Solve for x:
$(x+2)(x+3)=(x+1)(x+5)$
The given equation is \((x+2)(x+3) = (x+1)(x+5)\). We need to solve for the variable \(x\). This is an equation that involves products of linear binomials. By expanding both sides, we can simplify the expression to determine the value of \(x\).
To expand the left side, we use the FOIL method (First, Outer, Inner, Last):
\((x+2)(x+3) = x^2 + 3x + 2x + 6 = x^2 + 5x + 6\).
To expand the right side, we similarly use the FOIL method:
\((x+1)(x+5) = x^2 + 5x + x + 5 = x^2 + 6x + 5\).
Now we equate the two expanded forms:
\(x^2 + 5x + 6 = x^2 + 6x + 5\).
We can subtract \(x^2\) from both sides to eliminate the quadratic term:
\(5x + 6 = 6x + 5\).
Next, we isolate the \(x\) variable. Subtract \(5x\) from both sides:
\(6 = x + 5\).
Subtract 5 from both sides:
\(x = 1\).
Finally, we verify the solution by substituting \(x = 1\) back into the original equation:
Left side: \((1+2)(1+3) = 3 \times 4 = 12\).
Right side: \((1+1)(1+5) = 2 \times 6 = 12\).
Since both sides are equal to 12, the solution is correct. The final answer is \(\boxed{1}\).
This problem illustrates the application of algebraic expansion and balancing equations, which are fundamental in various real-life scenarios. For instance, if you are comparing two different pricing structures for a service—where one might have a base fee plus a rate (like \(x+2\) and \(x+3\)) and another has different parameters (like \(x+1\) and \(x+5\))—you can use this method to find the point at which both costs are identical. By setting the expressions equal and solving for \(x\), you determine the exact point where both options provide equal value, helping you make informed decisions when balancing costs or optimizing resources.