Grade 10 Mathematics Informal Test \# 6 Topic: Equations And Inequalities Question 1 1.1. Solve for \( x \) : 1.1.1. \( 15 x^{2}-8=14 x \) [4] 1.1.2. \( \quad 5^{x}=\frac{1}{125} \) 1.2. Mary gave one third of her money to Nathan and one fifth to Elias. Elias received R30 less than Nathan. How much money did Mary have originally? [4] Question 2 2.1. Solve for \( x \) : 2.1.1. \( \sqrt[x]{16}=2 \) 2.1.2. \( \frac{x}{x-3}-\frac{x}{2-x}=\frac{2 x^{2}-25}{x^{2}-5 x+6} \) 2.2. Solve for \( x \) without the use of a calculator: 2.1.1. \( x^{3}=9 x \) [5]

Solve the equation by following steps: - step0: Solve for \(x\): \(5^{x}=\frac{1}{125}\) - step1: Rewrite in exponential form: \(5^{x}=5^{-3}\) - step2: Set the exponents equal: \(x=-3\) Solve the equation \( x^{3}=9 x \). Solve the equation by following steps: - step0: Solve for \(x\): \(x^{3}=9x\) - step1: Move the expression to the left side: \(x^{3}-9x=0\) - step2: Factor the expression: \(x\left(x^{2}-9\right)=0\) - step3: Separate into possible cases: \(\begin{align}&x=0\\&x^{2}-9=0\end{align}\) - step4: Solve the equation: \(\begin{align}&x=0\\&x=3\\&x=-3\end{align}\) - step5: Rewrite: \(x_{1}=-3,x_{2}=0,x_{3}=3\) Solve the system of equations \( \frac{1}{3}m=\frac{1}{5}m+30; \frac{1}{3}m=\frac{1}{5}m+30 \). Solve the system of equations by following steps: - step0: Solution: \(\left\{ \begin{array}{l}\frac{1}{3}m=\frac{1}{5}m+30\\\frac{1}{3}m=\frac{1}{5}m+30\end{array}\right.\) - step1: Calculate: \(\left\{ \begin{array}{l}m=225\\m=225\end{array}\right.\) - step2: Find the intersection: \(m=225\) - step3: Check the solution: \(m=225\) Solve the equation \( 15 x^{2}-8=14 x \). Solve the quadratic equation by following steps: - step0: Solve by factoring: \(15x^{2}-8=14x\) - step1: Move the expression to the left side: \(15x^{2}-8-14x=0\) - step2: Factor the expression: \(\left(3x-4\right)\left(5x+2\right)=0\) - step3: Separate into possible cases: \(\begin{align}&3x-4=0\\&5x+2=0\end{align}\) - step4: Solve the equation: \(\begin{align}&x=\frac{4}{3}\\&x=-\frac{2}{5}\end{align}\) - step5: Rewrite: \(x_{1}=-\frac{2}{5},x_{2}=\frac{4}{3}\) Solve the equation \( \frac{x}{x-3}-\frac{x}{2-x}=\frac{2 x^{2}-25}{x^{2}-5 x+6} \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{x}{x-3}-\frac{x}{2-x}=\frac{2x^{2}-25}{x^{2}-5x+6}\) - step1: Find the domain: \(\frac{x}{x-3}-\frac{x}{2-x}=\frac{2x^{2}-25}{x^{2}-5x+6},x \in \left(-\infty,2\right)\cup \left(2,3\right)\cup \left(3,+\infty\right)\) - step2: Multiply both sides of the equation by LCD: \(\left(\frac{x}{x-3}-\frac{x}{2-x}\right)\left(-x+2\right)\left(x-3\right)=\frac{2x^{2}-25}{x^{2}-5x+6}\times \left(-x+2\right)\left(x-3\right)\) - step3: Simplify the equation: \(5x-2x^{2}=-2x^{2}+25\) - step4: Cancel equal terms: \(5x=25\) - step5: Divide both sides: \(\frac{5x}{5}=\frac{25}{5}\) - step6: Divide the numbers: \(x=5\) - step7: Check if the solution is in the defined range: \(x=5,x \in \left(-\infty,2\right)\cup \left(2,3\right)\cup \left(3,+\infty\right)\) - step8: Find the intersection: \(x=5\) Let's solve the problems step by step. ### Question 1 #### 1.1. Solve for \( x \): **1.1.1.** \( 15 x^{2}-8=14 x \) To solve this equation, we rearrange it into standard quadratic form: \[ 15x^2 - 14x - 8 = 0 \] Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 15 \), \( b = -14 \), and \( c = -8 \): The solutions are: \[ x_{1} = -\frac{2}{5}, \quad x_{2} = \frac{4}{3} \] **1.1.2.** \( 5^{x}=\frac{1}{125} \) We can rewrite \( \frac{1}{125} \) as \( 5^{-3} \): \[ 5^{x} = 5^{-3} \] Thus, we have: \[ x = -3 \] #### 1.2. Mary gave one third of her money to Nathan and one fifth to Elias. Let \( m \) be the total amount of money Mary had. According to the problem: - Nathan received \( \frac{1}{3}m \) - Elias received \( \frac{1}{5}m \) We know that Elias received R30 less than Nathan: \[ \frac{1}{5}m = \frac{1}{3}m - 30 \] To solve for \( m \), we can rearrange and solve the equation: \[ \frac{1}{3}m - \frac{1}{5}m = 30 \] Finding a common denominator (15): \[ \frac{5}{15}m - \frac{3}{15}m = 30 \] This simplifies to: \[ \frac{2}{15}m = 30 \] Multiplying both sides by \( \frac{15}{2} \): \[ m = 225 \] ### Question 2 #### 2.1. Solve for \( x \): **2.1.1.** \( \sqrt[x]{16}=2 \) This can be rewritten as: \[ 16^{\frac{1}{x}} = 2 \] Since \( 16 = 2^4 \): \[ (2^4)^{\frac{1}{x}} = 2 \] This simplifies to: \[ 2^{\frac{4}{x}} = 2^1 \] Thus, we have: \[ \frac{4}{x} = 1 \implies x = 4 \] **2.1.2.** \( \frac{x}{x-3}-\frac{x}{2-x}=\frac{2 x^{2}-25}{x^{2}-5 x+6} \) This equation can be solved, and the solution is: \[ x = 5 \] #### 2.2. Solve for \( x \) without the use of a calculator: **2.2.1.** \( x^{3}=9 x \) Rearranging gives: \[ x^{3} - 9x = 0 \] Factoring out \( x \): \[ x(x^{2} - 9) = 0 \] This gives us: \[ x = 0, \quad x^{2} - 9 = 0 \implies x = 3, \quad x = -3 \] Thus, the solutions are: \[ x_{1} = -3, \quad x_{2} = 0, \quad x_{3} = 3 \] ### Summary of Solutions: 1.1.1: \( x_{1} = -\frac{2}{5}, x_{2} = \frac{4}{3} \) 1.1.2: \( x = -3 \) 1.2: \( m = 225 \) 2.1.1: \( x = 4 \) 2.1.2: \( x = 5 \) 2.2.1: \( x_{1} = -3, x_{2} = 0, x_{3} = 3 \)