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The point (5,-3) is a solution to which system of equations? (1 point)

A. y=x-8 2x+y=7
B. y=2/3x+6 3y+6x-18=0
C. y=x+2 y=x+5
D. y=-1/2x+6 y=3x-1

3 answers

To determine which system of equations has the point (5, -3) as a solution, we need to substitute \( x = 5 \) and \( y = -3 \) into each equation in the systems provided and see which system holds true for both equations.

Option A:

  1. Equation 1: \( y = x - 8 \)
    Substitute \( x = 5 \):
    \( y = 5 - 8 = -3 \) (True)

  2. Equation 2: \( 2x + y = 7 \)
    Substitute \( x = 5 \) and \( y = -3 \):
    \( 2(5) + (-3) = 10 - 3 = 7 \) (True)

Both equations in Option A are satisfied.

Option B:

  1. Equation 1: \( y = \frac{2}{3}x + 6 \)
    Substitute \( x = 5 \):
    \( y = \frac{2}{3}(5) + 6 = \frac{10}{3} + 6 = \frac{10}{3} + \frac{18}{3} = \frac{28}{3} \) (Not true, since -3 is not equal to \(\frac{28}{3}\))

Since the first equation is not satisfied, we can discard this option.

Option C:

  1. Equation 1: \( y = x + 2 \)
    Substitute \( x = 5 \):
    \( y = 5 + 2 = 7 \) (Not true since -3 is not equal to 7)

Again, since the first equation is not satisfied, we can discard this option.

Option D:

  1. Equation 1: \( y = -\frac{1}{2}x + 6 \)
    Substitute \( x = 5 \):
    \( y = -\frac{1}{2}(5) + 6 = -\frac{5}{2} + 6 = -\frac{5}{2} + \frac{12}{2} = \frac{7}{2} \) (Not true since -3 is not equal to \(\frac{7}{2}\))

Since the first equation is not satisfied, we can discard this option.

Conclusion:

After checking all options, the only system that satisfies the point (5, -3) is:

Option A: \( y = x - 8 \) and \( 2x + y = 7 \)

How many solutions does the system of equations have?
y=-2x+12
4y+x-20=0 (1 point)
A. Infinite solutions
B. One solution
C. Two solutions
D. No solution

To determine how many solutions the system of equations has, we can analyze the two equations given:

  1. Equation 1: \( y = -2x + 12 \)

  2. Equation 2: \( 4y + x - 20 = 0 \)

First, let's rewrite Equation 2 in slope-intercept form (i.e., \(y = mx + b\)):

Starting with: \[ 4y + x - 20 = 0 \]

We can isolate \(y\): \[ 4y = -x + 20 \]

Now, divide by 4: \[ y = -\frac{1}{4}x + 5 \]

Now, we have both equations in the form of \(y = mx + b\):

  1. \( y = -2x + 12 \) (slope = -2)
  2. \( y = -\frac{1}{4}x + 5 \) (slope = -1/4)

Analysis

  • The slopes of the two lines are different (-2 and -1/4).
  • Since the slopes are not equal, the lines will intersect at one point.

Conclusion

The system of equations has one solution.

The correct answer is: B. One solution.