Answer :
To solve the given system of equations:
[tex]\[ \left\{ \begin{array}{c} 2.4x - y = -3.5 \\ x^2 + x + y = 6 \end{array} \right. \][/tex]
Let's analyze which points satisfy both equations:
### Equation 1:
[tex]\[ 2.4x - y = -3.5 \][/tex]
### Equation 2:
[tex]\[ x^2 + x + y = 6 \][/tex]
Check the candidate solutions provided:
1. Point: [tex]\( (-6.2, -4) \)[/tex]
- Substitute [tex]\( x = -6.2 \)[/tex] and [tex]\( y = -4 \)[/tex] into Equation 1:
[tex]\[ 2.4(-6.2) - (-4) = -14.88 + 4 = -10.88 \neq -3.5 \][/tex]
This point does not satisfy the first equation.
2. Point: [tex]\( (5, 0.6) \)[/tex]
- Substitute [tex]\( x = 5 \)[/tex] and [tex]\( y = 0.6 \)[/tex] into Equation 1:
[tex]\[ 2.4 \cdot 5 - 0.6 = 12 - 0.6 = 11.4 \neq -3.5 \][/tex]
This point does not satisfy the first equation.
3. Point: [tex]\( (-4, -6.2) \)[/tex]
- Substitute [tex]\( x = -4 \)[/tex] and [tex]\( y = -6.2 \)[/tex] into Equation 1:
[tex]\[ 2.4(-4) - (-6.2) = -9.6 + 6.2 = -3.4 \approx -3.5 \][/tex]
- Substitute [tex]\( x = -4 \)[/tex] and [tex]\( y = -6.2 \)[/tex] into Equation 2:
[tex]\[ (-4)^2 + (-4) + (-6.2) = 16 - 4 - 6.2 = 5.8 \neq 6 \][/tex]
This point does not satisfy the second equation.
4. Point: [tex]\( (0.6, 5) \)[/tex]
- Substitute [tex]\( x = 0.6 \)[/tex] and [tex]\( y = 5 \)[/tex] into Equation 1:
[tex]\[ 2.4 \cdot 0.6 - 5 = 1.44 - 5 = -3.56 \approx -3.5 \][/tex]
- Substitute [tex]\( x = 0.6 \)[/tex] and [tex]\( y = 5 \)[/tex] into Equation 2:
[tex]\[ (0.6)^2 + 0.6 + 5 = 0.36 + 0.6 + 5 = 5.96 \approx 6 \][/tex]
This point approximately satisfies both equations.
5. Point: [tex]\( (-2.4, 2.4) \)[/tex]
- Substitute [tex]\( x = -2.4 \)[/tex] and [tex]\( y = 2.4 \)[/tex] into Equation 1:
[tex]\[ 2.4(-2.4) - 2.4 = -5.76 - 2.4 = -8.16 \neq -3.5 \][/tex]
This point does not satisfy the first equation.
6. Point: [tex]\( (3.7, -12.8) \)[/tex]
- Substitute [tex]\( x = 3.7 \)[/tex] and [tex]\( y = -12.8 \)[/tex] into Equation 1:
[tex]\[ 2.4 \cdot 3.7 - (-12.8) = 8.88 + 12.8 = 21.68 \neq -3.5 \][/tex]
This point does not satisfy the first equation.
7. Point: [tex]\( (2.4, -2.4) \)[/tex]
- Substitute [tex]\( x = 2.4 \)[/tex] and [tex]\( y = -2.4 \)[/tex] into Equation 1:
[tex]\[ 2.4 \cdot 2.4 - (-2.4) = 5.76 + 2.4 = 8.16 \neq -3.5 \][/tex]
This point does not satisfy the first equation.
8. Point: [tex]\( (-12.8, 3.7) \)[/tex]
- Substitute [tex]\( x = -12.8 \)[/tex] and [tex]\( y = 3.7 \)[/tex] into Equation 1:
[tex]\[ 2.4(-12.8) - 3.7 = -30.72 - 3.7 = -34.42 \neq -3.5 \][/tex]
This point does not satisfy the first equation.
