A basketball team sells tickets that cost$10,$20, $30 the team has sold 3322 tickets overall. It has sold 191 more $20 tickets than $10 tickets. The total sales are 64, 180.How many tickets of each kind have been sold

1022 of $10 tickets, 1326 of $20 tickets, 948 of VIP $30 tickets

$10 tickets = 1,119

$20 tickets = 1,310

$30 tickets = 893

Step-by-step explanation:

x = $10 tickets, y = $20 tickets, z = $30 tickets

10x + 20y + 30z = 64,180

x + y + z = 3,322

y = 191 + x

therefore;

10x + 20(191+x) + 30z = 64,180     eqn 1

x + (191 + x) + z = 3,322         eqn 2

make z the subject of eqn 2

z = 3,322 - x - (191 + x)

z = 3,322 - 191 - x - x

z = 3131 - 2x

put z in eqn 1

10x + 20(191+x) + 30(3131 - 2x) = 64,180

10x + 3820 + 20x + 93,930 - 60x = 64,180

10x + 20x - 60x = 64,180 - 3820 - 93,930

-30x         =      -33,570

x      =   1,119

put x = 1,119 in z

z = 3131 - 2(1119)

z = 893

put x = 1,119 in y

y = 191 + 1119

  = 1,310

10(1119) + 20(1310) + 30(893) = 64,180

1119 + 1310 + 893 = 3,322

f is the midpoint of

6e14 is the answer.