Etermine whether the system of linear equations has one and only one solution, infinitely many solutions, or no solution. 3 2 x − 2y = 4 x + 1 3 y = 2 one and only one solution infinitely many solutions no solution find the solution, if one exists. (if there are infinitely many solutions, express x and y in terms of the parameter t. if there is no solution, enter no solution.) (x, y) =

(32/15, -2/5)

Step-by-step explanation:

(1) ³/₂x   -    2y =  4

(2)   x  +    ⅓y =  2

(3)  3x   -     4y =  8              Multiplied (1) by 2

(4)  3x  +       y =  6              Multiplied (2) by 3

                  5y = -2              Subtracted (3) from (4)

(5)                 y = -⅖             Divided each side by 5

       x + ⅓(-⅖) =  2              Substituted (5) into (2)

       x  -  ²/₁₅    =  2

       x              =  2 + ²/₁₅     Added ²/₁₅ to each side

      x              = 32/15

The system of equations has only one solution: (32/15, -2/5).

jwjwkwkwkkkdjebd b jdjejdndj he jenej

step-by-step explanation: so we look at the formula un = a + (n-1)d

un is the amount of seats

n is the nth term aka row number

a is the original sequence/row number

d is the difference between each term

so in this a=22 d=6 and n can remain as n because we are simply formjng a rule

lets put these in the equation

un=22+(n-1)(6)

this is your answer to part a

however

lets double check that this is correct, we'll try putting it into the second term

un=a+(n-1)d

u2=22+(2-1)(6)

u2=22+6

u2=28

as the question said the second row has 28 seats so now we know the formula we made is correct.

as for part b

un=a+(n-1)

its asking us to find a row with 100 seats.

remember un was the amount of seata so all we do is substitute this in.

un=100 a=22 d=6

un=a+(n-1)d

100=22+(n-1)(6)

so now we simplify to find n which is the row number.

100-22=22+(n-1)(6)-22

78=(n-1)(6)

78/6=(n-1)(6)/6

13=n-1

13+1=n-1+1

14=n

so now we know that in the 14th row there is 100 seats.

hope this and that you understood! its my first time using the app or anyone out on here : p its also me to revise so a bunch! ask more questions if you dont understand ☺