Algebraic expressions and equations worksheet with fractions and variables.

It looks like you've uploaded a list of algebraic expressions involving rational expressions (fractions with variables), and the task is likely to simplify each expression.

Since I can't see the image directly, but you've provided the text, I will go through each expression one by one and simplify them step-by-step, explaining the process.

We’ll simplify each rational expression by:

1. Finding a common denominator.
2. Combining the numerators.
3. Factoring where possible.
4. Canceling common factors.

Let’s begin:

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a)

$$
\frac{2x+1}{y} - \frac{3x+2}{2y}
$$

Common denominator: $2y$

$$
= \frac{2(2x+1)}{2y} - \frac{3x+2}{2y} = \frac{4x + 2 - (3x + 2)}{2y} = \frac{4x + 2 - 3x - 2}{2y} = \frac{x}{2y}
$$

✔ Answer: $\boxed{\frac{x}{2y}}$

---

b)

$$
\frac{2a - 3b}{a} + \frac{4a^2 - 5b^2}{ab}
$$

Common denominator: $ab$

$$
= \frac{b(2a - 3b)}{ab} + \frac{4a^2 - 5b^2}{ab} = \frac{2ab - 3b^2 + 4a^2 - 5b^2}{ab} = \frac{4a^2 + 2ab - 8b^2}{ab}
$$

Factor numerator: $4a^2 + 2ab - 8b^2$ → no obvious factorization.

So leave as:

✔ Answer: $\boxed{\frac{4a^2 + 2ab - 8b^2}{ab}}$

---

c)

$$
\frac{a + 3b}{(a - b)^2} + \frac{a - 3b}{a^2 - b^2}
$$

Note: $a^2 - b^2 = (a - b)(a + b)$, so common denominator: $(a - b)^2(a + b)$

First term: $\frac{a + 3b}{(a - b)^2}$ → multiply numerator and denominator by $(a + b)$

Second term: $\frac{a - 3b}{(a - b)(a + b)}$ → multiply numerator and denominator by $(a - b)$

Now:
$$
= \frac{(a + 3b)(a + b)}{(a - b)^2(a + b)} + \frac{(a - 3b)(a - b)}{(a - b)^2(a + b)}
$$

Add numerators:

Numerator:
$$
(a + 3b)(a + b) + (a - 3b)(a - b)
$$

Compute each:

- $(a + 3b)(a + b) = a^2 + ab + 3ab + 3b^2 = a^2 + 4ab + 3b^2$
- $(a - 3b)(a - b) = a^2 - ab - 3ab + 3b^2 = a^2 - 4ab + 3b^2$

Sum: $a^2 + 4ab + 3b^2 + a^2 - 4ab + 3b^2 = 2a^2 + 6b^2$

So:
$$
\frac{2a^2 + 6b^2}{(a - b)^2(a + b)} = \frac{2(a^2 + 3b^2)}{(a - b)^2(a + b)}
$$

✔ Answer: $\boxed{\frac{2(a^2 + 3b^2)}{(a - b)^2(a + b)}}$

---

d)

$$
\frac{5a^2 - b^2}{ab} - \frac{3a - 2b}{b}
$$

Common denominator: $ab$

Second term: $\frac{3a - 2b}{b} = \frac{a(3a - 2b)}{ab} = \frac{3a^2 - 2ab}{ab}$

Now:
$$
\frac{5a^2 - b^2 - (3a^2 - 2ab)}{ab} = \frac{5a^2 - b^2 - 3a^2 + 2ab}{ab} = \frac{2a^2 + 2ab - b^2}{ab}
$$

✔ Answer: $\boxed{\frac{2a^2 + 2ab - b^2}{ab}}$

---

e)

$$
\frac{x}{ac} - \frac{x}{bc} + \frac{x}{ab}
$$

Factor $x$: $x\left(\frac{1}{ac} - \frac{1}{bc} + \frac{1}{ab}\right)$

Common denominator: $abc$

$$
= x\left(\frac{b - a + c}{abc}\right) = \frac{x(b - a + c)}{abc}
$$

✔ Answer: $\boxed{\frac{x(b - a + c)}{abc}}$

---

f)

$$
\frac{a}{bc} + \frac{b}{ac} + \frac{c}{ba}
$$

Write all terms clearly:

$$
\frac{a}{bc} + \frac{b}{ac} + \frac{c}{ab}
$$

Common denominator: $abc$

$$
= \frac{a^2}{abc} + \frac{b^2}{abc} + \frac{c^2}{abc} = \frac{a^2 + b^2 + c^2}{abc}
$$

✔ Answer: $\boxed{\frac{a^2 + b^2 + c^2}{abc}}$

---

g)

$$
\frac{1}{x^4 y^3} + \frac{2}{x^3 y^4}
$$

Common denominator: $x^4 y^4$

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

✔ Answer: $\boxed{\frac{2x + y}{x^4 y^4}}$

---

h)

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

Note: $2x^2 - 2 = 2(x^2 - 1) = 2(x - 1)(x + 1)$

And $x^2 + 1$ doesn’t factor over reals.

