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
Accepted Solution
A:
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)