Use elimination to solve the linear-quadratic system of equations consisting of:
1. x2 + 12x - y + 36 = 0 and
2. 3x + 2y - 5 = 0
Step 1:
| Create an addition or subtraction problem with the
equations:
x2 + 12x - y + 36 = 0
+ 3x + 2y - 5 = 0
Step 2:
|
Decide which term to eliminate and then multiply the
equations by appropriate values.
In this example, the y term will be eliminated. Therefore, you must multiply Equation 1 by 2.
2x2 + 24x - 2y + 72 = 0
+ 3x + 2y - 5 = 0
Step 3:
| Solve the addition or subtraction problem.
2x2 + 24x - 2y + 72 = 0
+ 3x + 2y - 5 = 0
2x2 + 27x + + 67 = 0
Step 4:
| Expand and simplify the resulting equation.
2x2 + 27x + 67 = 0
Step 5:
| Solve.
In this case, use the quadratic formula:
| |
x = | _______
- b ± Öb2 - 4ac
| 2a |
| x = |
____________
-27 ± Ö272 - 4(2)(67)
| 2(2) |
| x = |
_________
-27 ± Ö729 - 536
| 4 |
| x = |
____
-27 ± Ö193
| 4 |
| x = | -27 ± 13.89
4
| x = | -40.89 and x = -13.11
4
4
| x = | -10.22 and x = -3.28
| | | | | | | | | | | |
Step 6:
| Substitute x = -10.22 and x = -3.28 into one of the original equations to find the corresponding y value.
| Let x = -10.22 | Let x = -3.28
Then:
3(-10.22) + 2y - 5 = 0 | Then:
3(-3.28) + 2y - 5 = 0
| -30.66 + 2y - 5 = 0 | -9.84 + 2y - 5 = 0
| 2y - 35.66 = 0 | 2y - 14.84 = 0
| 2y = 35.66 | 2y = 14.84
| y = 17.83 | y = 7.42
| | | | | |
| | | | | | |
| Thus the solutions to the system are (-10.22. 17.83) and (-3.28, 7.42).
| | |