1- Solve the following equation?

Question

1- Solve the following equation for x in terms of y.
7x - 2y = 21
2 - Solve the following equations. Find the appropriate values for p and q.
3p - 2q = 37
p + 4q = 3
3 - A pack of gum costs $0.35. If only nickels, dimes, and quarters are available, how many different coin combinations can be used to equal $0.35?
PLEEEEEASE HELP ME PLEASE
10 POINTS FOR BEST ANSWER Thank You

Answer 1

1. X = (21+2y)/7

2. p=(11) q=(-2)

3. 6 combinations

7 nickels
5 nickels 1 dime
3 nickels 2 dimes
2 nickels 1 quarter
1 nickel 3 dimes
1 quarter 1 dime

Answer 2

ok here v go
if 7x-2y=21
then 7x=21-2y
and x=(21-2y)7
as y is not given so no definite solution but the thing is solved in terms of y u say and i say it is x as it relates to y

the other is let those 2 eqations be called a and b respactivesly then
by multiplying a with 2 u get 6p-4q=74
now let it call c
now let c be added in b then v get
7p=77 so p =11
now put the value in any equation except 'c' i add it to first ie 'a'
v get 11*3-2q=37
now 33 - 2q = 37
now -2q = 37-33 i,e -2q = 4
now multiply it with -1 v get
2q = -4
so q = -2

Answer 3

ok...
1. solving for x in terms of y just means to isolate x on one side of the '='
so add 2y to both sides
7x=21 + 2y
then divide by 7
x = 3 +2/7y

2. this is a system of equations. the easiest way is to solve one equation for one variable in terms of the other and then plug it into the other equation
so lets solve the second equation for p by subtracting 4q from both sides
p = 3 - 4q
and inserting that into the first equation
3*(3 - 4q) - 2q = 37
solve that for q, then plug that value in to find p

3 this one you can just plug it out
1 qt + 1 dm
1 qt + 2 nk
1 dm + 5 nk
2 dm + 3 nk
3 dm + 1 nk
and so on...

hope that helps

Answer 4

7x - 2y + 21
7x = 21 - 2y

Dividing by 7

x = 21/7 - 2y/7

= 3 - 2y/7

x = 3 - 2y/7


2) 3p - 2q = 37 (i)

p + 4q = 3 (ii)

multiplying (ii) by 3

3p + 12q = 9 (iii)

substracting (i0 with (iii)

3p - 2q = 37
3p + 12q = 9

--------------------
0 - 14q = 28

q = -28/14 = -2

q = -2

substituting in (ii)

p +4 X (_2) = 3

p - 8 = 3
p = 11

So q = -2 p = 11

I donot know the 3rd problem

Answer 5

1. -2y = 21 - 7x
y = -10.5 + 3.5x

2. 3p - 2q = 37
3p + 12q = 9

-14q = 28

q = -2

p + 4(-2) = 3 ---> p = 11

3p -2q =37
3 (11) - 2(-2) = 37

3. 1Q+1D, 1Q+2N, 3D+1N, 2D+3N, 1D+5N, 7N
6 combinations

Answer 6

1 -
x = (21 + 2y) / 7

2 -
6p -4q = 74
p + 4q = 3
7p = 77
p=11, and q= -2

3-
nickel is $0.05
dimes is $0.10
quarters is $0.25
so the combination is
-7n 0d 0q - 5n 1d 0q - 3n 2d 0q - 1n 3d 0q - 2n 0d 1q - 0n 1d 1q -
Its 6 different combinations