After evaluating each candidate point, the point [tex]\( (0.6, 5) \)[/tex] approximately satisfies both equations. Thus, the approximate solution to the system of equations, rounded to the nearest tenth, is:
[tex]\[ \boxed{(0.6, 5)} \][/tex]
[tex]\[ \left\{ \begin{array}{c} 2.4x - y = -3.5 \\ x^2 + x + y = 6 \end{array} \right. \][/tex]
Let's analyze which points satisfy both equations:
### Equation 1:
[tex]\[ 2.4x - y = -3.5 \][/tex]
### Equation 2:
[tex]\[ x^2 + x + y = 6 \][/tex]
Check the candidate solutions provided:
1. Point: [tex]\( (-6.2, -4) \)[/tex]
- Substitute [tex]\( x = -6.2 \)[/tex] and [tex]\( y = -4 \)[/tex] into Equation 1:
[tex]\[ 2.4(-6.2) - (-4) = -14.88 + 4 = -10.88 \neq -3.5 \][/tex]
This point does not satisfy the first equation.
2. Point: [tex]\( (5, 0.6) \)[/tex]
- Substitute [tex]\( x = 5 \)[/tex] and [tex]\( y = 0.6 \)[/tex] into Equation 1:
[tex]\[ 2.4 \cdot 5 - 0.6 = 12 - 0.6 = 11.4 \neq -3.5 \][/tex]
This point does not satisfy the first equation.
3. Point: [tex]\( (-4, -6.2) \)[/tex]
- Substitute [tex]\( x = -4 \)[/tex] and [tex]\( y = -6.2 \)[/tex] into Equation 1:
[tex]\[ 2.4(-4) - (-6.2) = -9.6 + 6.2 = -3.4 \approx -3.5 \][/tex]
- Substitute [tex]\( x = -4 \)[/tex] and [tex]\( y = -6.2 \)[/tex] into Equation 2:
[tex]\[ (-4)^2 + (-4) + (-6.2) = 16 - 4 - 6.2 = 5.8 \neq 6 \][/tex]
This point does not satisfy the second equation.
4. Point: [tex]\( (0.6, 5) \)[/tex]
- Substitute [tex]\( x = 0.6 \)[/tex] and [tex]\( y = 5 \)[/tex] into Equation 1:
[tex]\[ 2.4 \cdot 0.6 - 5 = 1.44 - 5 = -3.56 \approx -3.5 \][/tex]
- Substitute [tex]\( x = 0.6 \)[/tex] and [tex]\( y = 5 \)[/tex] into Equation 2:
[tex]\[ (0.6)^2 + 0.6 + 5 = 0.36 + 0.6 + 5 = 5.96 \approx 6 \][/tex]
This point approximately satisfies both equations.
5. Point: [tex]\( (-2.4, 2.4) \)[/tex]
- Substitute [tex]\( x = -2.4 \)[/tex] and [tex]\( y = 2.4 \)[/tex] into Equation 1:
[tex]\[ 2.4(-2.4) - 2.4 = -5.76 - 2.4 = -8.16 \neq -3.5 \][/tex]
This point does not satisfy the first equation.
6. Point: [tex]\( (3.7, -12.8) \)[/tex]
- Substitute [tex]\( x = 3.7 \)[/tex] and [tex]\( y = -12.8 \)[/tex] into Equation 1:
[tex]\[ 2.4 \cdot 3.7 - (-12.8) = 8.88 + 12.8 = 21.68 \neq -3.5 \][/tex]
This point does not satisfy the first equation.
7. Point: [tex]\( (2.4, -2.4) \)[/tex]
- Substitute [tex]\( x = 2.4 \)[/tex] and [tex]\( y = -2.4 \)[/tex] into Equation 1:
[tex]\[ 2.4 \cdot 2.4 - (-2.4) = 5.76 + 2.4 = 8.16 \neq -3.5 \][/tex]
This point does not satisfy the first equation.
8. Point: [tex]\( (-12.8, 3.7) \)[/tex]
- Substitute [tex]\( x = -12.8 \)[/tex] and [tex]\( y = 3.7 \)[/tex] into Equation 1:
[tex]\[ 2.4(-12.8) - 3.7 = -30.72 - 3.7 = -34.42 \neq -3.5 \][/tex]
This point does not satisfy the first equation.
After evaluating each candidate point, the point [tex]\( (0.6, 5) \)[/tex] approximately satisfies both equations. Thus, the approximate solution to the system of equations, rounded to the nearest tenth, is:
[tex]\[ \boxed{(0.6, 5)} \][/tex]
