Sure, let's solve each equation to find the solutions of the form [tex]\( x = 0, y = a \)[/tex] and [tex]\( x = b, y = 0 \)[/tex].
### 1. Solve [tex]\( 5x - 2y = 10 \)[/tex]
#### When [tex]\( x = 0 \)[/tex]:
Substitute [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[ 5(0) - 2y = 10 \][/tex]
[tex]\[ -2y = 10 \][/tex]
[tex]\[ y = -5 \][/tex]
So, one solution is [tex]\( (0, -5) \)[/tex].
#### When [tex]\( y = 0 \)[/tex]:
Substitute [tex]\( y = 0 \)[/tex] into the equation:
[tex]\[ 5x - 2(0) = 10 \][/tex]
[tex]\[ 5x = 10 \][/tex]
[tex]\[ x = 2 \][/tex]
So, the other solution is [tex]\( (2, 0) \)[/tex].
### 2. Solve [tex]\( -4x + 3y = 12 \)[/tex]
#### When [tex]\( x = 0 \)[/tex]:
Substitute [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[ -4(0) + 3y = 12 \][/tex]
[tex]\[ 3y = 12 \][/tex]
[tex]\[ y = 4 \][/tex]
So, one solution is [tex]\( (0, 4) \)[/tex].
#### When [tex]\( y = 0 \)[/tex]:
Substitute [tex]\( y = 0 \)[/tex] into the equation:
[tex]\[ -4x + 3(0) = 12 \][/tex]
[tex]\[ -4x = 12 \][/tex]
[tex]\[ x = -3 \][/tex]
So, the other solution is [tex]\( (-3, 0) \)[/tex].
### 3. Solve [tex]\( 2x + 3y = 24 \)[/tex]
#### When [tex]\( x = 0 \)[/tex]:
Substitute [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[ 2(0) + 3y = 24 \][/tex]
[tex]\[ 3y = 24 \][/tex]
[tex]\[ y = 8 \][/tex]
So, one solution is [tex]\( (0, 8) \)[/tex].
#### When [tex]\( y = 0 \)[/tex]:
Substitute [tex]\( y = 0 \)[/tex] into the equation:
[tex]\[ 2x + 3(0) = 24 \][/tex]
[tex]\[ 2x = 24 \][/tex]
[tex]\[ x = 12 \][/tex]
So, the other solution is [tex]\( (12, 0) \)[/tex].
### Summary of Solutions
1. Equation [tex]\( 5x - 2y = 10 \)[/tex]:
- [tex]\( (0, -5) \)[/tex]
- [tex]\( (2, 0) \)[/tex]
2. Equation [tex]\( -4x + 3y = 12 \)[/tex]:
- [tex]\( (0, 4) \)[/tex]
- [tex]\( (-3, 0) \)[/tex]
3. Equation [tex]\( 2x + 3y = 24 \)[/tex]:
- [tex]\( (0, 8) \)[/tex]
- [tex]\( (12, 0) \)[/tex]
