Advanced Algebra Worksheets - Fill Online, Printable, Fillable ... - Free Printable

Let’s solve each equation one by one. We’ll go step by step, checking our work as we go.

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1. $\frac{1}{8} + \frac{2}{7} = \frac{17}{b}$

First, add the fractions on the left. Find a common denominator for 8 and 7 → that’s 56.

$\frac{1}{8} = \frac{7}{56}$
$\frac{2}{7} = \frac{16}{56}$
So, $\frac{7}{56} + \frac{16}{56} = \frac{23}{56}$

Now:
$\frac{23}{56} = \frac{17}{b}$

Cross-multiply:
$23b = 56 \cdot 17$

Calculate $56 \cdot 17$:
$56 \cdot 10 = 560$
$56 \cdot 7 = 392$
Total: $560 + 392 = 952$

So:
$23b = 952$
$b = \frac{952}{23}$

Let’s divide:
23 × 41 = 943
952 - 943 = 9 → so not whole? Wait — did I make a mistake?

Wait — let me double-check the addition:

$\frac{1}{8} + \frac{2}{7}$ → LCD is 56.

$\frac{1}{8} = \frac{7}{56}$ ✔
$\frac{2}{7} = \frac{16}{56}$ ✔
Sum: $\frac{23}{56}$ ✔

Then: $\frac{23}{56} = \frac{17}{b}$ → cross multiply: 23b = 56×17

56×17:
50×17=850
6×17=102
850+102=952 ✔

So b = 952 ÷ 23

Let me do exact division:
23 × 41 = 943
952 - 943 = 9 → so 41 and 9/23 → but maybe the problem expects fraction? Or did I misread?

Wait — perhaps the original problem was meant to have nice numbers? Let me check again.

Actually, maybe I should leave it as simplified fraction unless told otherwise.

But let’s see if 952 and 23 have common factors. 23 is prime. 23 doesn’t divide 952 evenly? 23×41=943, 23×42=966 → too big. So answer is $\frac{952}{23}$ or mixed number 41 ⁄₂₃.

But maybe I made an error in setup? Let me re-read the problem.

Problem says: $\frac{1}{8} + \frac{2}{7} = \frac{17}{b}$ — yes.

Perhaps the 17 is wrong? Or maybe it's correct and we just compute.

Alternatively, maybe the student is expected to solve for b algebraically without simplifying first? But no, this is straightforward.

I think we proceed with b = 952/23. But let’s keep going and come back if needed.

Actually — wait! Maybe I miscalculated 56×17?

56 × 17:

Breakdown:
56 × 10 = 560
56 × 7 = 392
560 + 392 = 952 — correct.

23 × 41 = 943 — correct.

So b = 952/23 — which is approximately 41.39, but since it’s algebra, we can leave as improper fraction.

But let’s check if the problem might have typo? No — we must solve as given.

So for now, Answer 1: $b = \frac{952}{23}$

But let’s hold off final answers until all are done — maybe I’ll spot pattern or error.

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2. $7 - \frac{1}{m} = 4$

Subtract 7 from both sides:

$-\frac{1}{m} = 4 - 7 = -3$

Multiply both sides by -1:

$\frac{1}{m} = 3$

Take reciprocal of both sides:

$m = \frac{1}{3}$

✔ Check: 7 - 1/(1/3) = 7 - 3 = 4 — correct.

Answer 2: $m = \frac{1}{3}$

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3. $\frac{1}{r-2} = \frac{7}{30}$

Cross-multiply:

$1 \cdot 30 = 7 \cdot (r - 2)$

30 = 7r - 14

Add 14 to both sides:

44 = 7r

Divide by 7:

$r = \frac{44}{7}$

✔ Check: r - 2 = 44/7 - 14/7 = 30/7 → 1/(30/7) = 7/30 — matches right side. Good.

Answer 3: $r = \frac{44}{7}$

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4. $\frac{1}{x} + \frac{x}{6} = \frac{5}{6}$

Get common denominator on left. LCD of x and 6 is 6x.

