use elimination to solve the below systems of equations. check your work by substituting the values back into the systems.

1. $\begin{aligned} x - y &= 1 \\ x + y &= 3 \end{aligned}$

2. $\begin{aligned} -x + y &= 1 \\ x + y &= 11 \end{aligned}$

3. $\begin{aligned} x + 4y &= 11 \\ x - 6y &= 11 \end{aligned}$

4. $\begin{aligned} -x + y &= 6 \\ x + 3y &= 18 \end{aligned}$

5. $\begin{aligned} 3x + 4y &= 19 \\ 3x + 6y &= 33 \end{aligned}$

6. $\begin{aligned} x + 4y &= -8 \\ x - 4y &= -8 \end{aligned}$

Question

use elimination to solve the below systems of equations. check your work by substituting the values back into the systems.

1. $\begin{aligned} x - y &= 1 \\ x + y &= 3 \end{aligned}$

2. $\begin{aligned} -x + y &= 1 \\ x + y &= 11 \end{aligned}$

3. $\begin{aligned} x + 4y &= 11 \\ x - 6y &= 11 \end{aligned}$

4. $\begin{aligned} -x + y &= 6 \\ x + 3y &= 18 \end{aligned}$

5. $\begin{aligned} 3x + 4y &= 19 \\ 3x + 6y &= 33 \end{aligned}$

6. $\begin{aligned} x + 4y &= -8 \\ x - 4y &= -8 \end{aligned}$

Accepted Answer

{ "quality": "clear", "question_count": 6, "skills_matched": [ "step_cot", "table_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Linear Equation Solving" ], "new_concepts": [ "Systems of Equations", "Elimination Method" ], "current_concepts": [ "Linear Equation Solving", "Systems of Equations", "Elimination Method" ] } ## Solve Question 1 Using the *Linear Equation Solving* knowledge point $$ \begin{aligned} &\begin{cases} x - y = 1 \ x + y = 3 \end{cases}\ &\text{Add the equations: } 2x = 4 \implies x = 2\ &\text{Substitute } x = 2 \text{ into the second equation: } 2 + y = 3 \implies y = 1\ &\text{Solution: } (2, 1) \end{aligned} $$ ## Solve Question 2 Using the *Linear Equation Solving* knowledge point $$ \begin{aligned} &\begin{cases} -x + y = 1 \ x + y = 11 \end{cases}\ &\text{Add the equations: } 2y = 12 \implies y = 6\ &\text{Substitute } y = 6 \text{ into the second equation: } x + 6 = 11 \implies x = 5\ &\text{Solution: } (5, 6) \end{aligned} $$ ## Solve Question 3 Using the *Linear Equation Solving* knowledge point $$ \begin{aligned} &\begin{cases} x + 4y = 11 \ x - 6y = 11 \end{cases}\ &\text{Subtract the second equation from the first: } 10y = 0 \implies y = 0\ &\text{Substitute } y = 0 \text{ into the first equation: } x + 4(0) = 11 \implies x = 11\ &\text{Solution: } (11, 0) \end{aligned} $$ ## Solve Question 4 Using the *Linear Equation Solving* knowledge point $$ \begin{aligned} &\begin{cases} -x + 3y = 6 \ -x + 3y = 18 \end{cases}\ &\text{Subtract the first equation from the second: } 0 = 12\ &\text{Since } 0 = 12 \text{ is a contradiction, there is no solution.}\ &\text{Solution: No solution} \end{aligned} $$ ## Solve Question 5 Using the *Linear Equation Solving* knowledge point $$ \begin{aligned} &\begin{cases} 3x + 4y = 19 \ 3x + 6y = 33 \end{cases}\ &\text{Subtract the first equation from the second: } 2y = 14 \implies y = 7\ &\text{Substitute } y = 7 \text{ into the first equation: } 3x + 4(7) = 19 \implies 3x + 28 = 19 \implies 3x = -9 \implies x = -3\ &\text{Solution: } (-3, 7) \end{aligned} $$ ## Solve Question 6 Using the *Linear Equation Solving* knowledge point $$ \begin{aligned} &\begin{cases} x + 4y = -8 \ x - 4y = -8 \end{cases}\ &\text{Add the equations: } 2x = -16 \implies x = -8\ &\text{Substitute } x = -8 \text{ into the first equation: } -8 + 4y = -8 \implies 4y = 0 \implies y = 0\ &\text{Solution: } (-8, 0) \end{aligned} $$ | No. | Answer | |---|---| | 1 | $(2, 1)$ | | 2 | $(5, 6)$ | | 3 | $(11, 0)$ | | 4 | No solution | | 5 | $(-3, 7)$ | | 6 | $(-8, 0)$ | { "subject": "Mathematics", "question_type": "Multi-part", "knowledge_point": [ "Mathematics", "Algebra", "Systems of Equations" ] }

No.	Answer
2	\((5, 6)\)
3	\((11, 0)\)
4	No solution
5	\((-3, 7)\)
6	\((-8, 0)\)

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