Math Problem Statement
) 2x + 8y = 20 2) x = 5 y = 2 2x + y = 10
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5x - 2y = 3 4) 2y + x = -15 y = 2x x = 3y
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4x + 7y = 19 6) y = 6x + 11 y = x + 9 2y - 4x = 14
Solution
Let's solve each system of equations individually.
1) System:
2x + 8y &= 20 \\ x &= 5 \\ y &= 2 \\ \end{aligned}$$ Substituting $$x = 5$$ and $$y = 2$$ into the first equation: $$2(5) + 8(2) = 20 \\ 10 + 16 = 26 \quad (\text{False})$$ So, this system has no solution as the first equation does not hold with the given values of $$x$$ and $$y$$. --- ### 2) System: $$\begin{aligned} x &= 5 \\ 2x + y &= 10 \end{aligned}$$ Substitute $$x = 5$$ into the second equation: $$2(5) + y = 10 \\ 10 + y = 10 \\ y = 0$$ So, the solution for this system is $$x = 5$$ and $$y = 0$$. --- ### 3) System: $$\begin{aligned} 5x - 2y &= 3 \\ y &= 2x \end{aligned}$$ Substitute $$y = 2x$$ into the first equation: $$5x - 2(2x) = 3 \\ 5x - 4x = 3 \\ x = 3$$ Substitute $$x = 3$$ back into $$y = 2x$$: $$y = 2(3) = 6$$ So, the solution for this system is $$x = 3$$ and $$y = 6$$. --- ### 4) System: $$\begin{aligned} 2y + x &= -15 \\ x &= 3y \end{aligned}$$ Substitute $$x = 3y$$ into the first equation: $$2y + 3y = -15 \\ 5y = -15 \\ y = -3$$ Substitute $$y = -3$$ back into $$x = 3y$$: $$x = 3(-3) = -9$$ So, the solution for this system is $$x = -9$$ and $$y = -3$$. --- ### 5) System: $$\begin{aligned} 4x + 7y &= 19 \\ y &= x + 9 \end{aligned}$$ Substitute $$y = x + 9$$ into the first equation: $$4x + 7(x + 9) = 19 \\ 4x + 7x + 63 = 19 \\ 11x + 63 = 19 \\ 11x = 19 - 63 \\ 11x = -44 \\ x = -4$$ Substitute $$x = -4$$ back into $$y = x + 9$$: $$y = -4 + 9 = 5$$ So, the solution for this system is $$x = -4$$ and $$y = 5$$. --- ### 6) System: $$\begin{aligned} y &= 6x + 11 \\ 2y - 4x &= 14 \end{aligned}$$ Substitute $$y = 6x + 11$$ into the second equation: $$2(6x + 11) - 4x = 14 \\ 12x + 22 - 4x = 14 \\ 8x + 22 = 14 \\ 8x = 14 - 22 \\ 8x = -8 \\ x = -1$$ Substitute $$x = -1$$ back into $$y = 6x + 11$$: $$y = 6(-1) + 11 = -6 + 11 = 5$$ So, the solution for this system is $$x = -1$$ and $$y = 5$$. --- ### Summary of Solutions: 1. No solution. 2. $$x = 5$$, $$y = 0$$ 3. $$x = 3$$, $$y = 6$$ 4. $$x = -9$$, $$y = -3$$ 5. $$x = -4$$, $$y = 5$$ 6. $$x = -1$$, $$y = 5$$ --- Would you like more details on any of these solutions, or have any questions? Here are 5 related questions to expand on this topic: 1. How do you solve systems of equations graphically? 2. What are the different methods for solving systems of equations? 3. Can a system of linear equations have infinite solutions? 4. How can substitution and elimination methods be compared? 5. What does it mean for a system of equations to be consistent or inconsistent? **Tip**: Always check your solutions by substituting them back into the original equations to verify their correctness!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Algebra
Linear Equations
Substitution Method
Elimination Method
Formulas
Substitution: Replace one variable with an equivalent expression from another equation
Elimination: Add or subtract equations to eliminate a variable
Theorems
System of Linear Equations
Consistency and Inconsistency in Equations
Suitable Grade Level
Grades 9-12