🧮 algebra

1. The problem is to find the sum of all natural numbers from 1 to a given number $n$. 2. The formula to find the sum of the first $n$ natural numbers is:

1. **State the problem:** Simplify the expression $(2p^2q)(2pq^2)$. 2. **Recall the multiplication rule for exponents:** When multiplying like bases, add the exponents: $a^m \cdot

1. **State the problem:** Solve for $t$ in the equation $$\frac{7}{12} \times \frac{t}{5} \times \frac{1}{3} = \frac{14}{45}.$$\n\n2. **Write the equation clearly:** $$\frac{7}{12}

1. **Problem:** Simplify the expression $$\frac{1}{35} x n^{9n - 12 \times 27^n + 1}$$. 2. **Formula and rules:**

1. **State the problem:** We need to find the missing number in the equation $$8 \div x = \frac{14}{5}$$ where $x$ is the missing number. 2. **Recall the division rule:** Dividing

1. **State the problem:** Solve for the unknown denominator $x$ in the equation $$4 \div \frac{23}{x} = \frac{20}{23}.$$\n\n2. **Recall the division of fractions rule:** Dividing b

1. **State the problem:** Solve the equation $4x = \sqrt{8x + 3}$ for $x$. 2. **Formula and rules:** To solve equations involving square roots, isolate the square root term and the

1. **State the problem:** Solve the equation $$2x^{\frac{2}{3}} - x^{\frac{1}{3}} = 3$$ for $x$. 2. **Introduce substitution:** Let $y = x^{\frac{1}{3}}$. Then $x^{\frac{2}{3}} = y

1. **State the problem:** Solve the equation $$\frac{2}{3}2x - \frac{1}{3} = x = 3$$ for $x$. 2. **Clarify the equation:** The expression seems ambiguous. Assuming the problem is t

1. **State the problem:** Solve the system of equations: $$\frac{2}{3} = 3x + \frac{1}{3}$$

1. **State the problem:** Solve the quadratic equation $$(5n + 1)^2 + 5(5n + 1) - 6 = 0$$ for $n$. 2. **Rewrite the equation:** Let $x = 5n + 1$. Then the equation becomes:

1. **State the problem:** Solve the equation $$x^{\frac{2}{3}} - 2x^{\frac{1}{3}} - 3 = 0$$ for $x$. 2. **Use substitution:** Let $$y = x^{\frac{1}{3}}$$. Then $$y^2 = x^{\frac{2}{

1. **State the problem:** Solve the quadratic equation $$2x^3 - 2x^1 3 - 8 = 0$$. It appears there might be a formatting issue, so we interpret the equation as $$2x^3 - 2x - 8 = 0$

1. **Problem statement:** Find the value of $k$ for which the quadratic equation $$(2k - 1)x^2 + 4x + k = 0$$ has: A. Two real roots

1. **State the problem:** Solve the quartic equation $$z^4 - 20z^2 + 64 = 0$$ for $z$. 2. **Rewrite the equation:** Let $u = z^2$. Then the equation becomes a quadratic in $u$:

1. **State the problem:** Solve the equation $$9x^4 + 26x^2 = 3$$ for $x$. 2. **Rewrite the equation:** Move all terms to one side to set the equation to zero:

1. **State the problem:** Solve the equation $$p^4 - 625 = 0$$ for $p$. 2. **Rewrite the equation:** Recognize that $625 = 5^4$, so the equation becomes $$p^4 - 5^4 = 0$$.

1. **State the problem:** Solve the equation $$\frac{2}{x^2 - 4x + 3} = \frac{2x}{x - 1} - \frac{x}{x - 3}$$ for $x$. 2. **Factor the quadratic denominator:** Note that $$x^2 - 4x

1. **State the problem:** Simplify the expression $$\frac{7x^2 - 4x + 3}{2x} \times \frac{x}{x - 3}$$ and solve for $$x$$. 2. **Rewrite the expression:**

1. **State the problem:** Solve the equation $$\frac{8}{x} + \frac{1}{x+2} = -2$$ for $x$. 2. **Find a common denominator:** The denominators are $x$ and $x+2$. The common denomina

1. **State the problem:** Solve the equation $$\frac{2}{x} + \frac{3}{x-3} = 1$$ for $x$. 2. **Find a common denominator:** The denominators are $x$ and $x-3$. The common denominat