Common denominator: $2(x^2 + 1)(x^2 - 1)$

But perhaps better to write:

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

No easy simplification unless we force it. Let's proceed:

Multiply first term by $2(x^2 - 1)$, second by $x^2 + 1$

Common denominator: $2(x^2 + 1)(x^2 - 1)$

Numerator:

$$
(x+1)\cdot 2(x^2 - 1) + (x+2)(x^2 + 1)
$$

Compute:

1. $2(x+1)(x^2 - 1) = 2(x+1)(x-1)(x+1) = 2(x+1)^2(x-1)$ → wait, simpler:

$(x+1)(x^2 - 1) = (x+1)(x-1)(x+1) = (x+1)^2(x-1)$? No:

Wait: $x^2 - 1 = (x-1)(x+1)$, so:

$(x+1)(x^2 - 1) = (x+1)(x-1)(x+1) = (x+1)^2(x-1)$ — yes.

But let's compute numerically:

First part: $2(x+1)(x^2 - 1) = 2(x+1)(x-1)(x+1) = 2(x+1)^2(x-1)$

But actually, just expand:

- $(x+1)(x^2 - 1) = x(x^2 - 1) + 1(x^2 - 1) = x^3 - x + x^2 - 1 = x^3 + x^2 - x - 1$

Then times 2: $2x^3 + 2x^2 - 2x - 2$

Second part: $(x+2)(x^2 + 1) = x(x^2 + 1) + 2(x^2 + 1) = x^3 + x + 2x^2 + 2 = x^3 + 2x^2 + x + 2$

Now sum:

$$
(2x^3 + 2x^2 - 2x - 2) + (x^3 + 2x^2 + x + 2) = 3x^3 + 4x^2 - x
$$

Denominator: $2(x^2 + 1)(x^2 - 1) = 2(x^4 - 1)$

So:
$$
\frac{3x^3 + 4x^2 - x}{2(x^4 - 1)}
$$

Factor numerator: $x(3x^2 + 4x - 1)$ — doesn’t factor nicely.

So final form:

✔ Answer: $\boxed{\frac{3x^3 + 4x^2 - x}{2(x^4 - 1)}}$

---

i)

$$
\frac{a+1}{a^2 - a} + \frac{a+2}{2a^2 - 2}
$$

Simplify denominators:

- $a^2 - a = a(a - 1)$
- $2a^2 - 2 = 2(a^2 - 1) = 2(a - 1)(a + 1)$

So:

$$
\frac{a+1}{a(a - 1)} + \frac{a+2}{2(a - 1)(a + 1)}
$$

Common denominator: $2a(a - 1)(a + 1)$

First term: $\frac{(a+1) \cdot 2(a+1)}{2a(a - 1)(a + 1)} = \frac{2(a+1)^2}{2a(a - 1)(a + 1)}$

Wait, better:

Multiply first term by $2(a+1)/[2(a+1)]$, second by $a/a$

So:

- First: $\frac{(a+1) \cdot 2(a+1)}{2a(a - 1)(a + 1)} = \frac{2(a+1)^2}{2a(a - 1)(a + 1)}$
- Second: $\frac{(a+2) \cdot a}{2a(a - 1)(a + 1)} = \frac{a(a+2)}{2a(a - 1)(a + 1)}$

Now add:

Numerator: $2(a+1)^2 + a(a+2)$

Expand:

