Question

Which equation matches the solution set of
$\{ 5\} $
$\frac {3x}{5}+4=7$
$2x-20=30$
$x^{2}-10=15$
$x+x-9=1+x$

Answer 1

To determine which equation matches the solution set \(\{ 5 \}\), we will solve each of the given equations and see which one has the solution \(x = 5\).

1. **First equation:**
\[
\frac{3x}{5} + 4 = 7
\]
Subtract 4 from both sides:
\[
\frac{3x}{5} = 3
\]
Multiply both sides by 5:
\[
3x = 15
\]
Divide by 3:
\[
x = 5
\]

2. **Second equation:**
\[
2x - 20 = 30
\]
Add 20 to both sides:
\[
2x = 50
\]
Divide by 2:
\[
x = 25
\]

3. **Third equation:**
\[
x^2 - 10 = 15
\]
Add 10 to both sides:
\[
x^2 = 25
\]
Taking the square root of both sides gives:
\[
x = 5 \quad \text{or} \quad x = -5
\]

4. **Fourth equation:**
\[
x + x - 9 = 1 + x
\]
Combine like terms on the left:
\[
2x - 9 = 1 + x
\]
Subtract \(x\) from both sides:
\[
x - 9 = 1
\]
Add 9 to both sides:
\[
x = 10
\]

Now, we summarize the solutions of each equation:
- The first equation has the solution \(x = 5\).
- The second equation has the solution \(x = 25\).
- The third equation has solutions \(x = 5\) and \(x = -5\).
- The fourth equation has the solution \(x = 10\).

The equations that match the solution set \(\{ 5 \}\) are:
- The first equation: \( \frac{3x}{5} + 4 = 7 \)
- The third equation: \( x^2 - 10 = 15 \)

Both of these equations provide \(5\) as a solution. However, since the problem states "which equation matches the solution set \(\{ 5 \}\)", it is likely seeking the simplest case.

Thus, the equation that directly leads to the solution \(x = 5\) is:

\[
\boxed{\frac{3x}{5} + 4 = 7}
\]