Question

solve the system.
y = -2|x + 1| + 3
y - x = 1
select all points that are a solution to the system.
a. (-2, 1)
b. (-1, 2)
c. (0, 1)
d. (2, -3)
e. (-4, -3)
f. (-3, -4)

Explanation:

Step1: Express y from the linear equation

From \( y - x = 1 \), we get \( y = x + 1 \).

Step2: Substitute y into the absolute - value equation

Substitute \( y = x + 1 \) into \( y=-2|x + 1|+3 \), we have:
\( x + 1=-2|x + 1|+3 \)
Let \( t=x + 1 \), then the equation becomes \( t=-2|t|+3 \)

Case 1: When \( t\geq0 \) (i.e., \( x + 1\geq0\) or \( x\geq - 1\))

The equation \( t=-2|t|+3 \) becomes \( t=-2t + 3 \)
Add \( 2t \) to both sides: \( t + 2t=-2t + 3+2t \), so \( 3t = 3 \), then \( t = 1 \)
Since \( t=x + 1 \), when \( t = 1 \), \( x+1=1 \), so \( x = 0 \)
And \( y=x + 1 \), so \( y=0 + 1=1 \)

Case 2: When \( t<0 \) (i.e., \( x + 1<0\) or \( x<-1\))

The equation \( t=-2|t|+3 \) becomes \( t=-2(-t)+3 \) (because \( |t|=-t \) when \( t < 0 \))
Simplify: \( t = 2t+3 \)
Subtract \( 2t \) from both sides: \( t-2t=2t + 3-2t \), so \( -t = 3 \), then \( t=-3 \)
Since \( t=x + 1 \), when \( t=-3 \), \( x + 1=-3 \), so \( x=-4 \)
And \( y=x + 1 \), so \( y=-4 + 1=-3 \)
We can also check each option by substituting the points into both equations:

• For option A \((-2,1)\):
• First equation: \( y=-2|x + 1|+3=-2|-2 + 1|+3=-2\times1 + 3=1 \)
• Second equation: \( y - x=1-(-2)=3

eq1\)? Wait, no, wait, \( y - x=1-(-2)=3\)? Wait, no, the second equation is \( y - x = 1 \), \( 1-(-2)=3
eq1 \). Wait, there is a mistake in the initial check. Wait, let's recalculate.
Wait, from \( y=x + 1 \), for \( x=-2 \), \( y=-2 + 1=-1
eq1 \). Wait, maybe the initial substitution method is better.
Wait, let's use the substitution method correctly.
We have \( y=x + 1 \) and \( y=-2|x + 1|+3 \)
Substitute \( y \):
\( x + 1=-2|x + 1|+3 \)
Let's check option A \((-2,1)\):
First equation: \( y=-2|-2 + 1|+3=-2\times1+3 = 1 \)
Second equation: \( y - x=1-(-2)=3
eq1 \). So there is a mistake. Wait, maybe the user made a mistake in the initial check. Let's do it properly.
Let's solve the system again.
From \( y=x + 1 \), substitute into \( y=-2|x + 1|+3 \):
\( x + 1=-2|x + 1|+3 \)
Let \( u=x + 1 \), then \( u=-2|u|+3 \)
If \( u\geq0 \): \( u=-2u + 3\Rightarrow3u = 3\Rightarrow u = 1\Rightarrow x=0,y = 1 \)
If \( u<0 \): \( u=-2(-u)+3\Rightarrow u = 2u+3\Rightarrow -u=3\Rightarrow u=-3\Rightarrow x=-4,y=-3 \)
Now check option A \((-2,1)\):
For \( y=-2|x + 1|+3 \): \( -2|-2 + 1|+3=-2\times1 + 3=1 \)
For \( y - x=1 \): \( 1-(-2)=3
eq1 \). So A is not a solution.
Option E \((-4,-3)\):
For \( y=-2|x + 1|+3 \): \( -2|-4 + 1|+3=-2\times3+3=-3 \)
For \( y - x=-3-(-4)=1 \). So E is a solution.
Option C \((0,1)\):
For \( y=-2|x + 1|+3 \): \( -2|0 + 1|+3=-2 + 3=1 \)
For \( y - x=1-0 = 1 \). So C is a solution. Wait, I missed this earlier.
Wait, when \( u = 1 \) (i.e., \( x = 0 \)), \( y=1 \), so \((0,1)\) is a solution.
Let's re - solve the equation \( t=-2|t|+3 \) (where \( t=x + 1 \))
Case 1: \( t\geq0 \)
\( t=-2t + 3\Rightarrow3t = 3\Rightarrow t = 1\Rightarrow x=0,y = 1 \) (so point C \((0,1)\) is a solution)
Case 2: \( t<0 \)
\( t=-2(-t)+3\Rightarrow t = 2t+3\Rightarrow -t=3\Rightarrow t=-3\Rightarrow x=-4,y=-3 \) (so point E \((-4,-3)\) is a solution)
Now check option B \((-1,2)\):
\( y=-2|-1 + 1|+3=3
eq2 \). So not a solution.
Option D \((2,-3)\):
\( y=-2|2 + 1|+3=-6 + 3=-3 \), \( y - x=-3-2=-5
eq1 \). Not a solution.
Option F \((-3,-4)\):
\( y=-2|-3 + 1|+3=-4 + 3=-1
eq-4 \). Not a solution.
Option A \((-2,1)\):
\( y - x=1-(-2)=3
eq1 \). Not a solution.

