Your SlideShare is downloading. ×
Quad-Linear Systems
Quad-Linear Systems
Quad-Linear Systems
Quad-Linear Systems
Quad-Linear Systems
Quad-Linear Systems
Quad-Linear Systems
Quad-Linear Systems
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Text the download link to your phone
Standard text messaging rates apply

Quad-Linear Systems

1,728

Published on

0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total Views
1,728
On Slideshare
0
From Embeds
0
Number of Embeds
1
Actions
Shares
0
Downloads
1
Comments
0
Likes
0
Embeds 0
No embeds

Report content
Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
No notes for slide

Transcript

  • 1. Systems of quadratic equations often have more than one solution. You can have no solutions and the graph would look like this: No points of intersection
  • 2. You can have one solution and the graph might look something like this: One intersection point.
  • 3. You could have two solutions and the graph might look something like this: Two points of intersection
  • 4. Solving a System of Non-Linear Equations The solution(s) are all (x, y) point(s) that make both equations true. Graph and find the intersection point(s) on the graphing calculator. Substitute the value of one variable from one equation into the other equation and solve for that variable, then the other. Eliminate one variable and solve for that variable, then the other.   Online extra help: Systems
  • 5. EXAMPLES: Solve each system of equations 1. Solve by graphing or substitution. *To solve by graphing, graph both equations on the same graph and find the intersection points.
  • 6. Substitution: (You can check using a calculator, but elimination is not a good idea) 2x + 1 = x² - 2 Substitute y = 2x + 1 in for y in the second equation 0 = x² - 2x – 3 Write in standard form 0 = (x – 3) (x + 1) Factor x - 3 = 0 or x + 1=0 Set each factor = 0 x = 3 x = -1 Solve each equation But these are not points of intersection – we need to find the y-coordinate. Substitute the x value in either equation to find the y. Solution 1: Solution 2: x = 3, use y = 2x + 1 x = -1, use y = 2x + 1 y = 2(3) + 1 = 7 y = 2(-1) + 1 = -1 (3, 7) is a solution (-1, -1) is a solution
  • 7. 2. Solve by graphing, elimination, or substitution. Elimination: (You can check using either of the other methods) 0 = 4x - 8 Subtract equation 1 from equation 2 and eliminate both y & x 2 x = 2 Solve for x But this is not a point, so find the y-coordinate by substitution. Solution: x = 2, use y = x² + x – 1 y = 2² + 2 – 1 = 5 (2, 5) is the only point where the quadratics intersect **You can check this point by graphing both equations on the same graph. (see next slide)

×