∑ Hey, KLPJDP615 ⊃
Answer:
x = 6 or x = -10
Step-by-step explanation:
<u><em>Given:</em></u>
<em>Solve for x.</em>
<em>1 + |2+x|= 9</em>
<em>O x = 4 or x = -8</em>
<em>O x = 7 or X = -11</em>
<em>O x = 5 or x = -9</em>
<em>x = 6 or X = -10 </em>
<u><em>Solve:</em></u>
<em>1 + |2+x|= 9</em>
<em>Subtract 1 from both sides:</em>
<em>1 + |2 +x| -1 = 9-1 </em>
<em>Simplify</em>
<em>|2 + x | = 8</em>
<em>Applying absolute value rule: If |u| = a, a > 0 then u = a or u = -a</em>
<em>2 + x = -8</em>
<em>2 + x = 8</em>
<u><em>Solving:</em></u>
<em>2 + x = -8</em>
<em>2 - 2 + x = -8 - 2</em>
<em>x = -10</em>
<u><em>Solving:</em></u>
<em>2 + x = 8</em>
<em>2 - 2 + x = 8 - 2</em>
<em>x = 6</em>
<em />
<em>Hence, x = 6 or x = -10</em>
<em />
<u><em>xcookiex12</em></u>
<em>8/26/2022</em>
Answer:
Mine was multiple choice and I got c
Step-by-step explanation:
QUESTION 1
The given system of equations are:


We equate the two equations to get:




When x=0,

The solution is (0,1)
QUESTION 2
The given equations are:

and

We equate both equations to get:

Group similar terms,



We put x=0 into any of the equations to find y.

The solution is (0,-1).
QUESTION 3
The given equations are:

and

We equate both equations:

Group similar terms:


This is not true.
Hence the system has no solution.