Answer :
Let's consider the three pairs of equations and determine their solutions step-by-step.
### Pair 1
#### Equations:
1. [tex]\( y = -\frac{2}{3} x + 2 \)[/tex]
2. [tex]\( x = -\frac{3}{2} y + 2 \)[/tex]
#### Converting to Standard Form:
1. [tex]\(-\frac{2}{3} x + y = 2\)[/tex]
2. [tex]\(x + \frac{3}{2} y = 2\)[/tex]
To solve these, we need to express them in a way that's easier to manipulate.
From the first equation:
[tex]\[ y = -\frac{2}{3} x + 2 \][/tex]
From the second equation:
[tex]\[ x = -\frac{3}{2} y + 2 \][/tex]
Solve one of the equations for [tex]\( x \)[/tex] or [tex]\( y \)[/tex] and substitute into the other equation:
[tex]\[ y = -\frac{2}{3} x + 2 \][/tex]
[tex]\[ x = -\frac{3}{2} y + 2 \][/tex]
Substituting [tex]\( y \)[/tex] from the first equation into the second equation:
[tex]\[ x = -\frac{3}{2} (-\frac{2}{3} x + 2) + 2 \][/tex]
[tex]\[ x = x \][/tex]
This system of equations has an infinite number of solutions because the equations are dependent (they are essentially the same line).
### Pair 2
#### Equations:
3. [tex]\( y = 3 x - 2 \)[/tex]
4. [tex]\( 9 x - 3 y = 6 \)[/tex]
#### Converting to Standard Form:
1. [tex]\(y - 3x = -2\)[/tex]
2. [tex]\(9 x - 3 y = 6 \)[/tex]
Multiply the first equation by 3:
[tex]\[ 3y - 9x = -6 \][/tex]
Notice that this is the same as the second equation but rearranged. Thus, these equations are dependent, representing the same line.
This system of equations has an infinite number of solutions.
### Pair 3
#### Equations:
5. [tex]\( y = 2 x - 3 \)[/tex]
6. [tex]\( x + 2 y = 4 \)[/tex]
#### Converting to Standard Form:
1. [tex]\(y - 2x = -3\)[/tex]
2. [tex]\(x + 2y = 4\)[/tex]
Let's solve these equations simultaneously.
From the first equation:
[tex]\[ y = 2x - 3 \][/tex]
Substitute [tex]\( y \)[/tex] from the first equation into the second equation:
[tex]\[ x + 2(2x - 3) = 4 \][/tex]
[tex]\[ x + 4x - 6 = 4 \][/tex]
[tex]\[ 5x - 6 = 4 \][/tex]
[tex]\[ 5x = 10 \][/tex]
[tex]\[ x = 2 \][/tex]
Now substitute [tex]\( x = 2 \)[/tex] back into the equation [tex]\( y = 2 x - 3 \)[/tex]:
[tex]\[ y = 2(2) - 3 \][/tex]
[tex]\[ y = 4 - 3 \][/tex]
[tex]\[ y = 1 \][/tex]
Therefore, the solution to this pair of equations is [tex]\((x, y) = (2, 1)\)[/tex].
### Conclusion
- The pair [tex]\((y = -\frac{2}{3} x + 2, x = -\frac{3}{2} y + 2)\)[/tex] has infinite number of solutions.
- The pair [tex]\((y = 3 x - 2, 9 x - 3 y = 6)\)[/tex] has infinite number of solutions.
- The pair [tex]\((y = 2 x - 3, x + 2 y = 4)\)[/tex] has the solution [tex]\((x, y) = (2, 1)\)[/tex].
### Pair 1
#### Equations:
1. [tex]\( y = -\frac{2}{3} x + 2 \)[/tex]
2. [tex]\( x = -\frac{3}{2} y + 2 \)[/tex]
#### Converting to Standard Form:
1. [tex]\(-\frac{2}{3} x + y = 2\)[/tex]
2. [tex]\(x + \frac{3}{2} y = 2\)[/tex]
To solve these, we need to express them in a way that's easier to manipulate.
From the first equation:
[tex]\[ y = -\frac{2}{3} x + 2 \][/tex]
From the second equation:
[tex]\[ x = -\frac{3}{2} y + 2 \][/tex]
Solve one of the equations for [tex]\( x \)[/tex] or [tex]\( y \)[/tex] and substitute into the other equation:
[tex]\[ y = -\frac{2}{3} x + 2 \][/tex]
[tex]\[ x = -\frac{3}{2} y + 2 \][/tex]
Substituting [tex]\( y \)[/tex] from the first equation into the second equation:
[tex]\[ x = -\frac{3}{2} (-\frac{2}{3} x + 2) + 2 \][/tex]
[tex]\[ x = x \][/tex]
This system of equations has an infinite number of solutions because the equations are dependent (they are essentially the same line).
### Pair 2
#### Equations:
3. [tex]\( y = 3 x - 2 \)[/tex]
4. [tex]\( 9 x - 3 y = 6 \)[/tex]
#### Converting to Standard Form:
1. [tex]\(y - 3x = -2\)[/tex]
2. [tex]\(9 x - 3 y = 6 \)[/tex]
Multiply the first equation by 3:
[tex]\[ 3y - 9x = -6 \][/tex]
Notice that this is the same as the second equation but rearranged. Thus, these equations are dependent, representing the same line.
This system of equations has an infinite number of solutions.
### Pair 3
#### Equations:
5. [tex]\( y = 2 x - 3 \)[/tex]
6. [tex]\( x + 2 y = 4 \)[/tex]
#### Converting to Standard Form:
1. [tex]\(y - 2x = -3\)[/tex]
2. [tex]\(x + 2y = 4\)[/tex]
Let's solve these equations simultaneously.
From the first equation:
[tex]\[ y = 2x - 3 \][/tex]
Substitute [tex]\( y \)[/tex] from the first equation into the second equation:
[tex]\[ x + 2(2x - 3) = 4 \][/tex]
[tex]\[ x + 4x - 6 = 4 \][/tex]
[tex]\[ 5x - 6 = 4 \][/tex]
[tex]\[ 5x = 10 \][/tex]
[tex]\[ x = 2 \][/tex]
Now substitute [tex]\( x = 2 \)[/tex] back into the equation [tex]\( y = 2 x - 3 \)[/tex]:
[tex]\[ y = 2(2) - 3 \][/tex]
[tex]\[ y = 4 - 3 \][/tex]
[tex]\[ y = 1 \][/tex]
Therefore, the solution to this pair of equations is [tex]\((x, y) = (2, 1)\)[/tex].
### Conclusion
- The pair [tex]\((y = -\frac{2}{3} x + 2, x = -\frac{3}{2} y + 2)\)[/tex] has infinite number of solutions.
- The pair [tex]\((y = 3 x - 2, 9 x - 3 y = 6)\)[/tex] has infinite number of solutions.
- The pair [tex]\((y = 2 x - 3, x + 2 y = 4)\)[/tex] has the solution [tex]\((x, y) = (2, 1)\)[/tex].