need help # Solving Systems of EquationsTwo systems of equations are given below. For each system, choose the best description of its solution. If applicable, give the solution.## System 1:\[-x - 2y = -2\]\[x + 2y = 2\]- ○ The system has no solution.- ⊙ The system has a unique solution: $(x, y) = ( \_ , \_ )$- ○ The system has infinitely many solutions. They must satisfy the following equation: $y = \_$Solution: The indicated answer for this system is that it has no solution.## System 2:\[x + 2y = 6\]\[-x - 2y = 6\]- ⊙ The system has no solution.- ○ The system has a unique solution: $(x, y) = ( \_ , \_ )$- ○ The system has infinitely many solutions. They must satisfy the following equation: $y = \_$Solution: The indicated answer for this system is that it has a unique solution.

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Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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# Solving Systems of Equations

Two systems of equations are given below. For each system, choose the best description of its solution. If applicable, give the solution.

## System 1:

\[
-x - 2y = -2
\]
\[
x + 2y = 2
\]

- ○ The system has no solution.
- ⊙ The system has a unique solution: $(x, y) = ( \_ , \_ )$
- ○ The system has infinitely many solutions. They must satisfy the following equation: $y = \_$

Solution: The indicated answer for this system is that it has no solution.

## System 2:

\[
x + 2y = 6
\]
\[
-x - 2y = 6
\]

- ⊙ The system has no solution.
- ○ The system has a unique solution: $(x, y) = ( \_ , \_ )$
- ○ The system has infinitely many solutions. They must satisfy the following equation: $y = \_$

Solution: The indicated answer for this system is that it has a unique solution.

Expert Solution

Step 1

$(1)$ The given equations are :

$- x - 2 y = - 2 x + 2 y = 2$

To find the solution of the equation ,

The coefficient in equation $1$ are ,

$a_{1} = - 1, b_{1} = - 2, c_{1} = - 2$

The coefficient in equation $2$ are ,

$a_{2} = 1, b_{2} = 2, c_{2} = 2$

Comparing the coefficient of the two equations ,

$\frac{a_{1}}{a_{2}} = \frac{- 1}{1} = - 1 \frac{b_{1}}{b_{2}} = \frac{- 2}{2} = - 1 \frac{c_{1}}{c_{2}} = \frac{- 2}{2} = - 1$

Hence , $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

Thus, the system of equation has the infinitely many solutions.

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