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$\text{Hello ZBRS7311,} \\ \text{This question is about system of two quadratic equations, and it asks about which values of c does this system have: } \\ \text{1-Exactly one real solution} \\ \text{2-More than one real solution} \\ \text{3- No real solutions} \\ \text{The strategy is to make use of the discriminant of a quadratic equation. Namely the quantity: } \\ \Delta = b^2-4ac \space\space\space \text{Where } \Delta \space\text{is called the discriminant}\\ \text{But the problem is, which quadratic equation shall we apply this on? } \\ \text{Well, the quadratic equation that satisifies both "constraints" I.e. the intersection of both equations, I.e. equate both equations} \\ \text{Hence, setting both equations equal to each other, (eliminating y), yields:} \\ \implies 6x^2-9x+c=5x^2-3x \\ \iff x^2-6x+c=0 \\ \text{Notice: a=1,b=-6,c=c} \\ \text{Now for part (a) of the question, to have one real solution, the discriminant must be equal to zero} \\ \iff (-6)^2-4(1)(c)=0 \implies c=9 \\\text{Thus, this is the value of c such that the system has exactly one pair of (x,y) I.e. one solution}\\ \text{Next, for part (b), we want the system to have multiple solutions, this will be satisifed if and only if the discriminant is greater than 0} \\ \iff (-6)^2-4(1)(c)>0 \implies 36-4c> 0 \implies c<9 \\ \text{Thus, whenever c<9, the system will have more than one real solution} \\ \text{For the last part, (c), for the system to have no solutions, the discriminant must be less than zero} \\ (-6)^2-4(1)(c)<0 \implies 36-4c<0 \implies c>9 \text{Therefore, to have no real solutions, c>9}$I hope this helps, and do not hesitate to ask further questions.