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deltamath back to home *** delta 2/2 - systems review due february 5 at…

Question

deltamath back to home *** delta 2/2 - systems review due february 5 at 10:30 pm grade: 92% ✓ substitution (level 1) ✓ substitution (level 1) ✓ substitution (level 1) ? solve linear system graphically (lev. 1) ✓ solve linear system graphically (graphs given) ✓ plotting points y = -2x - 7 y = \\(\frac{1}{2}x + 3\\) plot two lines by clicking the graph. click a line to delete it.

Explanation:

Step1: Find two points for \( y = -2x - 7 \)

  • When \( x = 0 \), \( y = -2(0) - 7 = -7 \), so the point is \( (0, -7) \).
  • When \( x = -1 \), \( y = -2(-1) - 7 = 2 - 7 = -5 \), so the point is \( (-1, -5) \).

Step2: Find two points for \( y = \frac{1}{2}x + 3 \)

  • When \( x = 0 \), \( y = \frac{1}{2}(0) + 3 = 3 \), so the point is \( (0, 3) \).
  • When \( x = 2 \), \( y = \frac{1}{2}(2) + 3 = 1 + 3 = 4 \), so the point is \( (2, 4) \).

Step3: Plot the lines

  • For \( y = -2x - 7 \), plot the points \( (0, -7) \) and \( (-1, -5) \) and draw the line through them.
  • For \( y = \frac{1}{2}x + 3 \), plot the points \( (0, 3) \) and \( (2, 4) \) and draw the line through them.

Step4: Find the intersection (solution)

The two lines intersect at \( x = -4 \), \( y = 1 \) (by solving the system algebraically or visually from the graph: substitute \( y \) from first equation into second: \( -2x - 7 = \frac{1}{2}x + 3 \), multiply by 2: \( -4x - 14 = x + 6 \), \( -5x = 20 \), \( x = -4 \), then \( y = -2(-4) -7 = 8 -7 = 1 \)).

Answer:

The solution to the system is \( x = -4 \), \( y = 1 \) (the intersection point of the two lines \( y = -2x - 7 \) and \( y = \frac{1}{2}x + 3 \) is \( (-4, 1) \)).