Answer :
To solve the system of equations using the substitution method, let's follow these steps:
1. Rewrite the Equations:
Given,
[tex]\[ 3y = -\frac{1}{2}x + 2 \quad \text{(Equation 1)} \][/tex]
[tex]\[ y = -x + 9 \quad \text{(Equation 2)} \][/tex]
2. Solve Equation 1 for [tex]\( y \)[/tex]:
Divide both sides of Equation 1 by 3:
[tex]\[ y = -\frac{1}{6}x + \frac{2}{3} \][/tex]
Now, we have:
[tex]\[ y = -\frac{1}{6}x + \frac{2}{3} \quad \text{(Equation 1')} \][/tex]
3. Set Equation 1' equal to Equation 2:
Since both expressions represent [tex]\( y \)[/tex], we can set them equal to each other:
[tex]\[ -\frac{1}{6}x + \frac{2}{3} = -x + 9 \][/tex]
4. Solve for [tex]\( x \)[/tex]:
To remove the fraction, multiply everything by 6:
[tex]\[ -x + 4 = -6x + 54 \][/tex]
Add [tex]\( 6x \)[/tex] to both sides:
[tex]\[ 5x + 4 = 54 \][/tex]
Subtract 4 from both sides:
[tex]\[ 5x = 50 \][/tex]
Divide by 5:
[tex]\[ x = 10 \][/tex]
5. Substitute [tex]\( x \)[/tex] back into Equation 2 to find [tex]\( y \)[/tex]:
[tex]\[ y = -10 + 9 = -1 \][/tex]
So, the solution should be the point [tex]\( (10, -1) \)[/tex].
6. Verify Solution by Checking Other Points:
We are given four points to check if they satisfy both equations:
- [tex]\( (3, 6) \)[/tex]:
[tex]\[ y = 6, \quad x = 3 \][/tex]
Substitute into Equation 2:
[tex]\[ 6 = -3 + 9 \implies 6 = 6 \quad \text{(Valid)} \][/tex]
Substitute into Equation 1':
[tex]\[ 6 = -\frac{1}{6}(3) + \frac{2}{3} \][/tex]
[tex]\[ 6 = -\frac{1}{2} + \frac{2}{3} \implies 6 = \frac{-1}{2} + \frac{2}{3} \implies 6 = \frac{4}{6} - \frac{3}{6} = \frac{1}{6} \quad \text{(Not Valid)} \][/tex]
Since the point [tex]\( (3, 6) \)[/tex] does not satisfy both equations, we move on to the next points:
- [tex]\( (10, -1) \)[/tex]:
[tex]\[ y = -1, \quad x = 10 \][/tex]
Substitute into Equation 2:
[tex]\[ -1 = -10 + 9 \implies -1 = -1 \quad \text{(Valid)} \][/tex]
Substitute into Equation 1':
[tex]\[ -1 = -\frac{1}{6}(10) + \frac{2}{3} \][/tex]
[tex]\[ -1 = -\frac{10}{6} + \frac{2}{3} \implies -1 = -\frac{5}{3} + \frac{2}{3} \implies -1 = -1 \quad \text{(Valid)} \][/tex]
So, the point [tex]\( (10, -1) \)[/tex] satisfies both equations.
- [tex]\( (-1, 8) \)[/tex]:
[tex]\[ y = 8, \quad x = -1 \][/tex]
Substitute into Equation 2:
[tex]\[ 8 = -(-1) + 9 \implies 8 = 1 + 9 \implies 8 = 10 \quad \text{(Not Valid)} \][/tex]
Since Equation 2 is not valid, there's no need to check Equation 1'.
- [tex]\( (20, -4) \)[/tex]:
[tex]\[ y = -4, \quad x = 20 \][/tex]
Substitute into Equation 2:
[tex]\[ -4 = -20 + 9 \implies -4 = -11 \quad \text{(Not Valid)} \][/tex]
Since Equation 2 is not valid, there's no need to check Equation 1'.
Given this analysis, none of the points satisfy both equations simultaneously. Therefore, no points from the given set of points (3,6), (10,-1), (-1,8), (20,-4) satisfy both equations at the same time.
