In school it is good practice to explain what steps your are applying. That way the teacher can see your way of thinking about manipulation and better understand you intention.

Given:

#2/x+p=3" "..................................Equation(1)#

#(5(7x+5))/3-23/2=13" "..............Equation(2)#

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#Equation(1)# has 2 unknowns so can not be solved directly. We need 1 equation with 1 unknown. That situation exists in #Equation(2)#

So we can solve for #x# in #Eqn(2)# and then substitute for #x# in #Eqn(1)#. Thus solving for #p#.

#color(brown)("Consider "Equation(2)->(5(7x+5))/3-23/2=13)#

Add #23/2# to both sides giving:

#(5(7x+5))/3=49/2#

Multiply both sides by #3/5#

#7x+5=3/5xx49/2#

#7x+5=147/10#

Subtract 5 from both sides:

#7x=97/10#

Divide both sides by 7

#color(blue)(x=97/70)#

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#color(brown)("Substitute for "x" in "Equation(1))#

#color(green)(2/color(red)(x)+p=3 color(white)("dddd") ->color(white)("dddd")(2 -:color(red)(97/70))+p=3 )#

#color(green)(color(white)("ddddddddddd.d")->color(white)("ddddddd")140/97color(white)("dd")+p=3)#

Subtract #140/97# from both sides

#color(blue)(color(white)("ddddddddddddd")->color(white)("dddddd")p=151/97)#

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#color(brown)("Check")#

#p=151/97#

#x=97/70#

Left hand side of #2/x+p=3#

#(color(white)(..)2color(white)(..))/(97/70) + 151/97 #

#140/97+151/97#

#291/97 ->3#

Thus #LHS=RHS=3#