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Solution 2.3:3e

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In this case, we see that the left-hand side contains the factor x+3, which we can take out to obtain

(x+3)(x−1)−(x+3)(2x−9)=(x+3)(x−1)−(2x−9)=(x+3)(x−1−2x+9)=(x+3)(−x+8).

This rewriting of the equation results in the new equation

(x+3)(−x+8)=0

which has the solutions x=−3 and x=8.

We check the solution x=8 by substituting it into the equation,

LHS=(8+3)(8−1)−(8+3)(28−9)=117−117=0=RHS.
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