Answer:
I is clear that, the linear equation [tex]5x+12=5x-7[/tex] has no solution.
Step-by-step explanation:
Checking the first option:
[tex]\frac{2}{3}\left(9x+6\right)=6x+4[/tex]
[tex]6x+4=6x+4[/tex]
[tex]\mathrm{Subtract\:}4\mathrm{\:from\:both\:sides}[/tex]
[tex]6x+4-4=6x+4-4[/tex]
[tex]6x=6x[/tex]
[tex]\mathrm{Subtract\:}6x\mathrm{\:from\:both\:sides}[/tex]
[tex]6x-6x=6x-6x[/tex]
[tex]0=0[/tex]
[tex]\mathrm{Both\:sides\:are\:equal}[/tex]
[tex]\mathrm{True\:for\:all}\:x[/tex]
Checking the 2nd option:
[tex]5x+12=5x-7[/tex]
[tex]\mathrm{Subtract\:}5x\mathrm{\:from\:both\:sides}[/tex]
[tex]5x+12-5x=5x-7-5x[/tex]
[tex]\mathrm{Simplify}[/tex]
[tex]12=-7[/tex]
[tex]\mathrm{The\:sides\:are\:not\:equal}[/tex]
[tex]\mathrm{No\:Solution}[/tex]
Checking the 3rd option:
[tex]4x+7=3x+7[/tex]
[tex]\mathrm{Subtract\:}7\mathrm{\:from\:both\:sides}[/tex]
[tex]4x+7-7=3x+7-7[/tex]
[tex]\mathrm{Simplify}[/tex]
[tex]4x=3x[/tex]
[tex]\mathrm{Subtract\:}3x\mathrm{\:from\:both\:sides}[/tex]
[tex]4x-3x=3x-3x[/tex]
[tex]\mathrm{Simplify}[/tex]
[tex]x=0[/tex]
Checking the 4th option:
[tex]-3\left(2x-5\right)=15-6x[/tex]
[tex]\mathrm{Subtract\:}15\mathrm{\:from\:both\:sides}[/tex]
[tex]-6x+15-15=15-6x-15[/tex]
[tex]\mathrm{Simplify}\[/tex]
[tex]-6x=-6x[/tex]
[tex]\mathrm{Add\:}6x\mathrm{\:to\:both\:sides}[/tex]
[tex]-6x+6x=-6x+6x[/tex]
[tex]\mathrm{Simplify}[/tex]
[tex]\mathrm{Both\:sides\:are\:equal}[/tex]
[tex]\mathrm{True\:for\:all}\:x[/tex]
Result:
Therefore, from the above calculations it is clear that, the linear equation
[tex]5x+12=5x-7[/tex] has no solution.
