Match each system on the left with the number of solutions that it has on the right. Answer options on the right may be used more than once.

[tex]\[
\begin{array}{l}
1. \quad x = y - 3 \\
2. \quad 2x - 2y = 6 \\
3. \quad 5x + 2y = -7 \\
4. \quad 10x + 4y = -14 \\
5. \quad y = 2x + 1 \\
6. \quad 4x - 2y = -2 \\
7. \quad 9x + 6y = 15 \\
8. \quad 2y = -1 + 3x \\
\end{array}
\][/tex]

A. \quad no solution

B. \quad one solution

C. \quad infinite number of solutions



Answer :

Let's analyze each pair of equations step by step to determine the number of solutions for each system.

### System 1:
Equations:
[tex]\[ \begin{cases} x = y - 3 \\ 2x - 2y = 6 \end{cases} \][/tex]

To solve this system:
1. Substitute [tex]\( x = y - 3 \)[/tex] into [tex]\( 2x - 2y = 6 \)[/tex].
2. This gives [tex]\( 2(y - 3) - 2y = 6 \)[/tex].
3. Simplify to get [tex]\( 2y - 6 - 2y = 6 \)[/tex].
4. This simplifies to [tex]\(-6 = 6\)[/tex].

This system is inconsistent because [tex]\(-6 \neq 6\)[/tex].

Result: no solution

### System 2:
Equations:
[tex]\[ \begin{cases} 5x + 2y = -7 \\ 10x + 4y = -14 \end{cases} \][/tex]

To solve this system:
1. Note that [tex]\( 10x + 4y = -14 \)[/tex] is just a multiple of [tex]\( 5x + 2y = -7 \)[/tex].
2. This implies both equations represent the same line.

This system is dependent because both equations are multiples of each other.

Result: infinite number of solutions

### System 3:
Equations:
[tex]\[ \begin{cases} y = 2x + 1 \\ 4x - 2y = -2 \end{cases} \][/tex]

To solve this system:
1. Substitute [tex]\( y = 2x + 1 \)[/tex] into [tex]\( 4x - 2y = -2 \)[/tex].
2. This gives [tex]\( 4x - 2(2x + 1) = -2 \)[/tex].
3. Simplify to get [tex]\( 4x - 4x - 2 = -2 \)[/tex].
4. This simplifies to [tex]\(-2 = -2\)[/tex].

This implies a consistent system with one solution.

Result: one solution

### System 4:
Equations:
[tex]\[ \begin{cases} 9x + 6y = 15 \\ 2y = -1 + 3x \end{cases} \][/tex]

To solve this system:
1. Solve [tex]\( 2y = -1 + 3x \)[/tex] for [tex]\( y \)[/tex].
2. This gives [tex]\( y = \frac{3x - 1}{2} \)[/tex].
3. Substitute [tex]\( y = \frac{3x - 1}{2} \)[/tex] into [tex]\( 9x + 6y = 15 \)[/tex].
4. This gives [tex]\( 9x + 6(\frac{3x - 1}{2}) = 15 \)[/tex].
5. Simplify to get [tex]\( 9x + 9x - 3 = 15 \)[/tex].
6. Combine like terms to get [tex]\( 18x - 3 = 15 \)[/tex].
7. This simplifies to [tex]\( 18x = 18 \)[/tex], hence [tex]\( x = 1 \)[/tex].
8. Substitute [tex]\( x = 1 \)[/tex] into [tex]\( 2y = -1 + 3x \)[/tex] to find [tex]\( y \)[/tex].
9. This gives [tex]\( 2y = -1 + 3(1) = 2 \)[/tex].
10. Hence, [tex]\( y = 1 \)[/tex].

This is a consistent system with one unique solution.

Result: one solution

Therefore, the matching of systems to solutions is:

- [tex]\( \begin{cases} x = y - 3 \\ 2x - 2y = 6 \end{cases} \)[/tex] → no solution
- [tex]\( \begin{cases} 5x + 2y = -7 \\ 10x + 4y = -14 \end{cases} \)[/tex] → infinite number of solutions
- [tex]\( \begin{cases} y = 2x + 1 \\ 4x - 2y = -2 \end{cases} \)[/tex] → one solution
- [tex]\( \begin{cases} 9x + 6y = 15 \\ 2y = -1 + 3x \end{cases} \)[/tex] → one solution