Answer :
To determine which equation gives the value [tex]\( x = 8 \)[/tex], let's solve each equation individually.
### Equation A
Solve [tex]\( \frac{5}{2} x + \frac{7}{2} = \frac{3}{4} x + 14 \)[/tex]:
1. Eliminate the fractions by multiplying everything by 4:
[tex]\[ 4 \left( \frac{5}{2} x + \frac{7}{2} \right) = 4 \left( \frac{3}{4} x + 14 \right) \][/tex]
This simplifies to:
[tex]\[ 10x + 14 = 3x + 56 \][/tex]
2. Move terms involving [tex]\( x \)[/tex] to one side and constants to the other side:
[tex]\[ 10x - 3x = 56 - 14 \][/tex]
[tex]\[ 7x = 42 \][/tex]
3. Solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{42}{7} = 6 \][/tex]
So, Equation A gives [tex]\( x = 6 \)[/tex].
### Equation B
Solve [tex]\( \frac{5}{4} x - 9 = \frac{3}{2} x - 12 \)[/tex]:
1. Eliminate the fractions by multiplying everything by 4:
[tex]\[ 4 \left( \frac{5}{4} x - 9 \right) = 4 \left( \frac{3}{2} x - 12 \right) \][/tex]
This simplifies to:
[tex]\[ 5x - 36 = 6x - 48 \][/tex]
2. Move terms involving [tex]\( x \)[/tex] to one side and constants to the other side:
[tex]\[ 5x - 6x = -48 + 36 \][/tex]
[tex]\[ -x = -12 \][/tex]
3. Solve for [tex]\( x \)[/tex]:
[tex]\[ x = 12 \][/tex]
So, Equation B gives [tex]\( x = 12 \)[/tex].
### Equation C
Solve [tex]\( \frac{5}{4} x - 2 = \frac{3}{2} x - 4 \)[/tex]:
1. Eliminate the fractions by multiplying everything by 4:
[tex]\[ 4 \left( \frac{5}{4} x - 2 \right) = 4 \left( \frac{3}{2} x - 4 \right) \][/tex]
This simplifies to:
[tex]\[ 5x - 8 = 6x - 16 \][/tex]
2. Move terms involving [tex]\( x \)[/tex] to one side and constants to the other side:
[tex]\[ 5x - 6x = -16 + 8 \][/tex]
[tex]\[ -x = -8 \][/tex]
3. Solve for [tex]\( x \)[/tex]:
[tex]\[ x = 8 \][/tex]
So, Equation C gives [tex]\( x = 8 \)[/tex].
### Equation D
Solve [tex]\( \frac{5}{2} x - 7 = \frac{3}{4} x + 14 \)[/tex]:
1. Eliminate the fractions by multiplying everything by 4:
[tex]\[ 4 \left( \frac{5}{2} x - 7 \right) = 4 \left( \frac{3}{4} x + 14 \right) \][/tex]
This simplifies to:
[tex]\[ 10x - 28 = 3x + 56 \][/tex]
2. Move terms involving [tex]\( x \)[/tex] to one side and constants to the other side:
[tex]\[ 10x - 3x = 56 + 28 \][/tex]
[tex]\[ 7x = 84 \][/tex]
3. Solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{84}{7} = 12 \][/tex]
So, Equation D gives [tex]\( x = 12 \)[/tex].
### Conclusion
After solving all the equations:
- Equation A gives [tex]\( x = 6 \)[/tex]
- Equation B gives [tex]\( x = 12 \)[/tex]
- Equation C gives [tex]\( x = 8 \)[/tex]
- Equation D gives [tex]\( x = 12 \)[/tex]
The correct equation that gives [tex]\( x = 8 \)[/tex] is Equation C:
[tex]\[ \frac{5}{4} x - 2 = \frac{3}{2} x - 4 \][/tex]
So, the correct answer is:
[tex]\[ \boxed{\text{C}} \][/tex]
### Equation A
Solve [tex]\( \frac{5}{2} x + \frac{7}{2} = \frac{3}{4} x + 14 \)[/tex]:
1. Eliminate the fractions by multiplying everything by 4:
[tex]\[ 4 \left( \frac{5}{2} x + \frac{7}{2} \right) = 4 \left( \frac{3}{4} x + 14 \right) \][/tex]
This simplifies to:
[tex]\[ 10x + 14 = 3x + 56 \][/tex]
2. Move terms involving [tex]\( x \)[/tex] to one side and constants to the other side:
[tex]\[ 10x - 3x = 56 - 14 \][/tex]
[tex]\[ 7x = 42 \][/tex]
3. Solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{42}{7} = 6 \][/tex]
So, Equation A gives [tex]\( x = 6 \)[/tex].
### Equation B
Solve [tex]\( \frac{5}{4} x - 9 = \frac{3}{2} x - 12 \)[/tex]:
1. Eliminate the fractions by multiplying everything by 4:
[tex]\[ 4 \left( \frac{5}{4} x - 9 \right) = 4 \left( \frac{3}{2} x - 12 \right) \][/tex]
This simplifies to:
[tex]\[ 5x - 36 = 6x - 48 \][/tex]
2. Move terms involving [tex]\( x \)[/tex] to one side and constants to the other side:
[tex]\[ 5x - 6x = -48 + 36 \][/tex]
[tex]\[ -x = -12 \][/tex]
3. Solve for [tex]\( x \)[/tex]:
[tex]\[ x = 12 \][/tex]
So, Equation B gives [tex]\( x = 12 \)[/tex].
### Equation C
Solve [tex]\( \frac{5}{4} x - 2 = \frac{3}{2} x - 4 \)[/tex]:
1. Eliminate the fractions by multiplying everything by 4:
[tex]\[ 4 \left( \frac{5}{4} x - 2 \right) = 4 \left( \frac{3}{2} x - 4 \right) \][/tex]
This simplifies to:
[tex]\[ 5x - 8 = 6x - 16 \][/tex]
2. Move terms involving [tex]\( x \)[/tex] to one side and constants to the other side:
[tex]\[ 5x - 6x = -16 + 8 \][/tex]
[tex]\[ -x = -8 \][/tex]
3. Solve for [tex]\( x \)[/tex]:
[tex]\[ x = 8 \][/tex]
So, Equation C gives [tex]\( x = 8 \)[/tex].
### Equation D
Solve [tex]\( \frac{5}{2} x - 7 = \frac{3}{4} x + 14 \)[/tex]:
1. Eliminate the fractions by multiplying everything by 4:
[tex]\[ 4 \left( \frac{5}{2} x - 7 \right) = 4 \left( \frac{3}{4} x + 14 \right) \][/tex]
This simplifies to:
[tex]\[ 10x - 28 = 3x + 56 \][/tex]
2. Move terms involving [tex]\( x \)[/tex] to one side and constants to the other side:
[tex]\[ 10x - 3x = 56 + 28 \][/tex]
[tex]\[ 7x = 84 \][/tex]
3. Solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{84}{7} = 12 \][/tex]
So, Equation D gives [tex]\( x = 12 \)[/tex].
### Conclusion
After solving all the equations:
- Equation A gives [tex]\( x = 6 \)[/tex]
- Equation B gives [tex]\( x = 12 \)[/tex]
- Equation C gives [tex]\( x = 8 \)[/tex]
- Equation D gives [tex]\( x = 12 \)[/tex]
The correct equation that gives [tex]\( x = 8 \)[/tex] is Equation C:
[tex]\[ \frac{5}{4} x - 2 = \frac{3}{2} x - 4 \][/tex]
So, the correct answer is:
[tex]\[ \boxed{\text{C}} \][/tex]
