Math Problem Statement

Solve the system of equations: x + 2y + 1 = 0 and 2x - 3y - 12 = 0 using substitution or elimination methods.

Solution

The image contains several systems of equations to solve using substitution and elimination methods. Let's work on one of them.

Let's solve problem 1:

1.\quad & x + 2y + 1 = 0 \quad & (i) \\ & 2x - 3y - 12 = 0 \quad & (ii) \end{aligned}$$ **Step 1: Solve one equation for one variable (using substitution).** From equation (i): $$x + 2y + 1 = 0 \quad \Rightarrow \quad x = -2y - 1 \quad \text{(iii)}$$ **Step 2: Substitute this expression for $$x$$ into equation (ii).** Substitute $$x = -2y - 1$$ into equation (ii): $$2(-2y - 1) - 3y - 12 = 0$$ Simplifying: $$-4y - 2 - 3y - 12 = 0$$ $$-7y - 14 = 0 \quad \Rightarrow \quad y = -2$$ **Step 3: Substitute $$y = -2$$ back into equation (iii) to find $$x$$.** $$x = -2(-2) - 1 = 4 - 1 = 3$$ Thus, the solution is $$x = 3$$ and $$y = -2$$. Would you like a solution for any other problem from the image? ### 5 related questions: 1. Can you solve another system of equations from this image using elimination instead? 2. How does substitution compare to elimination in terms of solving speed for larger systems? 3. How do you graphically represent the solution of a system of linear equations? 4. What happens if two lines in a system of equations are parallel? 5. How can we solve a system where both equations are quadratic? ### Tip: Always check your solutions by substituting both variables back into the original equations to confirm the results.

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Math Problem Analysis

Mathematical Concepts

Algebra
Systems of Linear Equations

Formulas

Substitution method
Elimination method

Theorems

System of Linear Equations Theorem

Suitable Grade Level

Grades 8-10