Algebra 2 Worksheets | Dynamically Created Algebra 2 Worksheets - Free Printable

Let’s solve each equation one by one. We’ll take it slow and check our work as we go.

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1)
Equation:
\[
\frac{1}{24x^2} = \frac{1}{12x^2} - \frac{1}{x}
\]

Step 1: Get all terms on one side to set equal to zero.
Subtract \(\frac{1}{12x^2}\) from both sides:

\[
\frac{1}{24x^2} - \frac{1}{12x^2} = -\frac{1}{x}
\]

Step 2: Combine left side. Common denominator is \(24x^2\):

\[
\frac{1 - 2}{24x^2} = -\frac{1}{x}
\Rightarrow \frac{-1}{24x^2} = -\frac{1}{x}
\]

Step 3: Multiply both sides by \(-1\):

\[
\frac{1}{24x^2} = \frac{1}{x}
\]

Step 4: Cross-multiply (or multiply both sides by \(24x^2 \cdot x\)):

Multiply both sides by \(24x^2\):

\[
1 = 24x
\Rightarrow x = \frac{1}{24}
\]

Check for restrictions: Denominators can’t be zero → \(x ≠ 0\). Our answer is fine.

✔ Answer: \(x = \frac{1}{24}\)

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2)
Equation:
\[
\frac{1}{n} = \frac{11}{6n} - 8
\]

Step 1: Move all terms to left side:

\[
\frac{1}{n} - \frac{11}{6n} + 8 = 0
\]

Step 2: Combine fractions. Common denominator is \(6n\):

\[
\frac{6}{6n} - \frac{11}{6n} + 8 = 0
\Rightarrow \frac{-5}{6n} + 8 = 0
\]

Step 3: Subtract 8:

\[
\frac{-5}{6n} = -8
\]

Step 4: Multiply both sides by \(6n\):

\[
-5 = -48n
\Rightarrow n = \frac{5}{48}
\]

Check: \(n ≠ 0\) — OK.

✔ Answer: \(n = \frac{5}{48}\)

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3)
Equation:
\[
\frac{y - 6}{10y^2} + \frac{12}{5y^2} = \frac{y + 7}{5y^2}
\]

Step 1: All denominators are multiples of \(10y^2\), so let’s use that as common denominator.

Rewrite second term: \(\frac{12}{5y^2} = \frac{24}{10y^2}\)

Right side: \(\frac{y+7}{5y^2} = \frac{2(y+7)}{10y^2} = \frac{2y + 14}{10y^2}\)

Now equation becomes:

\[
\frac{y - 6}{10y^2} + \frac{24}{10y^2} = \frac{2y + 14}{10y^2}
\]

Step 2: Combine left side:

\[
\frac{y - 6 + 24}{10y^2} = \frac{2y + 14}{10y^2}
\Rightarrow \frac{y + 18}{10y^2} = \frac{2y + 14}{10y^2}
\]

Step 3: Since denominators are same and non-zero (\(y ≠ 0\)), set numerators equal:

\[
y + 18 = 2y + 14
\]

Step 4: Solve:

\[
18 - 14 = 2y - y
\Rightarrow 4 = y
\]

✔ Answer: \(y = 4\)

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4)
Equation:
\[
\frac{1}{p} + \frac{10p + 7}{p^2 - 3p} = \frac{2p + 4}{p^2 - 3p}
\]

Note: \(p^2 - 3p = p(p - 3)\), so domain: \(p ≠ 0, 3\)

Step 1: Move all terms to left:

\[
\frac{1}{p} + \frac{10p + 7}{p(p - 3)} - \frac{2p + 4}{p(p - 3)} = 0
\]

Combine the two fractions with same denominator:

\[
\frac{1}{p} + \frac{(10p + 7) - (2p + 4)}{p(p - 3)} = 0
\Rightarrow \frac{1}{p} + \frac{8p + 3}{p(p - 3)} = 0
\]

