The price of petrol in january was y dollars per litre and Mr
Tan paid $72 to fill up his car with petrol. He could cover a
distance of 500 km with this amount of pertrol. In August,the price
of petrol increase by 20 cents. Mr Tan noticed that he was
getting 2 litres of petrol less after paying $76. Form an equation
in y and show that it reduces to 5y^2+11y-36=0
2)Simplify 6v^2 + 6u^2 + 12uv / 3vx +3ux + 4uw + 4vw
Amount of Petrol he topped up in January
= 72/y litres
Amount of Petrol he topped up in August
= ( ( 72/y ) - 2 ) litres
Amount paid per litre in August
= $( y+0.2 )
Amount of petrol topped up X Amount paid per litre = Total
Amount paid
( ( 72/y ) -2 ) X ( y+0.2 ) = 76
( 72 -2y )( y+0.2 ) = 76y
( 72-2y ) ( 5y+1 ) = 380y
360y - 10y^2 + 72 - 2y - 380y = 0
-10y^2 - 22y + 72 = 0
10y^2 + 22y - 72 = 0
5y^2 + 11y -
36 = 0
6v^2 + 6u^2 + 12uv / 3vx +3ux + 4uw + 4vw
= 6(v+u)^2 / [ 3x(v+u) + 4w(v+u) ]
= 6(v+u)^2 / (v+u)(3x+4w)
= 6(v+u) /
(3x+4w)