Answer:
Total amount Silas and Raja has at first = $252
Step-by-step explanation:
Let,
Amount Silas had = x
Amount Raja had = y
According to given statement;
x = 3y Eqn 1
x-84 = y+42 Eqn 2
Putting value of x from Eqn 1 in Eqn 2
3y - 84 = y+42
3y-y = 42+84
2y = 126
Dividing both sides by 2

Putting y = 63 in Eqn 1
x = 63(3)
x = 189
Total amount they had = 189 + 63 = $252
Hence,
Total amount Silas and Raja has at first = $252
How many three-fourths are in 2? 2 complete sets of three-fourths can be made and 2 of the 3 pieces need to make \frac34 are left over, so we have another \frac23 of a three-fourths. and we can see that there are 2 wholes with 4 fourths in each whole, so there are 2\times 4 fourths in 2.
Reduces to ,7=19. no solutions

first let's name a couples of variable
• the number of adults tickets sold: a
• the number of children tickets sold: c
From the problem we know
a + c = 128
and
$5.40c + $9.20a = $976.20
1) solve the equation to alpha
a+c-c = 128 -c
a+0=128-c
a=128-c
2) substitute (128 - c) for a in the second equation and solve to c
$5.40c + $9.20a = $976.20 become
$5.40c + $9.20(128 - c) = $976.20
$5.40c + ($9.20 × 128) - ($9.20 - c) = $976.20
$5.40c - $9.20c + $ 1177.6 = $976.20
($5.40 - $9.20)c +$1177.6 = $976.20
-$3.80c + $1177.6 = $9.76.20
-$3.80c + $1177.60 - $1177.60 = $976.20 - $1177.60
-$8.30c + 0 = $201.40
-$3.80c = - $201.40
-$3.80c. -$201.40
________. = _________
-$3.80. -$3.80
-$3.80c. -$201.40
________. = _________. - they are 4 cut the no
-$3.80. -$3.80
c = $201.40
________
3.80
c = 53