math help?

Question

i have a worksheet of math problems and i can do some of them and i cant do some..so if anyone could explain it better that would be great.... 4y+3 over 7..i have the answer i just dont know how to do it and 1/5 [4(k+2)-(3-k)]=4...i dont get them

Answer 1

for the first one, multiply by 7 to get it out of the denominator and then distribute. 7(4y+3)= 28y + 21, subtract 21, and divide by 28 to get y

Answer 2

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(4y+3)/7 = 0
multiply both sides by 7
so 4y + 3 = 0
subtract 3 from both sides
4y = -3
divide by 4
y = -3/4

1/5 [4(k+2)-(3-k)]=4
Multiply both sides by 5
4k +8 -3 +k = 20
subtract 5 from both sides
5k = 15
k = 3

Answer 3

(4y+3)/7 = 0
multiply both sides by 7
so 4y + 3 = 0
subtract 3 from both sides
4y = -3
divide by 4
y = -3/4

1/5 [4(k+2)-(3-k)]=4
Multiply both sides by 5
4k +8 -3 +k = 20
subtract 5 from both sides
5k = 15
k = 3

Answer 4

for the first one it must be equal to something or else you cannot solve for it and the second one you first times the 4 to the k and 2

1/5 [4k+8-3+k]=4
1/5 [5k+5]=4
k+1=4
k = 3

Answer 5

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