Algebraic fractions practice problems from a math worksheet.

Let’s solve each equation one by one. We’ll use cross-multiplication when we have fractions equal to other fractions — that means if $\frac{a}{b} = \frac{c}{d}$, then $a \cdot d = b \cdot c$. For equations with variables in denominators, we must also remember: the denominator cannot be zero, so we’ll check our answers at the end to make sure they don’t break that rule.

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Problem 20:
$\frac{6x + 8}{3} = \frac{x - 2}{4}$

Cross-multiply:
$4(6x + 8) = 3(x - 2)$

Multiply out:
$24x + 32 = 3x - 6$

Subtract $3x$ from both sides:
$21x + 32 = -6$

Subtract 32 from both sides:
$21x = -38$

Divide by 21:
$x = -\frac{38}{21}$

Check denominators: original has 3 and 4 — no variable there, so okay.

✔ Final answer for 20: $x = -\frac{38}{21}$

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Problem 21:
$\frac{7x}{8} = \frac{x - 5}{6}$

Cross-multiply:
$6 \cdot 7x = 8 \cdot (x - 5)$
$42x = 8x - 40$

Subtract $8x$:
$34x = -40$

Divide by 34:
$x = -\frac{40}{34} = -\frac{20}{17}$ (simplified)

Denominators are constants — safe.

✔ Final answer for 21: $x = -\frac{20}{17}$

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Problem 22:
$\frac{3x + 5}{7} = \frac{x - 1}{2}$

Cross-multiply:
$2(3x + 5) = 7(x - 1)$
$6x + 10 = 7x - 7$

Subtract $6x$:
$10 = x - 7$

Add 7:
$x = 17$

Check denominators: 7 and 2 — fine.

✔ Final answer for 22: $x = 17$

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Problem 23:
$\frac{4}{x + 1} = \frac{1}{x + 1}$

Wait — same denominator! So numerators must be equal? But 4 ≠ 1. That can’t happen unless... let’s think.

If two fractions are equal and have the same denominator, their numerators must be equal. Here, 4 = 1? No. So no solution?

But let’s do it properly.

Cross-multiply:
$4(x + 1) = 1(x + 1)$
$4x + 4 = x + 1$

Subtract $x$:
$3x + 4 = 1$

Subtract 4:
$3x = -3$

Divide by 3:
$x = -1$

BUT — plug back into original: denominator becomes $-1 + 1 = 0$ → division by zero! Not allowed.

So even though algebra gave us $x = -1$, it’s invalid.

✔ Final answer for 23: No solution

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Problem 24:
$\frac{3}{x + 4} = \frac{2}{x + 4}$

Same situation as #23.

Cross-multiply:
$3(x + 4) = 2(x + 4)$
$3x + 12 = 2x + 8$

Subtract $2x$:
$x + 12 = 8$

Subtract 12:
$x = -4$

Plug back: denominator = $-4 + 4 = 0$ → invalid.

✔ Final answer for 24: No solution

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Problem 25:
$\frac{2}{x - 5} = \frac{3}{x + 5}$

Cross-multiply:
$2(x + 5) = 3(x - 5)$
$2x + 10 = 3x - 15$

Subtract $2x$:
$10 = x - 15$

Add 15:
$x = 25$

Check denominators:
$x - 5 = 20$ → ok
$x + 5 = 30$ → ok

✔ Final answer for 25: $x = 25$

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Problem 26:
$\frac{1}{2x + 1} = \frac{2}{x - 1}$

Cross-multiply:
$1 \cdot (x - 1) = 2 \cdot (2x + 1)$
$x - 1 = 4x + 2$

Subtract $x$:
$-1 = 3x + 2$

Subtract 2:
$-3 = 3x$

Divide by 3:
$x = -1$

Check denominators:
$2x + 1 = 2(-1) + 1 = -1$ → ok
$x - 1 = -1 - 1 = -2$ → ok

✔ Final answer for 26: $x = -1$

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Problem 27:
$\frac{4}{x + 5} = \frac{6}{x - 3}$

Cross-multiply:
$4(x - 3) = 6(x + 5)$
$4x - 12 = 6x + 30$

Subtract $4x$:
$-12 = 2x + 30$

Subtract 30:
$-42 = 2x$

Divide by 2:
$x = -21$

Check denominators:
$x + 5 = -16$ → ok
$x - 3 = -24$ → ok

✔ Final answer for 27: $x = -21$

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Problem 28:
$\frac{4}{2x - 1} = \frac{2}{x + 4}$

Cross-multiply:
$4(x + 4) = 2(2x - 1)$
$4x + 16 = 4x - 2$

Subtract $4x$:
$16 = -2$

That’s false! Never true.

So no solution.

Wait — did I do that right?

Left: $4(x+4) = 4x + 16$
Right: $2(2x - 1) = 4x - 2$

Set equal: $4x + 16 = 4x - 2$

Subtract $4x$: $16 = -2$ → contradiction.

So no value of x makes this true.

Also, check if any x would make denominator zero:

$2x - 1 = 0$ → $x = 0.5$
$x + 4 = 0$ → $x = -4$

Our contradiction doesn’t depend on those — it’s always false.

✔ Final answer for 28: No solution

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Final Answer:
20. $x = -\frac{38}{21}$
21. $x = -\frac{20}{17}$
22. $x = 17$
23. No solution
24. No solution
25. $x = 25$
26. $x = -1$
27. $x = -21$
28. No solution