Algebra 2 help?

Please help me with these problem.



Solve the equation for y. Then find the value of y for the given value x.



1) 3xy-4x=19 ; x=2

2) 11y+2xy=9 ; x= -5



Solve. And explain how you got the answer.



3) A chess tutor charges a fee for the first lesson that is 1.5 times the fee for later lessons. You spend $315 for 10 lessons. How much does the first lesson cost? How much does a later lesson cost?



4) You buy some calla lilies and peonies at a flower store. Calla lilies cost $3.50 each and peonies cost $5.50 each. The total cost of 12 flowers is $52. how many calla lilies and how many peonies did you buy?

Algebra 2 help?
1) 3xy-4x=19 ; x=2

taking x common:

x(3y-4)=19

3y-4=19/x

3y=19/x + 4 =(19+4x)/x

y=(19+4x)/(x*3)=(19+4x)/3x

putting the value of x in (19+4x)/3x:

y=(19+8)/(3*2)

y=27/6

dividing by 3

y=9/2

y=4.5



2)11y+2xy=9 ; x= -5

taking y common:

y(11+2x)=9

y=9/(11+2x)

putting the value of x in y=9/(11+2x)

y=9/[11+(2*-5)]

y=9/[11+(-10)]

y=9/(11-10)

y=9/1

y=9



3) let the fee for 1 later lesson = $x

therefore, fee for first lesson = $1.5x

no. of later lessons = 10-1=9

total fee for all the lessons = fee for first lesson + fee for 9 later lessons=$1.5x+$9x

therefore, 1.5x+9x=315

10.5x=315

x=315/10.5=3150/105=30

therefore, cost of a later lesson = $x = $30



4) let the no. of calla lilies = x and no. of peonies = y

total no. of flowers = 12

therefore, x+y=12

total cost of flowers = $52

therefore, 3.50x+5.50y=52



solve these simultaneous equations:

x+y=12 and 3.50x+5.50y=52

by substitution method:

x+y=12

therefore, x=12-y

putting this value of x in 3.50x+5.50y=52

3.50(12-y)+5.50y=52

42-3.50y+5.50y=52

42+2y=52

2y=52-42

2y=10

y=10/2

y=5

putting this value of y in x+y=12

x+5=12

x=12-5

x=7

therefore, no. of calla lilies bought by you = x = 7

and no. of peonies bought by you = y = 5



:-):-)