Question
Part A: System of Linear Equations with No Solution
Consider the following system of linear equations:
2x+3y=6
4x+6y=12
To ensure that this system has no solution, I made sure that the second equation is a multiple of the first equation. Specifically, the second equation is simply the first equation multiplied by 2. This means that both equations represent the same line in the coordinate plane, which implies that they will overlap entirely. Because they are effectively the same line, they do not intersect at any distinct point and thus have no unique solution.
Part B: System of Linear Equations with One Solution
Let's modify one of the original equations to create a new system that has a unique solution. We will keep the first equation and change the second:
2x+3y=6 (This equation remains unchanged)
2x+y=4 (This is a new equation)
Now we can use substitution to solve for x and y.
Step 1: Solve the second equation for y:
y=4−2x
Step 2: Substitute y in the first equation:
2x+3(4−2x)=6
Step 3: Distribute and simplify:
2x+12−6x=6
Combine like terms:
−4x+12=6
Step 4: Solve for x:
−4x=6−12−4x=−6x=−6−4=32
Step 5: Substitute x back into the equation for y:
y=4−2(32)y=4−3=1
Solution:
The solution to the system is (x,y)=(32,1).
can you make this way less advanced use less words and remove all the fractions
Consider the following system of linear equations:
2x+3y=6
4x+6y=12
To ensure that this system has no solution, I made sure that the second equation is a multiple of the first equation. Specifically, the second equation is simply the first equation multiplied by 2. This means that both equations represent the same line in the coordinate plane, which implies that they will overlap entirely. Because they are effectively the same line, they do not intersect at any distinct point and thus have no unique solution.
Part B: System of Linear Equations with One Solution
Let's modify one of the original equations to create a new system that has a unique solution. We will keep the first equation and change the second:
2x+3y=6 (This equation remains unchanged)
2x+y=4 (This is a new equation)
Now we can use substitution to solve for x and y.
Step 1: Solve the second equation for y:
y=4−2x
Step 2: Substitute y in the first equation:
2x+3(4−2x)=6
Step 3: Distribute and simplify:
2x+12−6x=6
Combine like terms:
−4x+12=6
Step 4: Solve for x:
−4x=6−12−4x=−6x=−6−4=32
Step 5: Substitute x back into the equation for y:
y=4−2(32)y=4−3=1
Solution:
The solution to the system is (x,y)=(32,1).
can you make this way less advanced use less words and remove all the fractions
Answers
GPT-4o mini
**Part A: System of Linear Equations with No Solution**
Consider these equations:
1. \(2x + 3y = 6\)
2. \(4x + 6y = 12\)
The second equation is just the first one multiplied by 2. This means both equations represent the same line. Since they overlap completely, they don’t intersect at any distinct point, so there’s no unique solution.
---
**Part B: System of Linear Equations with One Solution**
Let's change one equation:
1. \(2x + 3y = 6\) (stays the same)
2. \(2x + y = 4\) (new equation)
Now, we will solve for \(x\) and \(y\).
1. From the second equation, solve for \(y\):
\(y = 4 - 2x\)
2. Substitute \(y\) into the first equation:
\(2x + 3(4 - 2x) = 6\)
3. Distribute:
\(2x + 12 - 6x = 6\)
4. Combine like terms:
\(-4x + 12 = 6\)
5. Solve for \(x\):
\(-4x = 6 - 12\)
\(-4x = -6\)
\(x = \frac{6}{4} = 1.5\)
6. Substitute \(x\) back into the equation for \(y\):
\(y = 4 - 2(1.5)\)
\(y = 4 - 3 = 1\)
**Solution:**
The solution to the system is \((x, y) = (1.5, 1)\).
Consider these equations:
1. \(2x + 3y = 6\)
2. \(4x + 6y = 12\)
The second equation is just the first one multiplied by 2. This means both equations represent the same line. Since they overlap completely, they don’t intersect at any distinct point, so there’s no unique solution.
---
**Part B: System of Linear Equations with One Solution**
Let's change one equation:
1. \(2x + 3y = 6\) (stays the same)
2. \(2x + y = 4\) (new equation)
Now, we will solve for \(x\) and \(y\).
1. From the second equation, solve for \(y\):
\(y = 4 - 2x\)
2. Substitute \(y\) into the first equation:
\(2x + 3(4 - 2x) = 6\)
3. Distribute:
\(2x + 12 - 6x = 6\)
4. Combine like terms:
\(-4x + 12 = 6\)
5. Solve for \(x\):
\(-4x = 6 - 12\)
\(-4x = -6\)
\(x = \frac{6}{4} = 1.5\)
6. Substitute \(x\) back into the equation for \(y\):
\(y = 4 - 2(1.5)\)
\(y = 4 - 3 = 1\)
**Solution:**
The solution to the system is \((x, y) = (1.5, 1)\).