Example

3(r+2s)=2t-4
a⋅(b-2)=3b
3(r+2s)=2t-4
a⋅(b-2)=3b

Question

$$\frac{(x-7)(x+3)}{(x-1)^{2}}\le0$$

Answer

-3<=x<1;x<=7

Solution


Replace the inequality sign with an equal sign, so that we can solve it like a normal equation.
\[\frac{(x-7)(x+3)}{{(x-1)}^{2}}=0\]
Multiply both sides by \({(x-1)}^{2}\).
\[(x-7)(x+3)=0\]
Solve for \(x\).
\[x=7,-3\]
Also note that \(x\) is undefined at \(1\).
\[x\ne 1\]
From the values of \(x\) above, we have these 4 intervals to test.
\[\begin{aligned}&x\le -3\\&-3\le x\le 1\\&1\le x\le 7\\&x\ge 7\end{aligned}\]
Pick a test point for each interval.
For the interval \(x\le -3\):
Let's pick \(x=-4\). Then, \(\frac{(-4-7)(-4+3)}{{(-4-1)}^{2}}\le 0\).After simplifying, we get \(0.44\le 0\), which is
false
.
Drop this interval.
.
For the interval \(-3\le x\le 1\):
Let's pick \(x=0\). Then, \(\frac{(0-7)(0+3)}{{(0-1)}^{2}}\le 0\).After simplifying, we get \(-21\le 0\), which is
true
.
Keep this interval.
.
For the interval \(1\le x\le 7\):
Let's pick \(x=2\). Then, \(\frac{(2-7)(2+3)}{{(2-1)}^{2}}\le 0\).After simplifying, we get \(-25\le 0\), which is
true
.
Keep this interval.
.
For the interval \(x\ge 7\):
Let's pick \(x=8\). Then, \(\frac{(8-7)(8+3)}{{(8-1)}^{2}}\le 0\).After simplifying, we get \(0.224490\le 0\), which is
false
.
Drop this interval.
.
Therefore,
\[\begin{aligned}&-3\le x\le 1\\&1\le x\le 7\end{aligned}\]
Notice the equation contains \({(x-1)}^{2}\) in the denominator. Since any denominator must not equal zero, the domain is restricted to \({(x-1)}^{2}\ne 0\). Solving for \(x\), we have:
\[x\ne 1\]
Add the domain restrictions.
\[\begin{aligned}&-3\le x<1\\&x\le 7\end{aligned}\]
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