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





:-):-)