There are 28 students in first period and 56 students in sceond period and 74 students in third period
<em><u>Solution:</u></em>
Let the number of students in first period be "x"
Let the number of students in second period be "y"
Let the number of students in third period be "z"
<em><u>There are 158 students registered for American History classes.</u></em>
Therefore,
x + y + z = 158 ---------- eqn 1
<em><u>There are twice as many students registered in second period as first period</u></em>
number of students in second period = twice of number of students in first period
y = 2x ------- eqn 2
<em><u>There are 10 less than three times as many students in third period as in first period</u></em>
number of students in third period = 3 times number of students in first period - 10
z = 3x - 10 ------ eqn 3
<em><u>Substitute eqn 2 and eqn 3 in eqn 1</u></em>
x + 2x + 3x - 10 = 158
6x = 168
<h3>x = 28</h3>
<em><u>Substitute x = 28 in eqn 2</u></em>
y = 2(28)
<h3>y = 56</h3>
<em><u>Substitute x = 28 in eqn 3</u></em>
z = 3(28) - 10
z = 84 - 10
<h3>z = 74</h3>
Thus there are 28 students in first period and 56 students in sceond period and 74 students in third period