- $2(a^2 + 2a + 1) = 2a^2 + 4a + 2$
- $a(a+2) = a^2 + 2a$

Sum: $2a^2 + 4a + 2 + a^2 + 2a = 3a^2 + 6a + 2$

Denominator: $2a(a - 1)(a + 1) = 2a(a^2 - 1)$

✔ Answer: $\boxed{\frac{3a^2 + 6a + 2}{2a(a^2 - 1)}}$

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j)

$$
\frac{2x - 3y}{x^2 y} - \frac{4x - 5y}{x y^2}
$$

Common denominator: $x^2 y^2$

First term: $\frac{(2x - 3y)y}{x^2 y^2} = \frac{2xy - 3y^2}{x^2 y^2}$

Second term: $\frac{(4x - 5y)x}{x^2 y^2} = \frac{4x^2 - 5xy}{x^2 y^2}$

Now subtract:

$$
\frac{2xy - 3y^2 - (4x^2 - 5xy)}{x^2 y^2} = \frac{2xy - 3y^2 - 4x^2 + 5xy}{x^2 y^2} = \frac{-4x^2 + 7xy - 3y^2}{x^2 y^2}
$$

✔ Answer: $\boxed{\frac{-4x^2 + 7xy - 3y^2}{x^2 y^2}}$

---

k)

$$
\frac{2a^2 + 3a - 5}{a^2 b} - \frac{1 - 4a}{ab}
$$

Common denominator: $a^2 b$

Second term: $\frac{1 - 4a}{ab} = \frac{a(1 - 4a)}{a^2 b} = \frac{a - 4a^2}{a^2 b}$

Now:

$$
\frac{2a^2 + 3a - 5 - (a - 4a^2)}{a^2 b} = \frac{2a^2 + 3a - 5 - a + 4a^2}{a^2 b} = \frac{6a^2 + 2a - 5}{a^2 b}
$$

✔ Answer: $\boxed{\frac{6a^2 + 2a - 5}{a^2 b}}$

---

l)

$$
\frac{1}{a - b} - \frac{1}{a + b}
$$

Common denominator: $(a - b)(a + b) = a^2 - b^2$

$$
= \frac{(a + b) - (a - b)}{a^2 - b^2} = \frac{a + b - a + b}{a^2 - b^2} = \frac{2b}{a^2 - b^2}
$$

✔ Answer: $\boxed{\frac{2b}{a^2 - b^2}}$

---

m)

$$
\frac{5}{x - y} - \frac{3}{2x - 2y} = \frac{5}{x - y} - \frac{3}{2(x - y)}
$$

Common denominator: $2(x - y)$

$$
= \frac{10 - 3}{2(x - y)} = \frac{7}{2(x - y)}
$$

✔ Answer: $\boxed{\frac{7}{2(x - y)}}$

---

n)

$$
\frac{4}{a - b} - \frac{1}{b - a}
$$

Note: $b - a = -(a - b)$

So:
$$
\frac{4}{a - b} - \frac{1}{-(a - b)} = \frac{4}{a - b} + \frac{1}{a - b} = \frac{5}{a - b}
$$

✔ Answer: $\boxed{\frac{5}{a - b}}$

---

o)

$$
\frac{4}{r + 2} - \frac{3}{r - 2} - \frac{7r}{r^2 - 4}
$$

Note: $r^2 - 4 = (r + 2)(r - 2)$

Common denominator: $(r + 2)(r - 2)$

Now:

- $\frac{4(r - 2)}{(r + 2)(r - 2)} = \frac{4r - 8}{(r + 2)(r - 2)}$
- $\frac{3(r + 2)}{(r + 2)(r - 2)} = \frac{3r + 6}{(r + 2)(r - 2)}$
- $\frac{7r}{(r + 2)(r - 2)}$

Now combine:

$$
\frac{4r - 8 - (3r + 6) - 7r}{(r + 2)(r - 2)} = \frac{4r - 8 - 3r - 6 - 7r}{(r + 2)(r - 2)} = \frac{-6r - 14}{(r + 2)(r - 2)}
$$

Factor numerator: $-2(3r + 7)$

So:
$$
\frac{-2(3r + 7)}{r^2 - 4}
$$

✔ Answer: $\boxed{\frac{-2(3r + 7)}{r^2 - 4}}$

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p)

$$
\frac{7a^2}{a^2 - 9} + \frac{5a}{a - 3} + \frac{a}{a + 3}
$$

Note: $a^2 - 9 = (a - 3)(a + 3)$

Common denominator: $(a - 3)(a + 3)$

Rewrite:

- $\frac{7a^2}{(a - 3)(a + 3)}$
- $\frac{5a(a + 3)}{(a - 3)(a + 3)} = \frac{5a^2 + 15a}{(a - 3)(a + 3)}$
- $\frac{a(a - 3)}{(a - 3)(a + 3)} = \frac{a^2 - 3a}{(a - 3)(a + 3)}$

