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Sagot :
To determine which systems of equations have no solution, we will analyze each system step by step to check for consistency. A system of linear equations has no solution if the equations describe parallel lines that never intersect.
### System 1
[tex]\[ \begin{aligned} x + 4y &= 23 \quad \text{(1)} \\ -3x &= 12y + 1 \quad \text{(2)} \end{aligned} \][/tex]
First, rewrite equation (2) to make it similar to equation (1):
[tex]\[ -3x - 12y = 1 \][/tex]
Now, let's put both equations in standard form:
[tex]\[ \begin{aligned} x + 4y &= 23 \\ -3x - 12y &= 1 \end{aligned} \][/tex]
Multiply equation (1) by 3:
[tex]\[ 3(x + 4y) = 3 \cdot 23 \implies 3x + 12y = 69 \][/tex]
Now we have:
[tex]\[ \begin{aligned} 3x + 12y &= 69 \\ -3x - 12y &= 1 \end{aligned} \][/tex]
Add these two equations:
[tex]\[ (3x - 3x) + (12y - 12y) = 69 + 1 \implies 0 = 70 \][/tex]
This is a contradiction. Therefore, System 1 has no solution.
### System 2
[tex]\[ \begin{aligned} 2x + 4y &= 22 \quad \text{(1)} \\ -x &= 2y - 11 \quad \text{(2)} \end{aligned} \][/tex]
Rewrite equation (2) to standard form:
[tex]\[ -x - 2y = -11 \][/tex]
Now, put the equations together:
[tex]\[ \begin{aligned} 2x + 4y &= 22 \\ -x - 2y = -11 \end{aligned} \][/tex]
Multiply equation (2) by 2:
[tex]\[ -2x - 4y = -22 \][/tex]
Now we have:
[tex]\[ \begin{aligned} 2x + 4y &= 22 \\ -2x - 4y &= -22 \end{aligned} \][/tex]
Add these two equations:
[tex]\[ (2x - 2x) + (4y - 4y) = 22 - 22 \implies 0 = 0 \][/tex]
This is an identity, meaning the system has infinitely many solutions. Therefore, System 2 has solutions.
### System 3
[tex]\[ \begin{aligned} 2x + y &= 15 \quad \text{(1)} \\ x &= 15 - 2y \quad \text{(2)} \end{aligned} \][/tex]
Rewrite equation (2) in a different form:
[tex]\[ x + 2y = 15 \][/tex]
We now see that both equations are actually the same:
[tex]\[ 2x + y = 15 \][/tex]
Since both equations describe the same line, the system has infinitely many solutions. Therefore, System 3 has solutions.
### System 4
[tex]\[ \begin{aligned} 2x + y &= 17 \quad \text{(1)} \\ -4x &= 2y - 34 \quad \text{(2)} \end{aligned} \][/tex]
Rewrite equation (2) in standard form:
[tex]\[ -4x - 2y = -34 \][/tex]
Divide the entire equation by -2:
[tex]\[ 2x + y = 17 \][/tex]
We now have:
[tex]\[ 2x + y = 17 \][/tex]
Both equations are identical, meaning the system has infinitely many solutions. Therefore, System 4 has solutions.
### System 5
[tex]\[ \begin{aligned} 3y &= 10 - x \quad \text{(1)} \\ 2x + 6y &= 7 \quad \text{(2)} \end{aligned} \][/tex]
Rewrite equation (1) to standard form:
[tex]\[ x + 3y = 10 \implies x = 10 - 3y \][/tex]
Substitute [tex]\(x = 10 - 3y\)[/tex] in equation (2):
[tex]\[ 2(10 - 3y) + 6y = 7 \implies 20 - 6y + 6y = 7 \implies 20 = 7 \][/tex]
This is a contradiction. The system has no solution. Therefore, System 5 has no solution.
### System 6
[tex]\[ \begin{aligned} y &= 13 - 2x \quad \text{(1)} \\ 4x - y &= -1 \quad \text{(2)} \end{aligned} \][/tex]
Substitute [tex]\(y = 13 - 2x\)[/tex] into equation (2):
[tex]\[ 4x - (13 - 2x) = -1 \implies 4x - 13 + 2x = -1 \implies 6x - 13 = -1 \implies 6x = 12 \implies x = 2 \][/tex]
Substitute [tex]\(x = 2\)[/tex] into [tex]\(y = 13 - 2x\)[/tex]:
[tex]\[ y = 13 - 2(2) = 13 - 4 = 9 \][/tex]
The solution [tex]\( (x, y) = (2, 9) \)[/tex] satisfies both equations. Therefore, System 6 has a solution.
