Multiplying (ii.) by -9, and subtracting the result from (i.), we get 104x+156-80x +240 +5x2 = 16x2+24x, x = 3. (2) (x+2) (c −3) + 2(2+cx) = 0, cx+2c-3x-6+4+2cx=0, 3Cx-3x=2— 26, 6. Let x= number of party, 120 then = number of shillings each had to pay, 120 x Then +2=what each had to pay = = 240+5 2x x2 + 100x= = 11900, x2 + 100x+(50)2 = 11900 + 2500 = 14400. x+50=120. (1) 2x-y=80. (2) 47x-18y = 2100. Then from (1), 36x-18y=1440, (2), 47x-18y=2100. Subtracting (1) from (2), 11x=660, x=60. From (1), 120-y=80, y = 40. PROBLEMS LEADING TO SIMPLE EQUATIONS OF THE FIRST DEGREE OF TWO OR MORE UNKNOWN QUANTITIES. 1. Let x = numerator, y = denominator of required fraction. 3. Let y=price per score in pence of small. Then y + 1 = price per score in pence of large. Also let x = number bought of each kind. Then price paid for small=x.y 20 19 99 large = *(y+1) |