Q: How many tablespoons are equivalent to 3 ounces?

The correct answer is:6 tablespoons Explanation A US fluid ounce is a measure of volume: $1\ \text{fl oz} = 2\ \text{tbsp}$. So multiply that conversion by 3. Steps: Use the conversion: $$1\ \text{fl oz} = 2\ \text{tbsp}$$ Multiply by 3: $$3\ \text{fl oz} = 3 \times 2\ \text{tbsp} = 6\ \text{tbsp}$$ Therefore, 3 ounces equals 6 tablespoons.

Q: How many pounds is 62 kg?

The correct answer is: $136.6866025546241\text{ lb}$ Explanation We use the conversion factor $1\text{ kg} = 2.2046226218487757\text{ lb}$. Steps: Multiply the mass by the conversion factor: $$62\text{ kg}\times 2.2046226218487757\frac{\text{lb}}{\text{kg}} = 136.6866025546241\text{ lb}$$ Round as desired: to two decimal places $$\approx 136.69\text{ lb}$$ (to one decimal place $$\approx 136.7\text{ lb}$$). Therefore, 62 kg ≈ 136.69 lb (exactly about 136.6866 lb).

Q: Convert 165 pounds into kilograms.

The correct answer is: $74.84274105\text{ kg}$ Explanation We convert pounds to kilograms using the exact conversion factor $1\text{ lb} = 0.45359237\text{ kg}$. Multiplying 165 pounds by that factor gives the mass in kilograms. Steps: Use the conversion factor: $1\text{ lb} = 0.45359237\text{ kg}$. Multiply: $$165\times 0.45359237 = 74.84274105$$ Attach the unit: $74.84274105\text{ kg}$. Therefore, 165 pounds equals $74.84274105\text{ kg}$ (commonly rounded to $74.84\text{ kg}$).

Q: What is 300 milliliters in fluid ounces?

The correct answer is: $10.14\text{ US fl oz (approx.)}$ Explanation We convert using the US fluid ounce definition: $1\text{ US fl oz} = 29.5735295625\text{ mL}$, so multiply mL by the factor $1/29.5735295625 \approx 0.0338140226$. Steps: Multiply: $$300\text{ mL}\times 0.0338140226\frac{\text{US fl oz}}{\text{mL}}=10.14420678\text{ US fl oz}$$ Round to a sensible precision: $$\approx10.14\text{ US fl oz}$$ If you meant imperial (UK) fluid ounces instead: $$300\text{ mL}\div28.4130625\frac{\text{mL}}{\text{imp fl oz}}\approx10.56\text{ imp fl oz}$$

Q: What is 103 kg in lbs?

The correct answer is: $227.08\text{ lb}$ Explanation To convert kilograms to pounds multiply by the conversion factor $2.2046226218\text{ lb/kg}$. For 103 kg, multiply 103 by that factor and round to a convenient number of decimal places. Steps: Conversion factor: $1\text{ kg} = 2.2046226218\text{ lb}$ Multiply: $103\text{ kg}\times 2.2046226218\text{ lb/kg} = 227.0761300454\text{ lb}$ Round: $227.0761300454\text{ lb}\approx 227.08\text{ lb}$ Therefore, $103\text{ kg}\approx 227.08\text{ lb}$.

Question

Algebra Equations | Formula, Types & Examples

Accepted Answer

Here’s a complete guide to algebraic equations: definitions, common types, key formulas, and worked examples.

Definition
An algebraic equation is a statement that two algebraic expressions are equal, typically containing one or more variables (unknowns). Solving an equation means finding all values of the variable(s) that make the equality true.

Key formulas and facts

Linear equation (one variable): $ax+b=0 \Rightarrow x=-\frac{b}{a}$
Quadratic standard form: $ax^2+bx+c=0$
Quadratic formula: $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Discriminant: $\Delta=b^2-4ac$ (if $\Delta>0$ two real roots; $\Delta=0$ one real root; $\Delta<0$ two complex roots)
Sum/difference of squares: $a^2-b^2=(a-b)(a+b)$
Sum/difference of cubes: $a^3\pm b^3=(a\pm b)(a^2\mp ab+b^2)$
Completing the square: $ax^2+bx+c=a\Big(x+\frac{b}{2a}\Big)^2-\frac{b^2-4ac}{4a}$
For exponential $a^{x}=b \Rightarrow x=\log_a b = \frac{\ln b}{\ln a}$
Log rules: $\log(AB)=\log A+\log B$, $\log\left(\frac{A}{B}ight)=\log A-\log B$, $\log(A^k)=k\log A$

Common types & how to solve them (with examples)

1) Linear equations
Example: Solve $3x-7=8$  Steps:

Add 7: $3x=15$  
Divide by 3: $x=5$

2) Quadratic equations
Example A (factorable): Solve $x^2-3x-4=0$  Steps:

Factor: $(x-4)(x+1)=0$  
Solutions: $x=4,\ -1$

Example B (use quadratic formula): Solve $2x^2-4x-3=0$  Steps:

Apply formula: $$x=\frac{-(-4)\pm\sqrt{(-4)^2-4\cdot2\cdot(-3)}}{2\cdot2}=\frac{4\pm\sqrt{16+24}}{4}=\frac{4\pm\sqrt{40}}{4}$$
Simplify: $x=\frac{4\pm2\sqrt{10}}{4}=\frac{2\pm\sqrt{10}}{2}$

3) Systems of linear equations (two variables)
Example: Solve\begin{cases}2x+3y=6\x-y=1\end{cases}Steps:

From second: $x=y+1$. Substitute into first: $2(y+1)+3y=6\Rightarrow5y+2=6\Rightarrow y=\frac{4}{5}$  
Then $x=y+1=\frac{4}{5}+1=\frac{9}{5}$

4) Rational equations
Example: Solve $\frac{1}{x}+\frac{2}{x+1}=3$ (note: $x
eq0,-1$)  Steps:

Multiply both sides by $x(x+1)$: $(x+1)+2x=3x(x+1)$  
Simplify: $3x+1=3x^2+3x\Rightarrow3x^2-1=0$  
Solve: $x^2=\frac{1}{3}\Rightarrow x=\pm\frac{1}{\sqrt{3}}$ (both valid if not excluded by domain)

5) Radical equations
Example: Solve $\sqrt{2x+1}=x-1$ (domain: $x-1\ge0\Rightarrow x\ge1$)  Steps:

Square both sides: $2x+1=(x-1)^2=x^2-2x+1$  
Rearrange: $0=x^2-4x\Rightarrow x(x-4)=0$  
Candidates: $x=0$ or $x=4$. Check domain and original: $x=0$ invalid (domain), $x=4$ valid.

Always check for extraneous roots after squaring.

6) Exponential and logarithmic equations
Example: Solve $2^{x+1}=8$  Steps:

$8=2^3$, so $2^{x+1}=2^3\Rightarrow x+1=3\Rightarrow x=2$

Example: Solve $5^x=20$  Steps:

Take logs: $x=\frac{\ln 20}{\ln 5}$

7) Absolute value equations
Example: $|2x-3|=5$  Steps:

Solve $2x-3=5 \Rightarrow x=4$ and $2x-3=-5\Rightarrow x=-1$

Tips and best practices

Always state domain restrictions (no division by zero, radicand nonnegative for real square roots).
Check solutions in the original equation especially for rational and radical cases.
Use factoring, completing the square, or the quadratic formula depending on the form.
For systems, use substitution, elimination, or matrices depending on size/complexity.

If you want, I can provide more worked examples for any specific type (e.g., cubic, inequalities, systems with 3 variables, or word-problem modeling). Which type would you like next?