Step 2: Common denominator is \(p(p - 3)\)

Write first term: \(\frac{1}{p} = \frac{p - 3}{p(p - 3)}\)

So:

\[
\frac{p - 3}{p(p - 3)} + \frac{8p + 3}{p(p - 3)} = 0
\Rightarrow \frac{p - 3 + 8p + 3}{p(p - 3)} = 0
\Rightarrow \frac{9p}{p(p - 3)} = 0
\]

Simplify numerator and denominator (cancel \(p\), but remember \(p ≠ 0\)):

\[
\frac{9}{p - 3} = 0
\]

Wait — this fraction equals zero only if numerator is zero. But numerator is 9 → never zero.

That means no solution? Let’s double-check.

Original simplification:

After combining: \(\frac{9p}{p(p - 3)} = 0\)

This is equivalent to \(\frac{9}{p - 3} = 0\) only if \(p ≠ 0\), which is already excluded.

But \(\frac{9}{p - 3} = 0\) has NO solution because a non-zero number over something can’t be zero.

Alternatively, think: \(\frac{9p}{p(p - 3)} = 0\) → numerator must be zero → \(9p = 0\) → \(p = 0\), but \(p = 0\) is not allowed (makes original denominators zero).

So — no valid solution.

✔ Answer: No solution

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5)
Equation:
\[
\frac{1}{d^2} = \frac{1}{25}
\]

Step 1: Take reciprocal of both sides (or cross-multiply):

\[
d^2 = 25
\]

Step 2: Square root both sides:

\[
d = \pm 5
\]

Check: \(d ≠ 0\) — both values are fine.

✔ Answer: \(d = 5\) or \(d = -5\)

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6)
Equation:
\[
\frac{6}{x - 3} - \frac{1}{x^2 + 6x - 27} = \frac{1}{x - 3}
\]

First, factor denominator: \(x^2 + 6x - 27 = (x + 9)(x - 3)\)

Domain: \(x ≠ 3, -9\)

Step 1: Move all terms to left:

\[
\frac{6}{x - 3} - \frac{1}{(x + 9)(x - 3)} - \frac{1}{x - 3} = 0
\]

Combine first and third term:

\[
\left( \frac{6}{x - 3} - \frac{1}{x - 3} \right) - \frac{1}{(x + 9)(x - 3)} = 0
\Rightarrow \frac{5}{x - 3} - \frac{1}{(x + 9)(x - 3)} = 0
\]

Step 2: Common denominator is \((x + 9)(x - 3)\)

Write first term: \(\frac{5}{x - 3} = \frac{5(x + 9)}{(x + 9)(x - 3)}\)

So:

\[
\frac{5(x + 9) - 1}{(x + 9)(x - 3)} = 0
\Rightarrow \frac{5x + 45 - 1}{(x + 9)(x - 3)} = 0
\Rightarrow \frac{5x + 44}{(x + 9)(x - 3)} = 0
\]

Set numerator = 0:

\[
5x + 44 = 0 \Rightarrow x = -\frac{44}{5}
\]

Check: Not 3 or -9 → OK.

✔ Answer: \(x = -\frac{44}{5}\)

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7)
Equation:
\[
\frac{1}{x - 5} + \frac{1}{x^2 - 11x + 30} = \frac{7}{x - 5}
\]

Factor denominator: \(x^2 - 11x + 30 = (x - 5)(x - 6)\)

Domain: \(x ≠ 5, 6\)

Step 1: Move all to left:

\[
\frac{1}{x - 5} + \frac{1}{(x - 5)(x - 6)} - \frac{7}{x - 5} = 0
\]

Combine first and third:

\[
\left( \frac{1 - 7}{x - 5} \right) + \frac{1}{(x - 5)(x - 6)} = 0
\Rightarrow \frac{-6}{x - 5} + \frac{1}{(x - 5)(x - 6)} = 0
\]

