Q. Solve for all values of x.x+6x+2=x1Answer: x=
Find Common Denominator: First, we need to find a common denominator to combine the fractions. The common denominator for (x+6) and x is x(x+6). We will multiply both sides of the equation by this common denominator to eliminate the fractions.Calculation: x+6x+2×x(x+6)=x1×x(x+6)
Multiply and Simplify: After multiplying both sides by the common denominator, we can simplify the equation.Calculation: (x+2)×x=1×(x+6)
Distribute and Simplify: Now, we distribute the x on the left side and simplify the right side.Calculation: x2+2x=x+6
Bring Terms Together: Next, we bring all terms to one side to set the equation to zero and solve for x.Calculation: x2+2x−x−6=0
Factor the Equation: Simplify the equation by combining like terms.Calculation: x2+x−6=0
Solve for x: To find the values of x, we set each factor equal to zero and solve for x.Calculation: x+3=0 or x−2=0
Check Valid Solutions: Solving each equation gives us the values of x.Calculation: x=−3 or x=2However, we must check these solutions in the original equation to ensure they do not make the denominator zero.
Check Valid Solutions: Solving each equation gives us the values of x.Calculation: x=−3 or x=2However, we must check these solutions in the original equation to ensure they do not make the denominator zero.Checking x=−3 in the original equation's denominators:(x+6) becomes (−3+6) which is 3, not zero.x becomes −3, which is not zero.So, x=−3 is a valid solution.
Check Valid Solutions: Solving each equation gives us the values of x.Calculation: x=−3 or x=2However, we must check these solutions in the original equation to ensure they do not make the denominator zero.Checking x=−3 in the original equation's denominators:(x+6) becomes (−3+6) which is 3, not zero.x becomes −3, which is not zero.So, x=−3 is a valid solution.Checking x=2 in the original equation's denominators:(x+6) becomes x=−32 which is x=−33, not zero.x becomes x=−35, which is not zero.So, x=2 is also a valid solution.
More problems from Simplify variable expressions using properties