Simplifying Algebraic Fractions - Minimally Different - Free Printable

Here are the step-by-step solutions for each problem. I have simplified the expressions by finding common denominators, factoring where possible, and canceling out terms.

a)
1. Simplify the numerator: $1 + \frac{1}{x+1} = \frac{x+1+1}{x+1} = \frac{x+2}{x+1}$.
2. Simplify the denominator: $x - \frac{4}{x} = \frac{x^2-4}{x} = \frac{(x-2)(x+2)}{x}$.
3. Divide the numerator by the denominator: $\frac{x+2}{x+1} \cdot \frac{x}{(x-2)(x+2)}$.
4. Cancel $(x+2)$: $\frac{x}{(x+1)(x-2)}$.

b)
1. Rewrite negative exponents as fractions: $\frac{\frac{1}{x} + \frac{1}{y}}{\frac{1}{x^2} - \frac{1}{y^2}}$.
2. Combine terms in numerator: $\frac{y+x}{xy}$.
3. Combine terms in denominator: $\frac{y^2-x^2}{x^2y^2} = \frac{(y-x)(y+x)}{x^2y^2}$.
4. Multiply by reciprocal: $\frac{x+y}{xy} \cdot \frac{x^2y^2}{(y-x)(y+x)}$.
5. Cancel $(x+y)$ and simplify powers: $\frac{xy}{y-x}$.

c)
1. Factor numerators and denominators: $\frac{(a-1)(a+1)}{a+1} \cdot \frac{1}{\frac{a-1}{a}}$.
2. Simplify first fraction to $(a-1)$.
3. Invert second fraction: $\frac{a}{a-1}$.
4. Multiply: $(a-1) \cdot \frac{a}{a-1} = a$.

d)
1. Simplify first bracket: $\frac{1}{z+1} - \frac{2z}{(z-1)(z+1)} = \frac{z-1-2z}{(z-1)(z+1)} = \frac{-(z+1)}{(z-1)(z+1)} = \frac{-1}{z-1}$.
2. Simplify second bracket: $\frac{1}{z} - 1 = \frac{1-z}{z} = -\frac{z-1}{z}$.
3. Multiply results: $\left(\frac{-1}{z-1}\right) \cdot \left(-\frac{z-1}{z}\right) = \frac{1}{z}$.

e)
1. Factor differences of squares: $\frac{(a-y)(a+y)}{a+b} \cdot \frac{(a-b)(a+b)}{y(a+y)} \cdot \frac{a^2-ay}{a-y}$.
2. Factor third term numerator: $a(a-y)$.
3. Expression becomes: $\frac{(a-y)(a+y)}{a+b} \cdot \frac{(a-b)(a+b)}{y(a+y)} \cdot \frac{a(a-y)}{a-y}$.
4. Cancel $(a+y)$, $(a+b)$, and one $(a-y)$ pair.
5. Remaining terms: $\frac{a(a-y)(a-b)}{y}$.

f)
1. Numerator: $\frac{1-x}{1-x+x^2} + \frac{1+x}{1+x+x^2}$. Common denom is product. Num becomes $(1-x)(1+x+x^2) + (1+x)(1-x+x^2) = (1-x^3) + (1+x^3) = 2$.
2. Denominator: $\frac{1+x}{1+x+x^2} - \frac{1-x}{1-x+x^2}$. Num becomes $(1+x)(1-x+x^2) - (1-x)(1-x+x^2)$... wait, let's use sum/diff of cubes properly.
* Top Num: $(1-x^3) + (1+x^3) = 2$.
* Bottom Num: $(1+x^3) - (1-x^3) = 2x^3$.
3. The complex fraction is $\frac{\frac{2}{D_1 D_2}}{\frac{2x^3}{D_1 D_2}} = \frac{2}{2x^3} = \frac{1}{x^3}$.

g)
1. Numerator: $a^4-b^4 = (a^2-b^2)(a^2+b^2)$. Divided by $a^2b^2$.
2. Denominator bracket: $1 + \frac{b^2}{a^2} = \frac{a^2+b^2}{a^2}$. Second part: $1 - \frac{2a}{b} + \frac{a^2}{b^2} = (\frac{b-a}{b})^2 = \frac{(a-b)^2}{b^2}$.
3. Denominator total: $\frac{a^2+b^2}{a^2} \cdot \frac{(a-b)^2}{b^2}$.
4. Divide Num by Denom: $\frac{(a-b)(a+b)(a^2+b^2)}{a^2b^2} \cdot \frac{a^2b^2}{(a^2+b^2)(a-b)^2}$.
5. Cancel terms: $\frac{a+b}{a-b}$.

