To solve the equation \( 15(2a - 2) = 5(a^2 - 1) \), follow these steps:
### Step 1: Expand both sides of the equation
First, let's expand both sides of the equation to get rid of the parentheses:
[tex]\[ 15(2a - 2) = 5(a^2 - 1) \][/tex]
Expanding, we get:
[tex]\[ 30a - 30 = 5a^2 - 5 \][/tex]
### Step 2: Bring all terms to one side of the equation to set it to zero
Next, we need to move all terms to one side of the equation so that we have a standard quadratic form:
[tex]\[ 5a^2 - 30a + 25 = 0 \][/tex]
### Step 3: Simplify the equation
Simplify the equation by dividing all terms by 5:
[tex]\[ a^2 - 6a + 5 = 0 \][/tex]
### Step 4: Factor the quadratic equation
Now, factor the quadratic equation:
[tex]\[ (a - 5)(a - 1) = 0 \][/tex]
### Step 5: Solve for \( a \)
Set each factor equal to zero and solve for \( a \):
[tex]\[ a - 5 = 0 \quad \text{or} \quad a - 1 = 0 \][/tex]
[tex]\[ a = 5 \quad \text{or} \quad a = 1 \][/tex]
Therefore, the solutions to the equation \( 15(2a - 2) = 5(a^2 - 1) \) are:
[tex]\[ a = 5 \quad \text{and} \quad a = 1 \][/tex]
### Solution to \( \frac{15}{a^2-1}=\frac{5}{2a-2} \):
To solve the equation \( \frac{15}{a^2 - 1} = \frac{5}{2a - 2} \):
1. Cross-multiply to solve for \( a \):
[tex]\[ 15(2a - 2) = 5(a^2 - 1) \][/tex]
2. Notice that this is the same equation as before, so the solutions will be:
[tex]\[ a = 5 \quad \text{and} \quad a = 1 \][/tex]
Thus, the solution to \( \frac{15}{a^2 - 1} = \frac{5}{2a - 2} \) is:
[tex]\[ a = 5 \quad \text{or} \quad a = 1 \][/tex]
### Extraneous Solution
According to the information provided, the extraneous solution is given as 7.
Therefore, the extraneous solution is \( a = 7 \).
### Summary:
- The solutions are \( a = 5 \) and \( a = 1 \).
- The extraneous solution is \( a = 7 \).
So, in the boxes, the solutions would be filled as:
[tex]\[ a = 5 \][/tex] and [tex]\[ a = 1 \][/tex]
and the extraneous solution as [tex]\( a = 7 \)[/tex].