### Conclusion
The systems of equations with no solutions are:
- System 1
- System 5
### System 1
[tex]\[ \begin{aligned} x + 4y &= 23 \quad \text{(1)} \\ -3x &= 12y + 1 \quad \text{(2)} \end{aligned} \][/tex]
First, rewrite equation (2) to make it similar to equation (1):
[tex]\[ -3x - 12y = 1 \][/tex]
Now, let's put both equations in standard form:
[tex]\[ \begin{aligned} x + 4y &= 23 \\ -3x - 12y &= 1 \end{aligned} \][/tex]
Multiply equation (1) by 3:
[tex]\[ 3(x + 4y) = 3 \cdot 23 \implies 3x + 12y = 69 \][/tex]
Now we have:
[tex]\[ \begin{aligned} 3x + 12y &= 69 \\ -3x - 12y &= 1 \end{aligned} \][/tex]
Add these two equations:
[tex]\[ (3x - 3x) + (12y - 12y) = 69 + 1 \implies 0 = 70 \][/tex]
This is a contradiction. Therefore, System 1 has no solution.
### System 2
[tex]\[ \begin{aligned} 2x + 4y &= 22 \quad \text{(1)} \\ -x &= 2y - 11 \quad \text{(2)} \end{aligned} \][/tex]
Rewrite equation (2) to standard form:
[tex]\[ -x - 2y = -11 \][/tex]
Now, put the equations together:
[tex]\[ \begin{aligned} 2x + 4y &= 22 \\ -x - 2y = -11 \end{aligned} \][/tex]
Multiply equation (2) by 2:
[tex]\[ -2x - 4y = -22 \][/tex]
Now we have:
[tex]\[ \begin{aligned} 2x + 4y &= 22 \\ -2x - 4y &= -22 \end{aligned} \][/tex]
Add these two equations:
[tex]\[ (2x - 2x) + (4y - 4y) = 22 - 22 \implies 0 = 0 \][/tex]
This is an identity, meaning the system has infinitely many solutions. Therefore, System 2 has solutions.
### System 3
[tex]\[ \begin{aligned} 2x + y &= 15 \quad \text{(1)} \\ x &= 15 - 2y \quad \text{(2)} \end{aligned} \][/tex]
Rewrite equation (2) in a different form:
[tex]\[ x + 2y = 15 \][/tex]
We now see that both equations are actually the same:
[tex]\[ 2x + y = 15 \][/tex]
Since both equations describe the same line, the system has infinitely many solutions. Therefore, System 3 has solutions.
### System 4
[tex]\[ \begin{aligned} 2x + y &= 17 \quad \text{(1)} \\ -4x &= 2y - 34 \quad \text{(2)} \end{aligned} \][/tex]
Rewrite equation (2) in standard form:
[tex]\[ -4x - 2y = -34 \][/tex]
Divide the entire equation by -2:
[tex]\[ 2x + y = 17 \][/tex]
We now have:
[tex]\[ 2x + y = 17 \][/tex]
Both equations are identical, meaning the system has infinitely many solutions. Therefore, System 4 has solutions.
### System 5
[tex]\[ \begin{aligned} 3y &= 10 - x \quad \text{(1)} \\ 2x + 6y &= 7 \quad \text{(2)} \end{aligned} \][/tex]
Rewrite equation (1) to standard form:
[tex]\[ x + 3y = 10 \implies x = 10 - 3y \][/tex]
Substitute [tex]\(x = 10 - 3y\)[/tex] in equation (2):
[tex]\[ 2(10 - 3y) + 6y = 7 \implies 20 - 6y + 6y = 7 \implies 20 = 7 \][/tex]
This is a contradiction. The system has no solution. Therefore, System 5 has no solution.
### System 6
[tex]\[ \begin{aligned} y &= 13 - 2x \quad \text{(1)} \\ 4x - y &= -1 \quad \text{(2)} \end{aligned} \][/tex]
Substitute [tex]\(y = 13 - 2x\)[/tex] into equation (2):
[tex]\[ 4x - (13 - 2x) = -1 \implies 4x - 13 + 2x = -1 \implies 6x - 13 = -1 \implies 6x = 12 \implies x = 2 \][/tex]
Substitute [tex]\(x = 2\)[/tex] into [tex]\(y = 13 - 2x\)[/tex]:
[tex]\[ y = 13 - 2(2) = 13 - 4 = 9 \][/tex]
The solution [tex]\( (x, y) = (2, 9) \)[/tex] satisfies both equations. Therefore, System 6 has a solution.
### Conclusion
The systems of equations with no solutions are:
- System 1
- System 5
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