h)
1. First bracket: Common denom $x^2-y^2$. Num: $2x(x-y) + y(x+y) + y^2 = 2x^2-2xy+xy+y^2+y^2 = 2x^2-xy+2y^2$? No, let's recheck.
* Term 1: $\frac{2x(x-y)}{x^2-y^2}$. Term 2: $\frac{y(x+y)}{x^2-y^2}$. Term 3: $\frac{y^2}{x^2-y^2}$.
* Sum: $2x^2-2xy + xy+y^2 + y^2 = 2x^2-xy+2y^2$. This doesn't factor nicely. Let's look closer.
* Actually, usually these simplify to single terms. Let's try specific values or re-read. Ah, $\frac{y^2}{y^2-x^2} = -\frac{y^2}{x^2-y^2}$.
* Correct Num: $2x(x-y) + y(x+y) - y^2 = 2x^2-2xy+xy+y^2-y^2 = 2x^2-xy = x(2x-y)$.
* So Bracket 1 is $\frac{x(2x-y)}{(x-y)(x+y)}$.
2. Second bracket: $\frac{1}{x+y} + \frac{x}{(x-y)(x+y)} = \frac{x-y+x}{(x-y)(x+y)} = \frac{2x-y}{(x-y)(x+y)}$.
3. Divide: $\frac{x(2x-y)}{(x-y)(x+y)} \cdot \frac{(x-y)(x+y)}{2x-y} = x$.

i)
1. Inside bracket: $\frac{s+1}{2-2s} = -\frac{s+1}{2(s-1)}$. And $\frac{s^2+3}{2s^2-2} = \frac{s^2+3}{2(s-1)(s+1)}$.
2. This looks messy. Let's simplify $2s - (\dots)$.
3. Let's evaluate the term being subtracted: $A = \left(\frac{2s-3}{s+1} - \frac{s+1}{2(1-s)} - \frac{s^2+3}{2(s-1)(s+1)}\right) \frac{s^3+1}{s(s-1)}$.
4. Common denom for inside parenthesis: $2(s+1)(s-1)$.
* Term 1: $2(s-1)(2s-3) = 2(2s^2-5s+3) = 4s^2-10s+6$.
* Term 2: $+(s+1)^2 = s^2+2s+1$ (sign change from $1-s$).
* Term 3: $-(s^2+3)$.
* Sum Num: $4s^2-10s+6 + s^2+2s+1 - s^2-3 = 4s^2-8s+4 = 4(s-1)^2$.
* Inside value: $\frac{4(s-1)^2}{2(s+1)(s-1)} = \frac{2(s-1)}{s+1}$.
5. Multiply by fraction: $\frac{2(s-1)}{s+1} \cdot \frac{(s+1)(s^2-s+1)}{s(s-1)} = \frac{2(s^2-s+1)}{s} = 2s - 2 + \frac{2}{s}$.
6. Full expression: $2s - (2s - 2 + \frac{2}{s}) = 2 - \frac{2}{s} = \frac{2s-2}{s}$.

j)
1. Inner bracket: $1 - \frac{2}{1+1/k} = 1 - \frac{2k}{k+1} = \frac{k+1-2k}{k+1} = \frac{1-k}{k+1}$.
2. Outer fraction: $\frac{\frac{k^2+1-k(k-1)}{k-1}}{\frac{k^2-1+k+1}{k+1}} = \frac{\frac{k^2+1-k^2+k}{k-1}}{\frac{k^2+k}{k+1}} = \frac{\frac{k+1}{k-1}}{\frac{k(k+1)}{k+1}} = \frac{k+1}{k-1} \cdot \frac{1}{k}$.
3. Wait, denominator simplifies to $\frac{k(k+1)}{k+1} = k$. So outer fraction is $\frac{k+1}{k(k-1)}$.
4. Multiply by inner bracket result: $\frac{k+1}{k(k-1)} \cdot \frac{-(k-1)}{k+1} = -\frac{1}{k}$.

k)
1. First bracket: $\frac{a^3-ab^2+b^3}{(a-b)^3} - \frac{b}{a-b} = \frac{a^3-ab^2+b^3 - b(a-b)^2}{(a-b)^3}$.
* Num: $a^3-ab^2+b^3 - b(a^2-2ab+b^2) = a^3-ab^2+b^3-a^2b+2ab^2-b^3 = a^3-a^2b+ab^2 = a(a^2-ab+b^2)$.
* Result 1: $\frac{a(a^2-ab+b^2)}{(a-b)^3}$.
2. Second bracket: $\frac{a^2-2ab+2b^2}{a^2-ab+b^2} - \frac{b}{a} = \frac{a(a^2-2ab+2b^2)-b(a^2-ab+b^2)}{a(a^2-ab+b^2)}$.
* Num: $a^3-2a^2b+2ab^2 - a^2b+ab^2-b^3 = a^3-3a^2b+3ab^2-b^3 = (a-b)^3$.
* Result 2: $\frac{(a-b)^3}{a(a^2-ab+b^2)}$.
3. Multiply Result 1 and Result 2: Everything cancels to $1$.

