1- Solve the following equation?

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

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

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

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

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