Sure! Let's break down each equation and determine which one has no solution.
Equation 1:
[tex]\[ -\frac{1}{2}(8x + 6) - 2x = -3(2x + 1) \][/tex]
Expanding and simplifying:
[tex]\[ -\frac{1}{2} \cdot 8x - \frac{1}{2} \cdot 6 - 2x = -3 \cdot 2x - 3 \cdot 1 \][/tex]
[tex]\[ -4x - 3 - 2x = -6x - 3 \][/tex]
Combining like terms:
[tex]\[ -6x - 3 = -6x - 3 \][/tex]
Since this equation simplifies to a true statement for any [tex]\( x \)[/tex], it has infinitely many solutions.
Equation 2:
[tex]\[ -4\left(\frac{1}{2}x + 2\right) = -2x - 8 + 4x \][/tex]
Expanding and simplifying:
[tex]\[ -4 \cdot \frac{1}{2}x - 4 \cdot 2 = -2x - 8 + 4x \][/tex]
[tex]\[ -2x - 8 = -2x - 8 + 4x \][/tex]
Combining like terms:
[tex]\[ -2x - 8 = 2x - 8 \][/tex]
Adding [tex]\( 2x \)[/tex] to both sides gives:
[tex]\[ -8 = -8 \][/tex]
Since this equation simplifies to a true statement for any [tex]\( x \)[/tex],
it also has infinitely many solutions.
Equation 3:
[tex]\[ 7(x + 2) - 3x = \frac{2}{3}(6x + 3) \][/tex]
Expanding and simplifying:
[tex]\[ 7x + 14 - 3x = \frac{2}{3}(6x + 3) \][/tex]
[tex]\[ 4x + 14 = 4x + 2 \][/tex]
Subtract [tex]\( 4x \)[/tex] from both sides:
[tex]\[ 14 = 2 \][/tex]
This is a contradiction, meaning the equation has no solution.
Equation 4:
[tex]\[ -2x + 5 - 3x + 12 = -5(x - 3) + 2 \][/tex]
Expanding and simplifying:
[tex]\[ -2x - 3x + 5 + 12 = -5x + 15 + 2 \][/tex]
[tex]\[ -5x + 17 = -5x + 17 \][/tex]
Since this simplifies to a true statement for any [tex]\( x \)[/tex], it has infinitely many solutions.
Conclusion:
Equation 3 has no solution.
