To solve for the variable [tex]\( x \)[/tex] given that the hypotenuse of a right triangle can be written in two equivalent forms, [tex]\( 4x + 30 \)[/tex] and [tex]\( 6x + 14 \)[/tex], we need to set the two expressions equal to each other and solve for [tex]\( x \)[/tex].
1. Set up the equation:
[tex]\[
4x + 30 = 6x + 14
\][/tex]
2. Isolate the variable [tex]\( x \)[/tex]:
Subtract [tex]\( 4x \)[/tex] from both sides to start simplifying the equation:
[tex]\[
30 = 2x + 14
\][/tex]
3. Subtract 14 from both sides:
[tex]\[
30 - 14 = 2x
\][/tex]
[tex]\[
16 = 2x
\][/tex]
4. Solve for [tex]\( x \)[/tex]:
Divide both sides by 2 to isolate [tex]\( x \)[/tex]:
[tex]\[
x = \frac{16}{2}
\][/tex]
[tex]\[
x = 8
\][/tex]
Thus, the value of [tex]\( x \)[/tex] is [tex]\( \boxed{8} \)[/tex].
