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Sorry I have changed your function (but I thinks it is not the problem). You must usee a dictionary to reuse the result

a=3
var('x')
eq1 = x+a==0
sol=solve([eq1],x,solution_dict=True)
x_sol=sol[0][x]
show(LatexExpr(r"x\_sol="+latex(x_sol)))
f(x)=7*x +10
show(LatexExpr(r"f(x\_sol)="+latex(f(x_sol))))

Sorry I have changed your function (but I thinks it is not the problem). You must usee use a dictionary to reuse the result

a=3
var('x')
eq1 = x+a==0
sol=solve([eq1],x,solution_dict=True)
x_sol=sol[0][x]
show(LatexExpr(r"x\_sol="+latex(x_sol)))
f(x)=7*x +10
show(LatexExpr(r"f(x\_sol)="+latex(f(x_sol))))

Sorry I have changed your function (but I thinks it is not the problem). You must use a dictionary to reuse the result

a=3
var('x')
eq1 = x+a==0
sol=solve([eq1],x,solution_dict=True)
x_sol=sol[0][x]
show(LatexExpr(r"x\_sol="+latex(x_sol)))
f(x)=7*x f(x)=5*x +10
show(LatexExpr(r"f(x\_sol)="+latex(f(x_sol))))