Question

solving a rational equation in the form of a propo
solve and check: \\(\frac{1}{x+3} = \frac{x+10}{x-2}\\)
from least to greatest, the solutions are \\(x = \square\\) and \\(x = \square\\).

Explanation:

Step1: Cross - multiply the equation

To solve the rational equation \(\frac{1}{x + 3}=\frac{x + 10}{x - 2}\), we use the cross - multiplication property of proportions. If \(\frac{a}{b}=\frac{c}{d}\) (where \(b
eq0\) and \(d
eq0\)), then \(a\times d=b\times c\).
Applying this property, we get \(1\times(x - 2)=(x + 3)\times(x + 10)\).

Step2: Expand both sides

First, expand the right - hand side using the distributive property \((a + b)(c + d)=ac+ad+bc+bd\).
The left - hand side is \(x - 2\).
The right - hand side: \((x + 3)(x + 10)=x\times x+x\times10+3\times x + 3\times10=x^{2}+10x+3x + 30=x^{2}+13x + 30\).
So our equation becomes \(x - 2=x^{2}+13x + 30\).

Step3: Rearrange into standard quadratic form

Subtract \(x\) and add \(2\) to both sides to get a quadratic equation in standard form \(ax^{2}+bx + c = 0\).
\(0=x^{2}+13x+30 - x + 2\), which simplifies to \(x^{2}+12x + 32 = 0\).

Step4: Factor the quadratic equation

We need to find two numbers that multiply to \(32\) and add up to \(12\). The numbers are \(8\) and \(4\) since \(8\times4 = 32\) and \(8 + 4=12\).
So, \(x^{2}+12x + 32=(x + 8)(x + 4)=0\).

Step5: Solve for \(x\)

Using the zero - product property, if \(ab = 0\), then either \(a = 0\) or \(b = 0\).
If \(x+8 = 0\), then \(x=-8\).
If \(x + 4=0\), then \(x=-4\).

Step6: Check the solutions

We need to check if these solutions make the original denominators zero.
For the denominator \(x + 3\):

• When \(x=-8\), \(x + 3=-8 + 3=-5

eq0\).

• When \(x=-4\), \(x + 3=-4 + 3=-1

eq0\).
For the denominator \(x - 2\):

• When \(x=-8\), \(x - 2=-8-2=-10

eq0\).

• When \(x=-4\), \(x - 2=-4-2=-6

eq0\).
Both solutions are valid.

Answer:

From least to greatest, the solutions are \(x=-8\) and \(x = - 4\).