l)
1. First bracket: Common denom $2(u-v)(u+v)$.
* Num: $(u+v)^2 - (v-u)(u+v) + 4v^2$? No, middle term denom is $2(v+u)$. Last is $u^2-v^2$.
* LCM: $2(u-v)(u+v)$.
* Term 1: $(u+v)^2 = u^2+2uv+v^2$.
* Term 2: $-(v-u)(u+v) = -(v^2-u^2) = u^2-v^2$.
* Term 3: $2(2v^2) = 4v^2$.
* Sum: $u^2+2uv+v^2 + u^2-v^2 + 4v^2 = 2u^2+2uv+4v^2 = 2(u^2+uv+2v^2)$.
* Bracket 1: $\frac{2(u^2+uv+2v^2)}{2(u^2-v^2)} = \frac{u^2+uv+2v^2}{u^2-v^2}$.
2. Second bracket: $\frac{1}{v} - \frac{1}{u} = \frac{u-v}{uv}$.
3. Multiply: $\frac{u^2+uv+2v^2}{(u-v)(u+v)} \cdot \frac{u-v}{uv} = \frac{u^2+uv+2v^2}{uv(u+v)}$.

m)
1. Left side: $\frac{1}{\frac{1+x}{x}} + \frac{1-\frac{1}{x}}{\frac{1}{x}} = \frac{x}{1+x} + (1-\frac{1}{x})x = \frac{x}{1+x} + x - 1 = \frac{x + x(1+x) - (1+x)}{1+x} = \frac{x+x+x^2-1-x}{1+x} = \frac{x^2+x-1}{1+x}$.
2. Right side: $\frac{x^{-1}}{1+x^{-1}} - \frac{1-x^{-1}}{x^{-1}} = \frac{1/x}{1+1/x} - \frac{1-1/x}{1/x} = \frac{1}{x+1} - x(1-\frac{1}{x}) = \frac{1}{x+1} - (x-1) = \frac{1 - (x-1)(x+1)}{x+1} = \frac{1-(x^2-1)}{x+1} = \frac{2-x^2}{x+1}$.
3. Divide Left by Right: $\frac{x^2+x-1}{x+1} \cdot \frac{x+1}{2-x^2} = \frac{x^2+x-1}{2-x^2}$.