Rewrite:

$\frac{6}{6x} + \frac{x^2}{6x} = \frac{5}{6}$

Combine left:

$\frac{6 + x^2}{6x} = \frac{5}{6}$

Cross-multiply:

6(6 + x²) = 5 · 6x

36 + 6x² = 30x

Bring all terms to one side:

6x² - 30x + 36 = 0

Divide entire equation by 6:

x² - 5x + 6 = 0

Factor:

(x - 2)(x - 3) = 0

So x = 2 or x = 3

Check for restrictions: x ≠ 0 (from original equation), and denominators ok.

Check x=2:

Left: 1/2 + 2/6 = 1/2 + 1/3 = 3/6 + 2/6 = 5/6 — good.

Check x=3:

1/3 + 3/6 = 1/3 + 1/2 = 2/6 + 3/6 = 5/6 — also good.

So two solutions.

Answer 4: $x = 2$ or $x = 3$

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5. $m + \frac{12}{m} = 7$

Multiply both sides by m (assuming m ≠ 0):

m² + 12 = 7m

Bring all to one side:

m² - 7m + 12 = 0

Factor:

(m - 3)(m - 4) = 0

So m = 3 or m = 4

Check:

m=3: 3 + 12/3 = 3 + 4 = 7 — good.

m=4: 4 + 12/4 = 4 + 3 = 7 — good.

Answer 5: $m = 3$ or $m = 4$

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6. $n + \frac{1}{n} = 2$

Multiply both sides by n (n ≠ 0):

n² + 1 = 2n

n² - 2n + 1 = 0

(n - 1)² = 0 → n = 1

Check: 1 + 1/1 = 2 — good.

Answer 6: $n = 1$

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7. $\frac{2x}{x+2} + \frac{4}{x+2} = \frac{4}{x+2}$

Notice same denominator. Combine left:

$\frac{2x + 4}{x+2} = \frac{4}{x+2}$

Since denominators equal and non-zero (x ≠ -2), set numerators equal:

2x + 4 = 4

2x = 0 → x = 0

Check: Plug x=0 into original:

Left: 0/(0+2) + 4/(0+2) = 0 + 2 = 2
Right: 4/(0+2) = 2 — good.

Also, x ≠ -2 — satisfied.

Answer 7: $x = 0$

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8. $\frac{x}{x-3} + \frac{5}{2} = \frac{4x}{2x - 6}$

Note: 2x - 6 = 2(x - 3). So rewrite right side:

$\frac{4x}{2(x - 3)} = \frac{2x}{x - 3}$

So equation becomes:

$\frac{x}{x-3} + \frac{5}{2} = \frac{2x}{x-3}$

Subtract $\frac{x}{x-3}$ from both sides:

$\frac{5}{2} = \frac{2x}{x-3} - \frac{x}{x-3} = \frac{x}{x-3}$

So:

$\frac{5}{2} = \frac{x}{x-3}$

Cross-multiply:

5(x - 3) = 2x

5x - 15 = 2x

5x - 2x = 15

3x = 15 → x = 5

Check restriction: x ≠ 3 — ok.

Plug x=5 into original:

Left: 5/(5-3) + 5/2 = 5/2 + 5/2 = 10/2 = 5
Right: 4*5 / (2*5 - 6) = 20 / (10-6) = 20/4 = 5 — good.

Answer 8: $x = 5$

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9. $\frac{3}{x(x-1)} = \frac{3}{x^2 - x}$

Note: x² - x = x(x - 1) — same denominator!

So left and right are identical: $\frac{3}{x(x-1)} = \frac{3}{x(x-1)}$

This is always true — except where undefined.

Undefined when denominator = 0 → x(x-1)=0 → x=0 or x=1

So solution: all real numbers except x=0 and x=1.

But typically in such problems, if identity, we say “all real numbers except excluded values”.

But let’s see what the problem asks — “solve each equation”. Since it’s an identity, every x except 0 and 1 works.

But sometimes they expect you to recognize it’s always true.

In many textbooks, they’d write: “All real numbers except x=0 and x=1”

But let’s confirm: Is there any x that makes it false? Only where undefined.

So technically, solution set is {x | x ≠ 0, x ≠ 1}

But perhaps for this level, they want to note it’s an identity.

Looking at other problems, likely they expect to state the restriction.

But let’s see — maybe I should write: All real numbers except 0 and 1.

But to be precise.

Alternatively, if we subtract both sides:

$\frac{3}{x(x-1)} - \frac{3}{x(x-1)} = 0$ → 0=0 — always true.

