Sure, let's solve for the constant [tex]\( d \)[/tex] that makes [tex]\( y = 20 \)[/tex] a solution to the given equation:
[tex]\[ \sqrt{\frac{1}{2} y - 1} = \frac{3}{4} y + d \][/tex]
We need to substitute [tex]\( y = 20 \)[/tex] into the equation and solve for [tex]\( d \)[/tex].
1. Substitute [tex]\( y = 20 \)[/tex] into the equation:
[tex]\[ \sqrt{\frac{1}{2} \cdot 20 - 1} = \frac{3}{4} \cdot 20 + d \][/tex]
2. Simplify the expression inside the square root:
[tex]\[ \frac{1}{2} \cdot 20 - 1 = 10 - 1 = 9 \][/tex]
So the left side of the equation becomes:
[tex]\[ \sqrt{9} \][/tex]
And we know [tex]\( \sqrt{9} = 3 \)[/tex].
3. Simplify the expression on the right side of the equation:
[tex]\[ \frac{3}{4} \cdot 20 + d = 15 + d \][/tex]
Now we have the simplified equation:
[tex]\[ 3 = 15 + d \][/tex]
4. Solve for [tex]\( d \)[/tex]:
[tex]\[ 3 = 15 + d \][/tex]
[tex]\[ d = 3 - 15 \][/tex]
[tex]\[ d = -12 \][/tex]
Therefore, the value for the constant [tex]\( d \)[/tex] that makes [tex]\( y = 20 \)[/tex] a solution to the equation is:
[tex]\[
d = -12
\][/tex]
