Let the cost of a toy car be 'x' and that of candy bar be 'y' 3x+8y=21...........................eq1 2x+12y=24.........................eq2 Multiply eq1 by 2 6x+16y=42.........................eq3 Multiply eq2 by 3 6x+36y=72--------------------eq4 eq4 - eq3 20y=30 y=30/20 y=1.5 substituting for y in eq1 we get 3x+8(1.5)=21 3x=21-12 3x=9 x=3 One toy car costs $3