So yes.

Answer 9: All real numbers except $x = 0$ and $x = 1$

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10. $\frac{16n}{n-3} = \frac{5}{n-3}$

Same denominator. Set numerators equal (since n ≠ 3):

16n = 5 → n = 5/16

Check: n ≠ 3 — yes.

Plug in: Left: 16*(5/16)/(5/16 - 3) = 5 / (5/16 - 48/16) = 5 / (-43/16) = 5 * (-16/43) = -80/43

Right: 5 / (5/16 - 48/16) = same — yes.

Wait — but according to equation: 16n/(n-3) = 5/(n-3)

If n ≠ 3, then 16n = 5 → n=5/16 — correct.

And both sides equal when plugged in.

Answer 10: $n = \frac{5}{16}$

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11. $\frac{7}{x+1} = \frac{2}{x^2 - 4}$

Note: x² - 4 = (x-2)(x+2)

So:

$\frac{7}{x+1} = \frac{2}{(x-2)(x+2)}$

Cross-multiply:

7(x-2)(x+2) = 2(x+1)

Left: 7(x² - 4) = 7x² - 28

Right: 2x + 2

So:

7x² - 28 = 2x + 2

Bring all to left:

7x² - 2x - 30 = 0

Use quadratic formula:

x = [2 ± √(4 + 840)] / 14 = [2 ± √844] / 14

Simplify √844: 844 ÷ 4 = 211 → √844 = 2√211

So x = [2 ± 2√211]/14 = [1 ± √211]/7

Check restrictions: x ≠ -1, 2, -2 — none of these make denominator zero in solution? √211 ≈ 14.5, so x≈ (1+14.5)/7≈15.5/7≈2.2, and (1-14.5)/7≈-13.5/7≈-1.9 — neither is -1,2,-2 — so ok.

But let me verify calculation.

From:

7x² - 2x - 30 = 0

Discriminant: b² - 4ac = (-2)^2 - 4*7*(-30) = 4 + 840 = 844 — yes.

√844 = √(4*211) = 2√211 — yes.

So x = [2 ± 2√211]/14 = [1 ± √211]/7

We can leave like that.

Answer 11: $x = \frac{1 \pm \sqrt{211}}{7}$

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12. $\frac{4}{x} + 6 = \frac{1}{x}$

Subtract 1/x from both sides:

$\frac{4}{x} - \frac{1}{x} + 6 = 0$

$\frac{3}{x} + 6 = 0$

$\frac{3}{x} = -6$

Multiply both sides by x:

3 = -6x

x = 3 / (-6) = -1/2

Check: x ≠ 0 — ok.

Plug in: Left: 4/(-1/2) + 6 = -8 + 6 = -2
Right: 1/(-1/2) = -2 — good.

Answer 12: $x = -\frac{1}{2}$

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Now, let’s compile all answers clearly.

Final Answers:

1. $b = \frac{952}{23}$
2. $m = \frac{1}{3}$
3. $r = \frac{44}{7}$
4. $x = 2$ or $x = 3$
5. $m = 3$ or $m = 4$
6. $n = 1$
7. $x = 0$
8. $x = 5$
9. All real numbers except $x = 0$ and $x = 1$
10. $n = \frac{5}{16}$
11. $x = \frac{1 \pm \sqrt{211}}{7}$
12. $x = -\frac{1}{2}$

For problem 1, although it’s messy, it’s correct based on given numbers.

Double-check problem 1 once more:

$\frac{1}{8} + \frac{2}{7} = \frac{7+16}{56} = \frac{23}{56} = \frac{17}{b}$ → b = 17 * 56 / 23 = 952 / 23 — yes.

No simplification possible.

So we’re good.

Final Answer:
1. $b = \frac{952}{23}$
2. $m = \frac{1}{3}$
3. $r = \frac{44}{7}$
4. $x = 2$ or $x = 3$
5. $m = 3$ or $m = 4$
6. $n = 1$
7. $x = 0$
8. $x = 5$
9. All real numbers except $x = 0$ and $x = 1$
10. $n = \frac{5}{16}$
11. $x = \frac{1 \pm \sqrt{211}}{7}$
12. $x = -\frac{1}{2}$