n)
1. First term: $\frac{a^3+b^3}{a+b} = a^2-ab+b^2$.
2. Second term: $\frac{2b}{a+b}$.
3. Third term: $-\frac{ab}{(a-b)(a+b)}$.
4. Common denom: $a+b$? No, third term has $a-b$. Let's combine first two first.
* $a^2-ab+b^2 + \frac{2b}{a+b} = \frac{(a^2-ab+b^2)(a+b)+2b}{a+b} = \frac{a^3+b^3+2b}{a+b}$. This doesn't simplify well.
* Let's use common denom $(a+b)(a-b) = a^2-b^2$ for all three?
* Term 1: $\frac{(a^2-ab+b^2)(a-b)(a+b)}{a+b}$? No.
* Let's just put everything over $a^2-b^2$.
* Term 1: $\frac{(a^2-ab+b^2)(a-b)}{1} \cdot \frac{a+b}{a+b}$? No.
* $\frac{a^3+b^3}{a+b} = \frac{(a^3+b^3)(a-b)}{a^2-b^2}$.
* $\frac{2b}{a+b} = \frac{2b(a-b)}{a^2-b^2}$.
* $-\frac{ab}{a^2-b^2}$.
* Num: $(a^3+b^3)(a-b) + 2b(a-b) - ab = a^4-a^3b+ab^3-b^4 + 2ab-2b^2 - ab = a^4-a^3b+ab^3-b^4+ab-2b^2$.
* This seems overly complex. Is there a typo in my reading?
* Let's re-evaluate Term 1 + Term 2: $a^2-ab+b^2 + \frac{2b}{a+b}$.
* Maybe group Term 2 and 3? $\frac{2b(a-b)-ab}{a^2-b^2} = \frac{2ab-2b^2-ab}{a^2-b^2} = \frac{ab-2b^2}{a^2-b^2}$.
* Add Term 1: $a^2-ab+b^2 + \frac{ab-2b^2}{a^2-b^2} = \frac{(a^2-ab+b^2)(a^2-b^2) + ab-2b^2}{a^2-b^2}$.
* Num: $a^4-a^2b^2-a^3b+ab^3+a^2b^2-b^4 + ab-2b^2 = a^4-a^3b+ab^3-b^4+ab-2b^2$.
* Does not factor cleanly. Let me re-read the image carefully.
* Image: $\frac{a^3+b^3}{a+b} : (a^2-b^2) + \frac{2b}{a+b} - \frac{ab}{a^2-b^2}$.
* Ah, the colon `:` means division. It is NOT three terms added.
* It is $\left[ \frac{a^3+b^3}{a+b} \right] \div \left[ (a^2-b^2) + \frac{2b}{a+b} - \frac{ab}{a^2-b^2} \right]$.
* Let's simplify the divisor (Denominator D).
* $D = (a-b)(a+b) + \frac{2b}{a+b} - \frac{ab}{(a-b)(a+b)}$.
* Common denom for D: $(a-b)(a+b)$.
* Num of D: $(a-b)^2(a+b)^2 + 2b(a-b) - ab$.
* $(a^2-b^2)^2 + 2ab-2b^2-ab = a^4-2a^2b^2+b^4+ab-2b^2$. Still ugly.
* Let's look at the structure again. Maybe the first part divides ONLY the first term of the bracket? No, standard order of operations.
* Let's try simplifying the divisor differently.
* $D = \frac{(a^2-b^2)^2(a+b) + 2b(a-b) - ab}{a^2-b^2}$? No.
* Let's assume the question implies: $\frac{a^3+b^3}{a+b} \div \left( (a^2-b^2) + \frac{2b}{a+b} - \frac{ab}{a^2-b^2} \right)$.
* Let's check if $a=2, b=1$.
* Num: $(8+1)/3 = 3$.
* Divisor: $(4-1) + 2/3 - 2/3 = 3$.
* Result: $3/3 = 1$.
* Let's check $a=3, b=1$.
* Num: $(27+1)/4 = 7$.
* Divisor: $(9-1) + 2/4 - 3/8 = 8 + 0.5 - 0.375 = 8.125 = 65/8$.
* Result: $7 / (65/8) = 56/65$. Not 1.
* Let's re-read the divisor expression.
* $(a^2-b^2) + \frac{2b}{a+b} - \frac{ab}{a^2-b^2}$.
* Common denom $a^2-b^2$.
* Num: $(a^2-b^2)^2 + 2b(a-b) - ab$.
* $= (a^4 - 2a^2b^2 + b^4) + 2ab - 2b^2 - ab = a^4 - 2a^2b^2 + b^4 + ab - 2b^2$.
* This does not seem to cancel with $a^2-ab+b^2$.
* Is it possible the last term is $\frac{ab}{a+b}$? No, looks like $a^2-b^2$.
* Is it possible the middle term is $\frac{2b}{a-b}$?
* If middle is $\frac{2b}{a-b}$, Num D: $(a^2-b^2)^2 + 2b(a+b) - ab = a^4... + 2ab+2b^2-ab$. Still messy.
* Let's look at similar problems online. Often the answer is 1 or simple fraction.
* Let's try to factor $a^4 - 2a^2b^2 + b^4 + ab - 2b^2$.
* Maybe I should leave it as the simplified fraction.
* Numerator of whole expression: $a^2-ab+b^2$.
* Denominator of whole expression: $\frac{a^4 - 2a^2b^2 + b^4 + ab - 2b^2}{a^2-b^2}$.
* Final Answer: $\frac{(a^2-ab+b^2)(a^2-b^2)}{a^4 - 2a^2b^2 + b^4 + ab - 2b^2}$.

*Self-Correction on N*: Looking really closely at crop 6.
The term is $\frac{ab}{a^2-b^2}$.
The term before is $\frac{2b}{a+b}$.
The term before is $(a^2-b^2)$.

Let's try one more algebraic manipulation on the divisor numerator:
$a^4 - 2a^2b^2 + b^4 + ab - 2b^2$.
If $b=0$, $a^4$. Denom $a^2$. Result $a^2$. Num $a^2$. Ratio 1.
If $a=b$, Divisor Num $a^4-2a^4+a^4+a^2-2a^2 = -a^2$. Denom 0. Undefined.

Actually, there is a chance the second term is $\frac{2b^2}{a+b}$? No.
I will provide the unsimplified combined fraction as it is mathematically correct based on the visual input.

Final Answer:
a) $\frac{x}{(x+1)(x-2)}$
b) $\frac{xy}{y-x}$
c) $a$
d) $\frac{1}{z}$
e) $\frac{a(a-y)(a-b)}{y}$
f) $\frac{1}{x^3}$
g) $\frac{a+b}{a-b}$
h) $x$
i) $\frac{2s-2}{s}$
j) $-\frac{1}{k}$
k) $1$
l) $\frac{u^2+uv+2v^2}{uv(u+v)}$
m) $\frac{x^2+x-1}{2-x^2}$
n) $\frac{(a^2-ab+b^2)(a^2-b^2)}{a^4 - 2a^2b^2 + b^4 + ab - 2b^2}$