Add numerators:

$$
7a^2 + (5a^2 + 15a) + (a^2 - 3a) = 7a^2 + 5a^2 + 15a + a^2 - 3a = 13a^2 + 12a
$$

So:
$$
\frac{13a^2 + 12a}{(a - 3)(a + 3)} = \frac{a(13a + 12)}{a^2 - 9}
$$

✔ Answer: $\boxed{\frac{a(13a + 12)}{a^2 - 9}}$

---

q)

$$
\frac{a - b}{2a + 2b} + \frac{a^2 + b^2}{a^2 - a}
$$

Simplify:

- $2a + 2b = 2(a + b)$
- $a^2 - a = a(a - 1)$

So:
$$
\frac{a - b}{2(a + b)} + \frac{a^2 + b^2}{a(a - 1)}
$$

No common factors, so leave as is or find common denominator:

Common denominator: $2a(a + b)(a - 1)$

But this is messy. Maybe not needed unless asked.

But since we are simplifying, perhaps leave in simplified form:

But both terms are already simplified.

So expression is:
$$
\frac{a - b}{2(a + b)} + \frac{a^2 + b^2}{a(a - 1)}
$$

This is already simplified.

✔ Answer: $\boxed{\frac{a - b}{2(a + b)} + \frac{a^2 + b^2}{a(a - 1)}}$ (no further simplification)

---

r)

$$
\frac{7x - 1}{2x^2 - 6x} + \frac{3x - 5}{x^2 - 9}
$$

Simplify:

- $2x^2 - 6x = 2x(x - 3)$
- $x^2 - 9 = (x - 3)(x + 3)$

Common denominator: $2x(x - 3)(x + 3)$

First term: $\frac{7x - 1}{2x(x - 3)} = \frac{(7x - 1)(x + 3)}{2x(x - 3)(x + 3)}$

Second term: $\frac{3x - 5}{(x - 3)(x + 3)} = \frac{2x(3x - 5)}{2x(x - 3)(x + 3)}$

Now numerator:

$$
(7x - 1)(x + 3) + 2x(3x - 5)
$$

Compute:

- $(7x - 1)(x + 3) = 7x(x) + 7x(3) -1(x) -1(3) = 7x^2 + 21x - x - 3 = 7x^2 + 20x - 3$
- $2x(3x - 5) = 6x^2 - 10x$

Sum: $7x^2 + 20x - 3 + 6x^2 - 10x = 13x^2 + 10x - 3$

Denominator: $2x(x - 3)(x + 3) = 2x(x^2 - 9)$

✔ Answer: $\boxed{\frac{13x^2 + 10x - 3}{2x(x^2 - 9)}}$

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s)

$$
\frac{3a - b}{3a^2 b} + \frac{a^2 + b^2}{2a^2 b^2} - \frac{a + b}{2ab^2}
$$

Common denominator: $6a^2 b^2$

Check:

- $3a^2 b$ → multiply by $2b$: $6a^2 b^2$
- $2a^2 b^2$ → multiply by 3: $6a^2 b^2$
- $2ab^2$ → multiply by $3a$: $6a^2 b^2$

Now:

1. $\frac{3a - b}{3a^2 b} = \frac{2b(3a - b)}{6a^2 b^2} = \frac{6ab - 2b^2}{6a^2 b^2}$

2. $\frac{a^2 + b^2}{2a^2 b^2} = \frac{3(a^2 + b^2)}{6a^2 b^2} = \frac{3a^2 + 3b^2}{6a^2 b^2}$

3. $\frac{a + b}{2ab^2} = \frac{3a(a + b)}{6a^2 b^2} = \frac{3a^2 + 3ab}{6a^2 b^2}$

Now combine:

Numerator:

$$
(6ab - 2b^2) + (3a^2 + 3b^2) - (3a^2 + 3ab) =
6ab - 2b^2 + 3a^2 + 3b^2 - 3a^2 - 3ab
$$

Simplify:

- $6ab - 3ab = 3ab$
- $-2b^2 + 3b^2 = b^2$
- $3a^2 - 3a^2 = 0$

So: $3ab + b^2 = b(3a + b)$

Denominator: $6a^2 b^2$

So:
$$
\frac{b(3a + b)}{6a^2 b^2} = \frac{3a + b}{6a^2 b}
$$

✔ Answer: $\boxed{\frac{3a + b}{6a^2 b}}$

---

t)

$$
\frac{3x}{4a^2 b} - \frac{7}{6ab^2} - \frac{5x}{2ab^2}
$$

Common denominator: $12a^2 b^2$

Now:

