Identify any two equations that form a system with one solution (different slopes)

Question

? I Hint: Make sure all equations are in Y=mx + b format first. J
1. y = 57 -4
2. y = -2x + 3
3. y = 3x +1
4. 2y = 6x + 2
5. у - 5х = 2
6. у -4х +1 = 0

Answer 1

To identify two equations that form a system with one solution (meaning the lines intersect at exactly one point), we need to ensure they have different slopes.

First, let's rewrite all equations in the format \(y = mx + b\):

1. \(y = 57 - 4\) simplifies to \(y = 53\). (This is a horizontal line with slope = 0.)
2. \(y = -2x + 3\) is already in the correct form. (Slope = -2.)
3. \(y = 3x + 1\) is already in the correct form. (Slope = 3.)
4. Starting from \(2y = 6x + 2\), divide by 2: \(y = 3x + 1\) (This has a slope of 3).
5. From \(y - 5x = 2\), we rearrange to get \(y = 5x + 2\) (Slope = 5).
6. From \(y - 4x + 1 = 0\), rearranging gives \(y = 4x - 1\) (Slope = 4).

Now we have the slopes for each equation:

1. \(y = 53\) (slope = 0)
2. \(y = -2x + 3\) (slope = -2)
3. \(y = 3x + 1\) (slope = 3)
4. \(y = 3x + 1\) (slope = 3)
5. \(y = 5x + 2\) (slope = 5)
6. \(y = 4x - 1\) (slope = 4)

For the system to have one solution, we need pairs of equations with different slopes.

Let's select:

• Equation 2: \(y = -2x + 3\) (slope = -2)
• Equation 5: \(y = 5x + 2\) (slope = 5)

These two equations have different slopes and will intersect at exactly one point, thus forming a system with one solution.

So, the two equations are:

1. \(y = -2x + 3\)
2. \(y = 5x + 2\)