Step 2: Common denominator: \((x - 5)(x - 6)\)

First term: \(\frac{-6}{x - 5} = \frac{-6(x - 6)}{(x - 5)(x - 6)}\)

So:

\[
\frac{-6(x - 6) + 1}{(x - 5)(x - 6)} = 0
\Rightarrow \frac{-6x + 36 + 1}{(x - 5)(x - 6)} = 0
\Rightarrow \frac{-6x + 37}{(x - 5)(x - 6)} = 0
\]

Set numerator = 0:

\[
-6x + 37 = 0 \Rightarrow x = \frac{37}{6}
\]

Check: Not 5 or 6 → OK.

✔ Answer: \(x = \frac{37}{6}\)

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8)
Equation:
\[
5 + \frac{x^2 - 32x - 5}{7x} = \frac{x + 3}{7x}
\]

Domain: \(x ≠ 0\)

Step 1: Subtract right side from both sides:

\[
5 + \frac{x^2 - 32x - 5}{7x} - \frac{x + 3}{7x} = 0
\]

Combine fractions:

\[
5 + \frac{x^2 - 32x - 5 - (x + 3)}{7x} = 0
\Rightarrow 5 + \frac{x^2 - 33x - 8}{7x} = 0
\]

Step 2: Write 5 as \(\frac{35x}{7x}\) to combine:

\[
\frac{35x}{7x} + \frac{x^2 - 33x - 8}{7x} = 0
\Rightarrow \frac{x^2 + 2x - 8}{7x} = 0
\]

Set numerator = 0:

\[
x^2 + 2x - 8 = 0
\Rightarrow (x + 4)(x - 2) = 0
\Rightarrow x = -4 \text{ or } x = 2
\]

Both ≠ 0 → OK.

✔ Answer: \(x = -4\) or \(x = 2\)

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9)
Equation:
\[
\frac{1}{16k^2} = \frac{1}{8k^2} - \frac{1}{k}
\]

Domain: \(k ≠ 0\)

Step 1: Move all to left:

\[
\frac{1}{16k^2} - \frac{1}{8k^2} + \frac{1}{k} = 0
\]

Combine first two terms:

Common denominator \(16k^2\):

\[
\frac{1 - 2}{16k^2} + \frac{1}{k} = 0
\Rightarrow \frac{-1}{16k^2} + \frac{1}{k} = 0
\]

Step 2: Write \(\frac{1}{k} = \frac{16k}{16k^2}\)

So:

\[
\frac{-1 + 16k}{16k^2} = 0
\Rightarrow -1 + 16k = 0
\Rightarrow k = \frac{1}{16}
\]

✔ Answer: \(k = \frac{1}{16}\)

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10)
Equation:
\[
\frac{1}{c} = \frac{5}{9c} - 6
\]

Domain: \(c ≠ 0\)

Step 1: Move all to left:

\[
\frac{1}{c} - \frac{5}{9c} + 6 = 0
\]

Combine fractions:

Common denominator \(9c\):

\[
\frac{9}{9c} - \frac{5}{9c} + 6 = 0
\Rightarrow \frac{4}{9c} + 6 = 0
\]

Step 2: Subtract 6:

\[
\frac{4}{9c} = -6
\]

Step 3: Multiply both sides by \(9c\):

\[
4 = -54c
\Rightarrow c = -\frac{4}{54} = -\frac{2}{27}
\]

✔ Answer: \(c = -\frac{2}{27}\)

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Final Answer:
1) \(x = \frac{1}{24}\)
2) \(n = \frac{5}{48}\)
3) \(y = 4\)
4) No solution
5) \(d = 5\) or \(d = -5\)
6) \(x = -\frac{44}{5}\)
7) \(x = \frac{37}{6}\)
8) \(x = -4\) or \(x = 2\)
9) \(k = \frac{1}{16}\)
10) \(c = -\frac{2}{27}\)