Question

ASAP HELP PLS NO WRONG ANSWERS------------

Answer 1

Answer:

d=2

Step-by-step explanation:

We are given equation:

sqrt(4y-3)=d-y

Squaring both sides gives:

4y-3=(d-y)^2

Applying the identity (x+y)^2=x^2+2xy+y^2 on right:

4y-3=d^2-2dy+y^2

Now let's y=7:

4(7)-3=d^2-2d(7)+(7)^2

Simplify:

25=d^2-14d+49

Subtract 25 on both sides:

0=d^2-14d+24

Factor left:

0=(d-12)(d-2) since -2+-12=-14 and -2(-12)=24

This gives d=12 or d=2.

The d that makes d-y or I mean d-7 negative will give us y=7 as extraneous

Since 2-7 is -5, then d=2 is what we are looking for.

Check:

sqrt(4y-3)=d-y

Set d=2: sqrt(4y-3)=2-y

Now solve for y:

Square both sides: 4y-3=4-4y+y^2

Subtract 4y and 3 on both sides: 0=7-8y+y^2

Reorder right side: 0=y^2-8y+7

Factor: 0=(y-7)(y-1) since -7+-1=-8 and -7(-1)=7

This gives y=7 or y=1.

Plugging in y=7 for a check to this equation gives:

sqrt(4×7-3)=2-7

Sqrt(25)=-5

5=-5 which is not true which is what we wanted

Answer 2

Answer

d = 12

Step-by-step explanation:

square root 4y - 3 = d - y where y = 7

Substitute y for 7

Square root 4(7) - 3 = d - (7)

Multiply 4 and 7

Square root 28 - 3 = d - (7)

Subtract 28 - 3

Square root 25 = d - (7)

Square root 25

5 = d - (7)

Add 7 to both sides

12 = d


Therefore, d = 12