Which equation matches the solution set of {5}?(1 point)

Question

Responses

A) x^2−10=15
x squared minus 10 equals 15

B) 2x−20=30
2 x minus 20 equals 30

C) x+x−9=1+x
x plus x minus 9 equals 1 plus x

D) 3x/5+4=7
Start Fraction 3 x over 5 End Fraction plus 4 equals 7

Answer 1

To find which equation has the solution set of {5}, we can substitute \( x = 5 \) into each equation and check if it holds true.

A) \( x^2 - 10 = 15 \)
Substituting \( x = 5 \):
\( 5^2 - 10 = 15 \)
\( 25 - 10 = 15 \)
\( 15 = 15 \) (True)

B) \( 2x - 20 = 30 \)
Substituting \( x = 5 \):
\( 2(5) - 20 = 30 \)
\( 10 - 20 = 30 \)
\( -10 = 30 \) (False)

C) \( x + x - 9 = 1 + x \)
Substituting \( x = 5 \):
\( 5 + 5 - 9 = 1 + 5 \)
\( 10 - 9 = 6 \)
\( 1 = 6 \) (False)

D) \( \frac{3x}{5} + 4 = 7 \)
Substituting \( x = 5 \):
\( \frac{3(5)}{5} + 4 = 7 \)
\( 3 + 4 = 7 \)
\( 7 = 7 \) (True)

The equations that are true for \( x = 5 \) are A) and D). However, to determine the complete solution set, we must ensure that no other values satisfy the equations.

• For equation A: \( x^2 - 10 = 15 \) can be rearranged to \( x^2 = 25 \), giving two solutions \( x = 5 \) or \( x = -5 \). Therefore, the solution set is not just {5}.

• For equation D: Rearranging gives \( \frac{3x}{5} = 3 \). Multiplying both sides by 5 results in \( 3x = 15\), then \( x = 5\). Thus, the only solution is \( x = 5 \).

Since we are seeking which equation's sole solution is {5}, the answer is:

D) \( \frac{3x}{5} + 4 = 7 \)