Answer: The set does not have a solution
Step-by-step explanation:
Adding Equations 1 & 3 we get 5x = 7. This gives x = 7/5
Putting this value of x in eq. 2 we get
-2y + 2z = -1-(7/5) or
2y - 2z = 12/5 or 5y - 5z = 6
Multiplying eq. 1 by 2 we get
4x + 2y - 2z = 6
adding this with eq. 2 we get 5x = 5 or x = 1
As the common solution for x from equations 1&3 does not satisfy eq. 1&2 it comes out that the three equations do not have a common solution.
Same can be verified by using different sets of two equations also.
Answer:
Step-by-step explanation:
I hope I helped you! :)
Answer:
Step 2
Step-by-step explanation:
Meredith made a fatal mistake or error in step 2;
Any quadratic problem will always have two solutions. Therefore since, we only have one solution here, it is wrong. The mistake was incurred in step 2;
From the given problem ;
<u> Wrong right</u>
2(x + 4)² = 242
(x + 4)² = 121 Step 1
x+4=11 Step 2 x + 4 = ± 11
x = 7 Step 3 x + 4 = 11 or x + 4 = -11
x = 11- 4 or x = -11 - 4
x = 7 or -15
