Thread: challenging math algebra question. (need some help) no answers

here goes : (any help appreciated)

a) when driven in town, a car runs at x kilometres on each litre of petrol
(i) find in terms of x , the number of litres of petrol used when the car is driven 200km in town
(ii) when driven in town , the car runs (x+4) kilometres on each litre of petrol to go 200km out of town than to go 200km in town. Use this information to write down an equation involving x , and show that is simplifies to :

x^2 + 4x - 160 = 0

b) solve the equation x^2 + 4x - 160 = 0

c)Calculate total volume of petrol used when the car is driven 40km in down and then 120km out of town.


ALL HELP APPRECIATED. WILL THANK ANYONE WHO EVEN OFFER SUGGESTIONS OR TIPS

looking for guides like a general formula that I can cross multiply or make them common denominator

Like an equation like this for me to make common demoniator

(x/number) +/- (x/number) = (x/number)

Originally Posted by coffeegirly99

here goes : (any help appreciated)

a) when driven in town, a car runs at x kilometres on each litre of petrol
(i) find in terms of x , the number of litres of petrol used when the car is driven 200km in town

number of Liters, V = 200/x

(ii) when driven in town , the car runs (x+4) kilometres on each litre of petrol to go 200km out of town than to go 200km in town. Use this information to write down an equation involving x , and show that is simplifies to :

x^2 + 4x - 160 = 0

the bold statement makes little sense to me ... I understand that in town the car gets x km/L and out of town the car gets (x+4) km/L ... recheck the exact wording.

b) solve the equation x^2 + 4x - 160 = 0







since x represents km,

c)Calculate total volume of petrol used when the car is driven 40km in down and then 120km out of town.

hope this helps