1. $\frac{3x}{4a^2 b} = \frac{3x \cdot 3b}{12a^2 b^2} = \frac{9xb}{12a^2 b^2}$

2. $\frac{7}{6ab^2} = \frac{7 \cdot 2a}{12a^2 b^2} = \frac{14a}{12a^2 b^2}$

3. $\frac{5x}{2ab^2} = \frac{5x \cdot 6a}{12a^2 b^2} = \frac{30ax}{12a^2 b^2}$

Now combine:

$$
\frac{9xb - 14a - 30ax}{12a^2 b^2}
$$

Group terms: $9xb - 30ax = 3x(3b - 10a)$

So:
$$
\frac{3x(3b - 10a) - 14a}{12a^2 b^2}
$$

No further simplification.

✔ Answer: $\boxed{\frac{3x(3b - 10a) - 14a}{12a^2 b^2}}$

---

u)

$$
\frac{5}{t - 3} - \frac{t - 2}{t^2 - 9} + \frac{t - 1}{2t + 6}
$$

Note:

- $t^2 - 9 = (t - 3)(t + 3)$
- $2t + 6 = 2(t + 3)$

Common denominator: $2(t - 3)(t + 3)$

Now:

1. $\frac{5}{t - 3} = \frac{5 \cdot 2(t + 3)}{2(t - 3)(t + 3)} = \frac{10(t + 3)}{2(t - 3)(t + 3)}$

2. $\frac{t - 2}{(t - 3)(t + 3)} = \frac{2(t - 2)}{2(t - 3)(t + 3)}$

3. $\frac{t - 1}{2(t + 3)} = \frac{(t - 1)(t - 3)}{2(t - 3)(t + 3)}$

Now numerator:

$$
10(t + 3) - 2(t - 2) + (t - 1)(t - 3)
$$

Compute:

- $10(t + 3) = 10t + 30$
- $-2(t - 2) = -2t + 4$
- $(t - 1)(t - 3) = t^2 - 3t - t + 3 = t^2 - 4t + 3$

Sum: $10t + 30 - 2t + 4 + t^2 - 4t + 3 = t^2 + (10t - 2t - 4t) + (30 + 4 + 3) = t^2 + 4t + 37$

So:
$$
\frac{t^2 + 4t + 37}{2(t - 3)(t + 3)} = \frac{t^2 + 4t + 37}{2(t^2 - 9)}
$$

✔ Answer: $\boxed{\frac{t^2 + 4t + 37}{2(t^2 - 9)}}$

---

v)

$$
\frac{5}{a + 2} + \frac{2a}{a^2 + 4a + 4} - \frac{4}{a - 2}
$$

Note: $a^2 + 4a + 4 = (a + 2)^2$

So:

$$
\frac{5}{a + 2} + \frac{2a}{(a + 2)^2} - \frac{4}{a - 2}
$$

Common denominator: $(a + 2)^2(a - 2)$

This will be messy, but let’s do it.

1. $\frac{5}{a + 2} = \frac{5(a + 2)(a - 2)}{(a + 2)^2(a - 2)} = \frac{5(a^2 - 4)}{(a + 2)^2(a - 2)}$

2. $\frac{2a}{(a + 2)^2} = \frac{2a(a - 2)}{(a + 2)^2(a - 2)}$

3. $\frac{4}{a - 2} = \frac{4(a + 2)^2}{(a + 2)^2(a - 2)}$

Now numerator:

$$
5(a^2 - 4) + 2a(a - 2) - 4(a + 2)^2
$$

Compute:

- $5a^2 - 20$
- $2a^2 - 4a$
- $4(a^2 + 4a + 4) = 4a^2 + 16a + 16$

Now:
$$
(5a^2 - 20) + (2a^2 - 4a) - (4a^2 + 16a + 16) = 5a^2 - 20 + 2a^2 - 4a - 4a^2 - 16a - 16
$$

Combine:

- $5a^2 + 2a^2 - 4a^2 = 3a^2$
- $-4a - 16a = -20a$
- $-20 - 16 = -36$

So numerator: $3a^2 - 20a - 36$

Denominator: $(a + 2)^2(a - 2)$

✔ Answer: $\boxed{\frac{3a^2 - 20a - 36}{(a + 2)^2(a - 2)}}$

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Due to length, I'll stop here for now. But if you'd like me to continue with w to Z, please confirm!

Let me know if you want explanations for all or only specific ones!