Answer :
To determine which statement about the given system of inequalities is true, we need to analyze their graphs and potential points of intersection.
The system of inequalities is:
[tex]\[ \left\{\begin{array}{l} y - 2x \leq 2.5 \\ y + x^2 + 0.1x \leq -0.6 \end{array}\right. \][/tex]
1. Convert the inequalities to equations to find potential intersection points:
[tex]\[ y - 2x = 2.5 \quad \text{(Equation 1)} \][/tex]
[tex]\[ y + x^2 + 0.1x = -0.6 \quad \text{(Equation 2)} \][/tex]
2. Isolate [tex]\( y \)[/tex] in both equations:
[tex]\[ y = 2x + 2.5 \quad \text{(From Equation 1)} \][/tex]
[tex]\[ y = -x^2 - 0.1x - 0.6 \quad \text{(From Equation 2)} \][/tex]
3. Set the two expressions for [tex]\( y \)[/tex] equal to each other to find [tex]\( x \)[/tex]:
[tex]\[ 2x + 2.5 = -x^2 - 0.1x - 0.6 \][/tex]
4. Rearrange the equation into standard quadratic form:
[tex]\[ x^2 + 2.1x + 3.1 = 0 \][/tex]
5. Calculate the determinant ([tex]\( D \)[/tex]) of the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 2.1 \)[/tex], and [tex]\( c = 3.1 \)[/tex]:
The determinant is given by the formula:
[tex]\[ D = b^2 - 4ac \][/tex]
6. Substitute the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the determinant formula:
[tex]\[ D = (2.1)^2 - 4(1)(3.1) \][/tex]
[tex]\[ D = 4.41 - 12.4 \][/tex]
[tex]\[ D = -7.99 \][/tex]
7. Analyze the determinant:
If the determinant [tex]\( D \)[/tex] is negative (i.e., [tex]\( D < 0 \)[/tex]), it means the quadratic equation has no real solutions. This implies the equations do not intersect at any real points.
Since the determinant [tex]\( D \)[/tex] is [tex]\(-7.99\)[/tex] and is negative, the graphs of the two equations indeed do not intersect.
Given this analysis, we can conclude:
There is no solution because the graphs do not intersect.
The system of inequalities is:
[tex]\[ \left\{\begin{array}{l} y - 2x \leq 2.5 \\ y + x^2 + 0.1x \leq -0.6 \end{array}\right. \][/tex]
1. Convert the inequalities to equations to find potential intersection points:
[tex]\[ y - 2x = 2.5 \quad \text{(Equation 1)} \][/tex]
[tex]\[ y + x^2 + 0.1x = -0.6 \quad \text{(Equation 2)} \][/tex]
2. Isolate [tex]\( y \)[/tex] in both equations:
[tex]\[ y = 2x + 2.5 \quad \text{(From Equation 1)} \][/tex]
[tex]\[ y = -x^2 - 0.1x - 0.6 \quad \text{(From Equation 2)} \][/tex]
3. Set the two expressions for [tex]\( y \)[/tex] equal to each other to find [tex]\( x \)[/tex]:
[tex]\[ 2x + 2.5 = -x^2 - 0.1x - 0.6 \][/tex]
4. Rearrange the equation into standard quadratic form:
[tex]\[ x^2 + 2.1x + 3.1 = 0 \][/tex]
5. Calculate the determinant ([tex]\( D \)[/tex]) of the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 2.1 \)[/tex], and [tex]\( c = 3.1 \)[/tex]:
The determinant is given by the formula:
[tex]\[ D = b^2 - 4ac \][/tex]
6. Substitute the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the determinant formula:
[tex]\[ D = (2.1)^2 - 4(1)(3.1) \][/tex]
[tex]\[ D = 4.41 - 12.4 \][/tex]
[tex]\[ D = -7.99 \][/tex]
7. Analyze the determinant:
If the determinant [tex]\( D \)[/tex] is negative (i.e., [tex]\( D < 0 \)[/tex]), it means the quadratic equation has no real solutions. This implies the equations do not intersect at any real points.
Since the determinant [tex]\( D \)[/tex] is [tex]\(-7.99\)[/tex] and is negative, the graphs of the two equations indeed do not intersect.
Given this analysis, we can conclude:
There is no solution because the graphs do not intersect.