Solve the equation: $2(x+1)+3=3(x-1)$
The given equation is \(2(x+1) + 3 = 3(x-1)\). This is a linear equation in one variable, \(x\). Our objective is to find the value of \(x\) that satisfies this equality. We will perform algebraic manipulations to isolate \(x\) on one side of the equation.
First, we expand the expressions inside the parentheses on both sides of the equation.
On the left side: \(2(x+1) + 3 = 2x + 2 + 3 = 2x + 5\).
On the right side: \(3(x-1) = 3x - 3\).
Now the equation is:
\[2x + 5 = 3x - 3\]
Next, we gather the terms involving \(x\) on one side and the constant terms on the other side. By subtracting \(2x\) from both sides, we get:
\[5 = 3x - 2x - 3\]
\[5 = x - 3\]
Now, add \(3\) to both sides of the equation:
\[5 + 3 = x\]
\[8 = x\]
We found that the solution to the equation is \(x = 8\). To verify, we substitute \(x = 8\) back into the original equation:
Left Side: \(2(8+1) + 3 = 2(9) + 3 = 18 + 3 = 21\).
Right Side: \(3(8-1) = 3(7) = 21\).
Since both sides equal 21, the solution \(x = 8\) is correct. The final answer is \(\boxed{8}\).
Solving linear equations is essential in everyday life. For instance, imagine you are comparing two membership plans for a gym. Plan A costs \(\$2\) per visit plus an initial fee of \(\$5\), while Plan B costs \(\$3\) per visit but with a \(\$3\) discount. If you want to know how many visits you would need to make for both plans to cost the same, you set up the equation \(2x + 5 = 3x - 3\), where \(x\) is the number of visits. In this case, at 8 visits, both plans cost \(\$21\).