Select the correct answer from the drop-down menu. consider the equations y = |x − 1| and y = 3x + 2. the approximate solution of this system of equations is

y = |x − 1|→y=x-1, if x-1>=0
                    y=-(x-1), if x-1<0

Then if x-1>=0→x>=1, y=x-1
         if x-1<0→x<1→y=-(x-1)=-x+1→y=-x+1

For x>=1
y=x-1
y=3x+2
Solving for x by equaliting's method:
y=y→3x+2=x-1→3x+2-x-2=x-1-x-2→2x=-3→2x/2=-3/2→
x=-3/2<1 No solution

For x<1
y=-x+1
y=3x+2
Solving for x by equaliting's method:
y=y→3x+2=-x+1→3x+2+x-2=-x+1+x-2→4x=-1→4x/4=-1/4→
x=-1/4<1  Ok

y=-x+1
y=-(-1/4)+1
y=1/4+1
y=(1+4)/4
y=5/4

The approximate solution of this system of equations is
(x,y)=(-1/4,5/4)=(-0.25,1.25)
 

a³ + 1

Step-by-step explanation:

Step 1: Factor out -3a

Step 2: Remove -3a

a³ + 1