solve and check:\n$\frac{1}{x + 3}=\frac{x + 10}{x - 2}$\nfrom least to greatest, the solutions are…

solve and check:\n$\frac{1}{x + 3}=\frac{x + 10}{x - 2}$\nfrom least to greatest, the solutions are $x=square$ and $x=square$.

solve and check:\n$\frac{1}{x + 3}=\frac{x + 10}{x - 2}$\nfrom least to greatest, the solutions are $x=square$ and $x=square$.

Answer

Explanation:

Step1: Cross - multiply

$(x - 2)= (x + 3)(x + 10)$

Step2: Expand the right - hand side

$x-2=x^{2}+10x + 3x+30$

Step3: Rearrange to form a quadratic equation

$x^{2}+10x + 3x+30-x + 2 = 0$ $x^{2}+12x+32 = 0$

Step4: Factor the quadratic equation

$(x + 4)(x + 8)=0$

Step5: Solve for x

$x+4 = 0$ gives $x=-4$; $x + 8=0$ gives $x=-8$

Answer:

$x=-8$ and $x=-4$