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#
