Let's solve each equation one by one to find the solution sets.
-
Equation 1: \(5x^2 - 10 = 15\)
Start by adding 10 to both sides:
\[ 5x^2 = 25 \]
Divide both sides by 5:
\[ x^2 = 5 \]
Taking the square root of both sides, we get:
\[ x = \sqrt{5} \quad \text{or} \quad x = -\sqrt{5} \]
So the solution set is \( { -\sqrt{5}, \sqrt{5} } \).
-
Equation 2: \(\frac{3x}{5} + 4 = 7\)
Subtract 4 from both sides:
\[ \frac{3x}{5} = 3 \]
Multiply both sides by 5:
\[ 3x = 15 \]
Divide by 3:
\[ x = 5 \]
So the solution set is \( { 5 } \).
-
Equation 3: \(2x - 20 + 30x + x - 9 = 1 + x\)
Combine like terms on the left side:
\[ (2x + 30x + x - x) - 20 - 9 = 1 \]
Simplify:
\[ 32x - 29 = 1 \]
Add 29 to both sides:
\[ 32x = 30 \]
Divide by 32:
\[ x = \frac{30}{32} = \frac{15}{16} \]
So the solution set is \( { \frac{15}{16} } \).
Now let’s summarize the solution sets:
- For \(5x^2 - 10 = 15\), the solution set is \( { -\sqrt{5}, \sqrt{5} } \).
- For \(\frac{3x}{5} + 4 = 7\), the solution set is \( { 5 } \).
- For \(2x - 20 + 30x + x - 9 = 1 + x\), the solution set is \( { \frac{15}{16} } \).
Please clarify if you are looking for something specific with these equations or if you wish to compare the solution sets.
1 answer