Solve the inequality ||x-5|+2|≤3

Mathematics

1 answer:

Fudgin [204]2 years ago

3 0

Step-by-step explanation:

||x-5|+2|≤3

-3 ≤ (|x-5| +2) ≤ 3

-5 ≤ |x-5| ≤ 1

|x-5| ≥ -5 => not required

|x-5| ≤ 1

-1 ≤ x-5 ≤ 1

4 ≤ x ≤ 6

the Solutions:

{4≤ x ≤ 6}

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$\frac{x}{2}-\frac{2}{x+1}=1$

We have 2 denominators that we need to get rid of. Whenever there are the denominators, all we have to do is multiply all whole equation with the denominators.

Our denominators are both 2 and x+1. Therefore, we multiply the whole equation by 2(x+1)

$\frac{x}{2}[2(x+1)]-\frac{2}{x+1}[2(x+1)] = 1[2(x+1)]$

Then shorten the fractions.

$\frac{x}{2}[2(x+1)]-\frac{2}{x+1}[2(x+1)] = 1[2(x+1)]\\x(x+1)-2(2)=1(2x+2)$

Distribute in all.

$x^2+x-4=2x+2$

We should get like this. Because the polynomial is 2-degree, I'd suggest you to move all terms to one place. Therefore, moving 2x+2 to another side and subtract.

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We are almost there. All we have to do is, solving for x by factoring. (Although there are more than just factoring but factoring this polynomial is faster.)

$(x-3)(x+2)=0\\x=3,-2$