Solve the equation [tex]$\frac{6}{x-2}-\frac{6}{x+1}=1$[/tex]. Show clear algebraic working.
Answer (1)
Multiply both sides by ( x − 2 ) ( x + 1 ) to get rid of fractions.
Expand and simplify the equation to x 2 − x − 20 = 0 .
Factor the quadratic equation to ( x − 5 ) ( x + 4 ) = 0 .
Solve for x , which gives x = 5 and x = − 4 . The final answer is 5 , − 4 .
Explanation
Problem Analysis We are given the equation x − 2 6 − x + 1 6 = 1 and we need to solve for x .
Eliminating Fractions To eliminate the fractions, we multiply both sides of the equation by ( x − 2 ) ( x + 1 ) . This gives us 6 ( x + 1 ) − 6 ( x − 2 ) = ( x − 2 ) ( x + 1 ) .
Expanding the Equation Expanding both sides of the equation, we have 6 x + 6 − 6 x + 12 = x 2 − x − 2 .
Simplifying Simplifying the equation, we get 18 = x 2 − x − 2 .
Rearranging Rearranging the equation into a quadratic equation, we have x 2 − x − 20 = 0 .
Factoring Factoring the quadratic equation, we get ( x − 5 ) ( x + 4 ) = 0 .
Solving and Checking Solving for x , we find two possible solutions: x = 5 or x = − 4 . We need to check for extraneous solutions by substituting these values back into the original equation to make sure the denominators are not zero. If x = 5 , the denominators are 5 − 2 = 3 and 5 + 1 = 6 , which are not zero. If x = − 4 , the denominators are − 4 − 2 = − 6 and − 4 + 1 = − 3 , which are also not zero. Therefore, both solutions are valid.
Final Answer Thus, the solutions to the equation are x = 5 and x = − 4 .
Examples
Imagine you are designing a rectangular garden where the length is x + 1 meters and the width is x − 2 meters. You want the difference between the area used for planting flowers (represented by x − 2 6 ) and the area used for planting herbs (represented by x + 1 6 ) to be exactly 1 square meter. Solving the equation x − 2 6 − x + 1 6 = 1 helps you determine the possible values of x that satisfy this condition, ensuring your garden design meets your specific area requirements. This problem demonstrates how algebraic equations can be used to model and solve practical design problems.
