Math Problem Statement

Solve 23,24,26

Solution

Let's solve the given problems from the image.

Problem 23:

Solve the system of equations using any method:

1) & \quad \frac{x}{2} + \frac{y}{3} = \frac{7}{6} \\ & \quad \frac{y}{2} - \frac{x}{3} = \frac{1}{6} \end{aligned}$$ #### Solution: 1. Multiply the first equation by 6 to eliminate the denominators: $$3x + 2y = 7 \quad \text{(Equation 1)}$$ 2. Multiply the second equation by 6 to eliminate the denominators: $$3y - 2x = 1 \quad \text{(Equation 2)}$$ 3. Now, we have the system: $$\begin{aligned} 3x + 2y & = 7 \quad \text{(Equation 1)} \\ -2x + 3y & = 1 \quad \text{(Equation 2)} \end{aligned}$$ 4. Solve this system using either substitution or elimination. I will use the elimination method. Multiply Equation 1 by 3 and Equation 2 by 2 to align the coefficients of $$y$$: $$\begin{aligned} 9x + 6y & = 21 \quad \text{(Equation 3)} \\ -4x + 6y & = 2 \quad \text{(Equation 4)} \end{aligned}$$ 5. Subtract Equation 4 from Equation 3: $$13x = 19 \quad \Rightarrow \quad x = \frac{19}{13}$$ 6. Substitute $$x = \frac{19}{13}$$ back into Equation 1: $$3\left(\frac{19}{13}\right) + 2y = 7 \quad \Rightarrow \quad \frac{57}{13} + 2y = 7$$ Simplify and solve for $$y$$: $$2y = 7 - \frac{57}{13} = \frac{91}{13} - \frac{57}{13} = \frac{34}{13} \quad \Rightarrow \quad y = \frac{34}{26} = \frac{17}{13}$$ Thus, the solution is: $$x = \frac{19}{13}, \quad y = \frac{17}{13}$$ ### Problem 24: Solve the inequality: $$4x - 3y \geq 11, \quad 3x + y = 5$$ #### Solution: 1. Solve the second equation for $$y$$: $$y = 5 - 3x \quad \text{(Substitute this into the first inequality)}$$ 2. Substitute $$y = 5 - 3x$$ into the inequality: $$4x - 3(5 - 3x) \geq 11$$ 3. Simplify and solve: $$4x - 15 + 9x \geq 11 \quad \Rightarrow \quad 13x \geq 26 \quad \Rightarrow \quad x \geq 2$$ Substitute $$x = 2$$ into $$y = 5 - 3x$$: $$y = 5 - 3(2) = -1$$ The solution is $$x \geq 2$$, and the point of intersection $$(x, y) = (2, -1)$$. ### Problem 26: Solve the system of equations using matrices or any method: $$\begin{aligned} 3x + 4y & = 11 \\ 3x + y & = 5 \end{aligned}$$ #### Solution: 1. Subtract the second equation from the first: $$4y - y = 11 - 5 \quad \Rightarrow \quad 3y = 6 \quad \Rightarrow \quad y = 2$$ 2. Substitute $$y = 2$$ back into the second equation: $$3x + 2 = 5 \quad \Rightarrow \quad 3x = 3 \quad \Rightarrow \quad x = 1$$ Thus, the solution is: $$x = 1, \quad y = 2$$ --- Would you like further details on any of these solutions or have any questions? Here are five related questions to consider: 1. How would the solution change in Problem 23 if the equations were modified? 2. Can inequalities like in Problem 24 be graphed, and what do their graphs represent? 3. How can systems of equations be solved using matrix methods? 4. What are the implications of the solution in Problem 26 in a real-world context? 5. How can you check if the solutions to these systems of equations are correct? **Tip:** Always verify your solutions by substituting them back into the original equations to ensure accuracy.

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Math Problem Analysis

Mathematical Concepts

Systems of Equations
Inequalities
Matrix Methods

Formulas

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Theorems

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Suitable Grade Level

High School

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