begin{cases} x + y = 7 \\ x - y = -1 \end{cases}$$

\)

Step 1: Add the two equations

\( (x + y)+(x - y)=7+( - 1) \)
\( 2x=6 \)
\( x = 3 \)

Step 2: Find \( y \) using \( x = 3 \)

Substitute \( x = 3 \) into \( x + y = 7 \):
\( 3 + y = 7 \)
\( y = 4 \)
Solution: \( (3, 4) \) (matches the purple point at \( x = 3,y = 4 \))

To "drag the system to the point", match each system to its solution point using the above calculations:

• \(

$$\begin{cases} 3x + 2y = 10 \\ 6x - y = 10 \end{cases}$$

\) → \( (2, 2) \)

• \(

$$\begin{cases} y = 3x - 7 \\ y = 2x - 5 \end{cases}$$

\) → \( (2, -1) \)

• \(

$$\begin{cases} y = 6 \\ x = -5 \end{cases}$$

\) → \( (-5, 6) \)

• \(

$$\begin{cases} 4x - 3y = 5 \\ 3x + 2y = -9 \end{cases}$$

\) → \( (-1, -3) \)

• \(

$$\begin{cases} 2x - y = -13 \\ y = x + 9 \end{cases}$$

\) → \( (-4, 5) \)

• \(

$$\begin{cases} x + y = 7 \\ x - y = -1 \end{cases}$$

\) → \( (3, 4) \)

Answer:

To solve each system of equations, we can use methods like substitution or elimination. Let's go through each system:

System 1: \(

$$\begin{cases} 3x + 2y = 10 \\ 6x - y = 10 \end{cases}$$

\)

Step 1: Solve the second equation for \( y \)

From \( 6x - y = 10 \), we get \( y = 6x - 10 \).

Step 2: Substitute \( y = 6x - 10 \) into the first equation

Substitute into \( 3x + 2y = 10 \):
\( 3x + 2(6x - 10) = 10 \)
\( 3x + 12x - 20 = 10 \)
\( 15x = 30 \)
\( x = 2 \)

Step 3: Find \( y \) using \( x = 2 \)

Substitute \( x = 2 \) into \( y = 6x - 10 \):
\( y = 6(2) - 10 = 12 - 10 = 2 \)
Solution: \( (2, 2) \) (matches the point at \( x=2, y=2 \))

System 2: \(

$$\begin{cases} y = 3x - 7 \\ y = 2x - 5 \end{cases}$$

\)

Step 1: Set the two equations equal (since both equal \( y \))

\( 3x - 7 = 2x - 5 \)

Step 2: Solve for \( x \)

\( 3x - 2x = -5 + 7 \)
\( x = 2 \)

Step 3: Find \( y \) using \( x = 2 \)

Substitute \( x = 2 \) into \( y = 2x - 5 \):
\( y = 2(2) - 5 = 4 - 5 = -1 \)
Solution: \( (2, -1) \) (matches the point at \( x=2, y=-1 \))

System 3: \(

$$\begin{cases} y = 6 \\ x = -5 \end{cases}$$

\)
The solution is the point where \( x = -5 \) and \( y = 6 \), so \( (-5, 6) \) (matches the left - most purple point).

System 4: \(

$$\begin{cases} 4x - 3y = 5 \\ 3x + 2y = -9 \end{cases}$$

\)

Step 1: Eliminate a variable. Multiply the first equation by 2 and the second by 3

First equation * 2: \( 8x - 6y = 10 \)
Second equation * 3: \( 9x + 6y = -27 \)

Step 2: Add the two new equations

\( (8x - 6y)+(9x + 6y)=10+( - 27) \)
\( 17x=-17 \)
\( x = - 1 \)

Step 3: Find \( y \) using \( x = -1 \)

Substitute \( x = -1 \) into \( 4x - 3y = 5 \):
\( 4(-1)-3y = 5 \)
\( -4 - 3y = 5 \)
\( -3y = 9 \)
\( y=-3 \)
Solution: \( (-1, -3) \) (matches the point at \( x = - 1,y=-3 \))

System 5: \(

$$\begin{cases} 2x - y = -13 \\ y = x + 9 \end{cases}$$

\)

Step 1: Substitute \( y = x + 9 \) into \( 2x - y = -13 \)

\( 2x-(x + 9)=-13 \)
\( 2x - x-9=-13 \)
\( x-9=-13 \)
\( x=-13 + 9=-4 \)

Step 2: Find \( y \) using \( x = -4 \)

Substitute \( x = -4 \) into \( y = x + 9 \):
\( y=-4 + 9 = 5 \)
Solution: \( (-4, 5) \) (matches the point at \( x=-4,y = 5 \))

System 6: \(

$$\begin{cases} x + y = 7 \\ x - y = -1 \end{cases}$$

\)

Step 1: Add the two equations

\( (x + y)+(x - y)=7+( - 1) \)
\( 2x=6 \)
\( x = 3 \)

Step 2: Find \( y \) using \( x = 3 \)

Substitute \( x = 3 \) into \( x + y = 7 \):
\( 3 + y = 7 \)
\( y = 4 \)
Solution: \( (3, 4) \) (matches the purple point at \( x = 3,y = 4 \))

To "drag the system to the point", match each system to its solution point using the above calculations:

• \(

$$\begin{cases} 3x + 2y = 10 \\ 6x - y = 10 \end{cases}$$

\) → \( (2, 2) \)

• \(

$$\begin{cases} y = 3x - 7 \\ y = 2x - 5 \end{cases}$$

\) → \( (2, -1) \)

• \(

$$\begin{cases} y = 6 \\ x = -5 \end{cases}$$

\) → \( (-5, 6) \)

• \(

$$\begin{cases} 4x - 3y = 5 \\ 3x + 2y = -9 \end{cases}$$

\) → \( (-1, -3) \)

• \(

$$\begin{cases} 2x - y = -13 \\ y = x + 9 \end{cases}$$

\) → \( (-4, 5) \)

• \(

$$\begin{cases} x + y = 7 \\ x - y = -1 \end{cases}$$

\) → \( (3, 4) \)