Download presentation

Presentation is loading. Please wait.

Published byJayson Willis Modified over 8 years ago

1
SOLVING SYSTEMS ALGEBRAICALLY SECTION 3-2

2
SOLVING BY SUBSTITUTION 1) 3x + 4y = 12 STEP 1 : SOLVE ONE EQUATION FOR ONE OF THE VARIABLES 2) 2x + y = 10 2x + y = 10 - 2x a) Y = - 2x + 10

3
SUBSTITUTE Step 2: substitute -2x + 10 for y Step 3: Substitute the value of x that you equation 2, and solve for x. found into equation a). Solve for y: 3x + 4y = 12 2x + y = 10 The solution is (5.6, -1.2)

4
NOW YOU TRY ONE 1) X + 3y = 5 2) -2x -4y = - 5

5
SOLVING BY ELIMINATION SOLVE : 4x + 2y = 9 4x + 2y = 9 One equation has 4x - 4x + 3y = 16- 4x + 3y = 16 and the other has -4x Add to eliminate x 4x + 2y = 9 Choose one of the equations Substitute for y Solution is (, )

6
-2x + 8y = -8 ONE EQUATION HAS 8y 5x – 8y = 20 AND THE OTHER HAS – 8y ADD TO ELIMINATE Y Choose one of the original equations Substitute 4 for x Solution (, ) WHAT IS THE SOLUTION FOR -2x + 8y + -8 5x – 8y = 20

7
SOLVING AN EQUIVALENT SYSTEM 2x + 7y = 4 By multiplying the first 3x 3x + 5y = -5equation by 3 and the second by -2. The x terms become opposite. Add them and solve for y

8
YOU TRY 3x + 7y = 15 5x + 2y = -4

Similar presentations

© 2024 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google