So the correct solutions are C \((0,1)\) and E \((-4,-3)\)

Wait, there was a mistake in the initial check. Let's correct it.

Step1: Express \( y \) from the linear equation

From \( y - x=1 \), we derive \( y=x + 1 \).

Step2: Substitute \( y=x + 1 \) into the absolute - value equation

Substitute \( y=x + 1 \) into \( y=-2|x + 1|+3 \), we get:
\( x + 1=-2|x + 1|+3 \)
Let \( t = x + 1 \), then the equation becomes \( t=-2|t|+3 \)

Case 1: When \( t\geq0 \) (i.e., \( x\geq - 1\))

The equation \( t=-2|t|+3 \) simplifies to \( t=-2t + 3 \) (since \( |t| = t \) for \( t\geq0 \))
Add \( 2t \) to both sides: \( t + 2t=-2t+3 + 2t\), so \( 3t = 3 \), and \( t = 1 \)
Since \( t=x + 1 \), \( x+1 = 1\), so \( x = 0 \). Then \( y=x + 1=0 + 1=1 \). So the point \((0,1)\) (option C) is a solution.

Case 2: When \( t<0 \) (i.e., \( x<-1\))

The equation \( t=-2|t|+3 \) simplifies to \( t=-2(-t)+3 \) (since \( |t|=-t \) for \( t < 0 \))
Simplify: \( t = 2t+3 \)
Subtract \( 2t \) from both sides: \( t-2t=2t + 3-2t\), so \( -t = 3 \), and \( t=-3 \)
Since \( t=x + 1 \), \( x + 1=-3\), so \( x=-4 \). Then \( y=x + 1=-4 + 1=-3 \). So the point \((-4,-3)\) (option E) is a solution.
We can also verify each option by substituting the coordinates into both equations:

• Option A \((-2,1)\):
• For \( y=-2|x + 1|+3\): \( y=-2|-2 + 1|+3=-2\times1 + 3 = 1 \) (satisfies the first equation)
• For \( y - x=1\): \( 1-(-2)=3

eq1 \) (does not satisfy the second equation). So not a solution.

• Option B \((-1,2)\):
• For \( y=-2|x + 1|+3\): \( y=-2|-1 + 1|+3=3

eq2 \) (does not satisfy the first equation). So not a solution.

• Option C \((0,1)\):
• For \( y=-2|x + 1|+3\): \( y=-2|0 + 1|+3=-2 + 3 = 1 \) (satisfies the first equation)
• For \( y - x=1\): \( 1-0 = 1 \) (satisfies the second equation). So it is a solution.
• Option D \((2,-3)\):
• For \( y=-2|x + 1|+3\): \( y=-2|2 + 1|+3=-6 + 3=-3 \) (satisfies the first equation)
• For \( y - x=1\): \( -3-2=-5

eq1 \) (does not satisfy the second equation). So not a solution.

• Option E \((-4,-3)\):
• For \( y=-2|x + 1|+3\): \( y=-2|-4 + 1|+3=-6 + 3=-3 \) (satisfies the first equation)
• For \( y - x=1\): \( -3-(-4)=1 \) (satisfies the second equation). So it is a solution.
• Option F \((-3,-4)\):
• For \( y=-2|x + 1|+3\): \( y=-2|-3 + 1|+3=-4 + 3=-1

eq-4 \) (does not satisfy the first equation). So not a solution.

Answer:

C. \((0, 1)\), E. \((-4, -3)\)