How do you solve #2/(x+1) + 5/(x-2)=-2#?

Question

Eva Adamson · Accepted Answer

#x = 1/2, x = -3#

Since there are two terms on the left side of this equation, which contains fractions, we might not be able to cross multiply.

However we can get rid of the denominators by multiplying through by the LCM of the denominators which is #color(red)((x+1)(x-2))#

#color(red)(cancel((x+1))(x-2)xx)2/cancel((x+1)) + color(red)((x+1)cancel((x-2))xx)5/cancel(x-2)=-2color(red)((x+1)(x-2))#

What's left to do is:

#2(x-2) + 5(x+1) = -2(x+1)(x-2)#

#2x-4 +5x+5 = -2(x^2 -x -2)#

#7x+1 = -2x^2+2x+4#

#2x^2 +5x -3=0 " make the quadratic" = 0#

#(2x-1)(x+3) = 0" factorise"#

#2x -1 =0" " rArr x = 1/2#

#x+3 = 0 " "rArr x = -3#

Abigail Allen · Answer

undefined To solve the equation 2/(x+1) + 5/(x-2) = -2, you can follow these steps:

1. Find a common denominator for the fractions on the left side of the equation. In this case, the common denominator is (x+1)(x-2).

2. Multiply each term by the common denominator to eliminate the fractions. This gives you: 2(x-2) + 5(x+1) = -2(x+1)(x-2).

3. Simplify the equation by distributing and combining like terms. This results in: 2x - 4 + 5x + 5 = -2x^2 + 2.

4. Rearrange the equation to bring all terms to one side and set it equal to zero. This gives you: -2x^2 + 7x + 1 = 0.

5. Solve the quadratic equation by factoring, completing the square, or using the quadratic formula. In this case, the quadratic equation does not factor easily, so you can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a).

6. Substitute the values of a, b, and c from the quadratic equation into the quadratic formula and simplify to find the solutions for x.

By following these steps, you can find the solutions for the given equation.