Math Problem Statement

Which system of equations has the same solution as this one?

8x+5y=-203x–y=5

Solution

To determine which system of equations has the same solution as the given system:

1) \ 8x + 5y &= -20 \\ 2) \ 3x - y &= 5 \end{aligned}$$ We need to find another system of equations that would have the same solution set. Let’s solve this system step-by-step. ### Step 1: Solve one equation for one variable From the second equation $$3x - y = 5$$, solve for $$y$$: $$y = 3x - 5$$ ### Step 2: Substitute into the first equation Substitute $$y = 3x - 5$$ into the first equation $$8x + 5y = -20$$: $$8x + 5(3x - 5) = -20$$ Simplify: $$8x + 15x - 25 = -20$$ $$23x - 25 = -20$$ Solve for $$x$$: $$23x = -20 + 25$$ $$23x = 5$$ $$x = \frac{5}{23}$$ ### Step 3: Solve for $$y$$ Now substitute $$x = \frac{5}{23}$$ back into $$y = 3x - 5$$: $$y = 3\left(\frac{5}{23}\right) - 5 = \frac{15}{23} - 5 = \frac{15}{23} - \frac{115}{23} = \frac{-100}{23}$$ Thus, the solution to the system is $$x = \frac{5}{23}$$ and $$y = \frac{-100}{23}$$. ### Step 4: Identify another system with the same solution Any system that can be algebraically manipulated to yield the same solution will work. For example, multiplying both equations by some constant or adding a multiple of one equation to the other will not change the solution. Here's one such system that will have the same solution: $$\begin{aligned} 1) \ 16x + 10y &= -40 \\ 2) \ 3x - y &= 5 \end{aligned}$$ This new system is derived by multiplying the original first equation by 2, which preserves the same solution. Would you like more details or have any questions? --- Here are 5 related questions to further your understanding: 1. How do you verify that two systems of equations have the same solution? 2. What are the methods to solve a system of linear equations? 3. Can you explain why multiplying an equation by a constant doesn't change its solution? 4. How does substitution work when solving systems of equations? 5. What are other methods, like elimination, to solve systems of equations? **Tip**: When solving systems of equations, substitution is useful when one equation is easily solved for a variable, but elimination is often faster if both equations have similar forms.

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Math Problem Analysis

Mathematical Concepts

Linear Equations
Systems of Equations
Substitution Method

Formulas

Substitution: Solve one equation for one variable, then substitute into the other
Simplification of linear expressions

Theorems

Algebraic Manipulation: Multiplying both sides of an equation by a constant doesn't change its solution
Consistency of Systems: Systems derived from algebraic manipulation of the same equations have the same solutions

Suitable Grade Level

Grades 9-10