1. Rewrite the Equations:
Given,
[tex]\[ 3y = -\frac{1}{2}x + 2 \quad \text{(Equation 1)} \][/tex]
[tex]\[ y = -x + 9 \quad \text{(Equation 2)} \][/tex]
2. Solve Equation 1 for [tex]\( y \)[/tex]:
Divide both sides of Equation 1 by 3:
[tex]\[ y = -\frac{1}{6}x + \frac{2}{3} \][/tex]
Now, we have:
[tex]\[ y = -\frac{1}{6}x + \frac{2}{3} \quad \text{(Equation 1')} \][/tex]
3. Set Equation 1' equal to Equation 2:
Since both expressions represent [tex]\( y \)[/tex], we can set them equal to each other:
[tex]\[ -\frac{1}{6}x + \frac{2}{3} = -x + 9 \][/tex]
4. Solve for [tex]\( x \)[/tex]:
To remove the fraction, multiply everything by 6:
[tex]\[ -x + 4 = -6x + 54 \][/tex]
Add [tex]\( 6x \)[/tex] to both sides:
[tex]\[ 5x + 4 = 54 \][/tex]
Subtract 4 from both sides:
[tex]\[ 5x = 50 \][/tex]
Divide by 5:
[tex]\[ x = 10 \][/tex]
5. Substitute [tex]\( x \)[/tex] back into Equation 2 to find [tex]\( y \)[/tex]:
[tex]\[ y = -10 + 9 = -1 \][/tex]
So, the solution should be the point [tex]\( (10, -1) \)[/tex].
6. Verify Solution by Checking Other Points:
We are given four points to check if they satisfy both equations:
- [tex]\( (3, 6) \)[/tex]:
[tex]\[ y = 6, \quad x = 3 \][/tex]
Substitute into Equation 2:
[tex]\[ 6 = -3 + 9 \implies 6 = 6 \quad \text{(Valid)} \][/tex]
Substitute into Equation 1':
[tex]\[ 6 = -\frac{1}{6}(3) + \frac{2}{3} \][/tex]
[tex]\[ 6 = -\frac{1}{2} + \frac{2}{3} \implies 6 = \frac{-1}{2} + \frac{2}{3} \implies 6 = \frac{4}{6} - \frac{3}{6} = \frac{1}{6} \quad \text{(Not Valid)} \][/tex]
Since the point [tex]\( (3, 6) \)[/tex] does not satisfy both equations, we move on to the next points:
- [tex]\( (10, -1) \)[/tex]:
[tex]\[ y = -1, \quad x = 10 \][/tex]
Substitute into Equation 2:
[tex]\[ -1 = -10 + 9 \implies -1 = -1 \quad \text{(Valid)} \][/tex]
Substitute into Equation 1':
[tex]\[ -1 = -\frac{1}{6}(10) + \frac{2}{3} \][/tex]
[tex]\[ -1 = -\frac{10}{6} + \frac{2}{3} \implies -1 = -\frac{5}{3} + \frac{2}{3} \implies -1 = -1 \quad \text{(Valid)} \][/tex]
So, the point [tex]\( (10, -1) \)[/tex] satisfies both equations.
- [tex]\( (-1, 8) \)[/tex]:
[tex]\[ y = 8, \quad x = -1 \][/tex]
Substitute into Equation 2:
[tex]\[ 8 = -(-1) + 9 \implies 8 = 1 + 9 \implies 8 = 10 \quad \text{(Not Valid)} \][/tex]
Since Equation 2 is not valid, there's no need to check Equation 1'.
- [tex]\( (20, -4) \)[/tex]:
[tex]\[ y = -4, \quad x = 20 \][/tex]
Substitute into Equation 2:
[tex]\[ -4 = -20 + 9 \implies -4 = -11 \quad \text{(Not Valid)} \][/tex]
Since Equation 2 is not valid, there's no need to check Equation 1'.
Given this analysis, none of the points satisfy both equations simultaneously. Therefore, no points from the given set of points (3,6), (10,-1), (-1,8), (20,-4) satisfy both equations at the same time.
