If \( a(x+1)+(x-1)=7 x \) for all ral mamber \( x \),

To solve the equation \( a(x+1) + (x-1) = 7x \) for all real numbers \( x \), we can start by expanding and rearranging the equation. 1. Distribute \( a \) in the left-hand side: \[ ax + a + x - 1 = 7x \] 2. Combine like terms: \[ (a + 1)x + (a - 1) = 7x \] 3. For the equation to hold for all \( x \), the coefficients of \( x \) on both sides must be equal, and the constant terms must also be equal. This gives us two equations: - Coefficient of \( x \): \[ a + 1 = 7 \] - Constant term: \[ a - 1 = 0 \] 4. Solve the first equation: \[ a + 1 = 7 \implies a = 7 - 1 = 6 \] 5. Solve the second equation: \[ a - 1 = 0 \implies a = 1 \] Now we have two different values for \( a \) from the two equations. To find a consistent solution, we can substitute \( a = 6 \) into the second equation: \[ 6 - 1 = 5 \quad \text{(not equal to 0)} \] This indicates that the two equations are inconsistent. Thus, we need to check if both conditions can be satisfied simultaneously. Since they cannot, we conclude that the only solution for \( a \) that satisfies the equation for all \( x \) is: \[ \boxed{6} \] This means that the equation \( a(x+1) + (x-1) = 7x \) holds true for all \( x \) when \( a = 6 \).